POCKET GUIDE. Chemistry NCEA Level 2. External Standards. WORKING TO EXCELLENCE Bonding C2.4 AS Organics C2.5 AS Reactivity C2.

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1 Factor (if increased) Surface Area Concentration Temperaturee Reaction rates Increase the frequency of collisions Increase the energy of particles Increase the frequency of collisions Increased Reaction Rate Chemistry NCEA Level 2 External Standards POCKET GUIDE Catalyst Lowers the activation energy pathway More (%) collisions are successful Le Chatelier s Principle When a change is applied to a system at equilibrium, the system responds so that the effects of the change are minimised Change in conditions Direction of change in equilibrium position Concentration - increase products In the reverse direction - decrease products In the forward direction - increase reactants In the forward direction - decrease reactants In the reverse direction Pressure Increase In the direction with the least no. of moles of gas Decrease In the direction with the greater no. of moles of gas Temperature Increase In the direction of the endothermic reaction Decrease In the direction of the exothermic reaction Catalyst added No change in equilibrium position or in K c Equilibrium is reached more quickly (ie reaction rate changes) WORKING TO EXCELLENCE Bonding C2.4 AS Organics C2.5 AS Reactivity C2.6 AS Writing Excellence answers to Reaction Rates of Acids questions Writing Excellence answers to Molecule Polarity questions Reaction Rates of Acids QUESTION Question: The ph values of mol L 1 solutions of two acids, HA and HB, are given below. (i) Compare the relative strengths of the two acids, HA(aq) and HB(aq), using the information given above. Your answer should include equations and calculations. (ii) Predict and compare, with reasons, what would be observed when two 5 g samples of calcium carbonate chips, CaCO3(s), are reacted, separately, with excess HA and HB. Molecule Polarity QUESTION Question: The Lewis structures for two molecules are shown. Ammonia, NH3, is polar, and borane, BH3, is non-polar. Justify this statement. 1. Write an equation for HA [Remembering H3O + must be produced] HA + H2O A + H3O + 2. Calculate H3O + for HA ph = 1.0 [H3O + ] = mol L 1 3. For HA link concentration of ions formed to level of dissociation AND compare to concentration of acid (are they the same?) 4. Write an equation for HB [Remembering H3O + must be produced] 5. Calculate H3O + for HB HA is a strong acid since it fully dissociates, as shown by concentration of hydronium ions in HA solution same as original concentration of HA (both mol L 1 ). HB + H2O B + H3O + ph = 2.2 [H3O + ] = mol L 1 1. For the first molecule (name) state the types of bonds present (name atoms) and state whether they are polar (form a dipole) or non-polar due to electronegativity. 2. link electronegativity differences to sharing of electrons for your bond 3. state the shape of your molecule and link to being symmetrical or not and result in dipoles cancelling (or not) Each N-H bond in NH3 is polar / forms a dipole because the N and H atoms have different electronegativities. The electronegativity differences of the atoms means there is uneven sharing of the valence electrons (the electrons will spend more time around N than H causing N to be slightly negative and H to be slightly positive) The shape of the molecule (due to the presence of one non-bonding electron pair) is trigonal pyramidal which is asymmetrical, so the dipoles / bond polarities do not cancel. 6. For HB link concentration of ions formed to level of dissociation AND compare to concentration of acid (are they the same?) 7. For HA link observation of reaction to concentration of ions 8. then For HA link collision frequency to rate of reaction 9. For HB link observation of reaction to concentration of ions 10. then For HB link collision frequency to rate of reaction HB is a weak acid since it only partially dissociates; as shown by the concentration of hydronium ions in HB solution concentration is only mol L 1. Expect reaction to be more vigorous; rapidly produces gas / bubbles (CO2) since the concentration of hydrogen ions is high, there will be more frequent collisions resulting in a faster rate of reaction. Expect a slower reaction, taking longer to produce the same volume of gas since the concentration of hydrogen ions is low, there will be less frequent collisions resulting in a slower rate of reaction. 4. link to final polarity of molecule The resulting NH3 molecule overall is polar. 5. For the second molecule (name) Each B-H bond in BH3 is polar / forms a dipole because the B and H atoms have state the types of bonds present different electronegativities. (name atoms) and state whether they are polar (form a dipole) or nonpolar due to electronegativity. 6. link electronegativity differences to sharing of electrons for your bond 7. state the shape of your molecule and link to being symmetrical or not and result in dipoles cancelling (or not) The electronegativity differences of the atoms means there is uneven sharing of the valence electrons (the electrons will spend more time around B than H causing B to be slightly negative and H to be slightly positive) The shape of the molecule is trigonal planar which is symmetrical, so the dipoles / bond polarities cancel. 8. link to final polarity of molecule The resulting BH3 molecule overall is non-polar. 30 3

2 Writing Excellence answers to Molecule shapes and bond angle questions Writing Excellence answers to ph calculations questions Molecule shapes and bond angle QUESTION Question: Carbon atoms can bond with different atoms to form many different compounds. The following table shows the Lewis structure for two molecules containing carbon as the central atom, CCl4 and COCl2. These molecules have different bond angles and shapes. Evaluate the Lewis structure of each molecule to determine why they have different bond angles and shapes. In your answer you should include: The approximate bond angle in each molecule The shape of each molecule Factors that determine the shape and bond angle for each molecule. 1. for first molecule (name) state number of regions of negative charge around the central atom (name central atom) 2. state the Valence shell electron pair repulsion (VSEPR) theory 3. state the base arrangement of negative regions and the bond angle they form 4. state the number of bonded and non-bonded regions AND the final shape of the first molecule In each CCl4 molecule, there are four negative electron clouds / regions around the central C atom. These regions of negative charge repel each other as far away from each other as possible around the central C atom in a tetrahedral (base) arrangement, resulting in a o bond angle All of these regions of electrons are bonding, without any non-bonding regions, so the final shape of the molecule is tetrahedral. 5. for second molecule (name) state In each COCl2 molecule, there are three negative electron clouds / regions number of regions of negative charge around the central C atom. around the central atom (name central atom) 6. state the Valence shell electron pair repulsion (VSEPR) theory 7. state the base arrangement of negative regions and the bond angle they form 8. state the number of bonded and non-bonded regions AND the final shape of the second molecule 9. compare differences in bond angle linked to number of regions of negative charge. These regions of negative charge repel each other as far away from each other as possible around the central C atom in a triangular / trigonal planar (base) shape, resulting in a 120 bond angle. All of these regions of electrons are bonding, without any non-bonding regions, so the final shape of the molecule is trigonal planar. Both molecules have no non-bonding pairs but because CCl4 has 4 regions of negative charge around the central atom compared to the 3 regions that COCl2 has, then CCl4 has a smaller bond angle of o compared to the 120 bond angle of COCl ph calculations QUESTION 1 Question: In a solution of potassium hydroxide, KOH, the ph is found to be (i) Calculate the hydronium ion concentration, [H3O + ], and the hydroxide ion concentration, [OH ], in the solution. Kw = (ii) Calculate the ph of a mol L 1 sodium hydroxide, NaOH, solution. STEP 1. Calculate H3O + for KOH STEP 2. Calculate OH - for KOH ( Kw = ) STEP 1. Calculate poh for NaOH poh = -log[oh - ] [H3O + ] = 1.58 x moll -1 [OH - ] = moll -1 poh = -log[oh - ] poh = 3.60 (3sgf) STEP 2. Calculate ph for NaOH ph = 14 - poh ph = 14 - poh ph = ph = 10.4 (3sgf) ph calculations QUESTION 2 Question: (i) A solution of nitric acid, HNO3(aq), has a hydronium ion, H3O +, concentration of mol L 1. Determine, by calculation, the ph and the concentration of hydroxide ions, OH, in this solution. Kw = (ii) Determine the hydroxide ion concentration, [OH ], of a solution of potassium hydroxide, KOH(aq), with a ph of STEP 1. Calculate ph for HNO3 ph = -log[h3o + ] (3sgf) STEP 2. Calculate OH - for HNO3 ( Kw = ) STEP 1. Calculate H3O + for KOH STEP 2. Calculate OH - for KOH ( Kw = ) ph = -log[h3o + ] ph = 1.61 [OH - ] = 4.12 x moll -1 [H3O + ] = 1.58 x moll -1 [OH - ] = 6.31 x 10-3 moll -1 Writing Excellence answers to Solids Melting Point questions Writing Excellence answers to Ions and Conductivity questions Solids Melting Point QUESTION Question: Explain why chlorine is a gas at room temperature, but copper chloride is a solid at room temperature. In your answer, you should refer to the particles and the forces between the particles in both s. (you will need to fill in the chart below correctly as part of the question and use the terms in your answer) Ions and Conductivity QUESTION Question: Some properties of three aqueous solutions A, B and C, of equal concentration are shown in the table below. It is known that the solutions are NH3(aq), HCl(aq) and NH4Cl(aq) Justify the identification of all three solutions. refer to both ph and electrical conductivity of the solutions link your answers to appropriate chemical equations. Substance Type of Type of particle Attractive forces between particles Cl 2 (s) chlorine Molecular Molecules Weak intermolecular forces CuCl 2(s) copper chloride Ionic Ion Ionic bonds / electrostatic attraction 1. For the first (name) 2. describe the structure of this type of using the terms above 3. explain how the bonding relates to the energy required to break bonds of your 4. link to the observation (state at room temperature) in your question for the first 5. For the second (name) 6. describe the structure of this type of using the terms above 7. explain how the bonding relates to the energy required to break bonds of your 8. link to the observation (state at room temperature) in your question for the first Chlorine is a molecular composed of chlorine molecules held together by weak intermolecular forces The weak intermolecular forces do not require much heat energy to break, so the boiling point is low (lower than room temperature); therefore chlorine is a gas at room temperature. Copper chloride is an ionic. It is composed of a lattice of positive copper ions and negative chloride ions held together by electrostatic attraction (ionic bonds) between these positive and negative ions. These are strong forces, therefore they require considerable energy to disrupt them and melt the copper chloride; hence copper chloride is a solid at room temperature. 1. Identify each solution as either A, B or C by linking to being a weak or strong acid or base and also to the ph 2. State requirements for conductivity 3. Solution A (ph 5.15) weak acid salt. Equation 1. [A salt will first dissociate fully into ions] Write equation AND link ions formed to conductivity and level of dissociation 4. Solution A (ph 5.15) weak acid salt. Equation 2.[One of the products of dissociation will further react as an acid] Write equation AND link ions formed to conductivity and level of dissociation (must form H3O + ions) 5. Solution B (ph 11.6) weak base. Write equation AND link ions formed to conductivity and level of dissociation (must form OH - ions) 6. Solution C (ph 1.05) strong acid. Write equation AND link ions formed to conductivity and level of dissociation (must form H3O + ions) Solution A with a ph of 5.15 is a weak acid (salt) and is NH4Cl(aq) Solution B with a ph of 11.6 is a weak base and is NH3(aq) Solution C with a ph of 1.05 is a strong acid and is HCl(aq) In order to conduct electricity there needs to be the presence of free moving charged particles. The more charged particles there are available the better conductivity there will be. Ions in solution provide the charged particle. NH 4Cl NH Cl NH4Cl(aq) is solution A: good conductor of electricity it fully dissociates in solution into ammonium and chloride ions, which conduct electricity. NH H 2O NH 3 + H 3O + Its ph (5.15) is that of a weak acid, as the ammonium ion is a weak acid and partially dissociates in water, producing hydronium ions. NH 3 + H 2O NH OH NH3(aq) is solution B: its ph (11.6) is that of a weak base as NH3 so it partially dissociates in water, producing hydroxide ions. It is a poor conductor of electricity as it is only partially dissociated into ions in water. The remaining NH3 molecules are neutral and do not conduct electricity. HCl + H 2O H 3O + + Cl HCl(aq) is solution C: low ph (1.05) is that of a strong acid, HCl fully dissociates in water, producing hydronium ions. It is a good conductor of electricity as it fully dissociates into ions in solution which conduct electricity. 4 29

3 Writing Excellence answers to Equilibrium Concentration questions Equilibrium Concentration QUESTION Question: When acid is added to a yellow solution of chromate ions, CrO4 2 (aq), the following equilibrium is established. 2CrO4 2 (aq) + 2H + (aq) Cr2O7 2 (aq) + H2O(l) yellow orange Analyse this equilibrium using equilibrium principles to explain the effect on the colour of the solution when: (i) more dilute acid is added AND when (ii) dilute base is added: 1. State the equilibrium principle When a change is made to a system that is at equilibrium, the system responds to reduce the effect of that change. 2. Describe the factor in your question AND Link the principle to how the system responds to increasing or decreasing concentration of reactants 3. Generally, explain which side of the equation is favoured (relate to reactants or products) by increasing or decreasing concentration 4. Specifically, for your reaction explain how you are increasing the concentration of reactants, AND link the direction of reaction that would be favoured 5. Describe how the system shift by increasing the concentration of reactants would affect which s are made AND final observation. 6. Specifically, for your reaction explain how you are decreasing the concentration of reactants, AND link the direction of reaction that would be favoured 7. Describe how the system shift by decreasing the concentration of reactants would affect which s are made AND final observation. The factor in the question above is concentration of reactants. If there is an increase in concentration of reactants, the system responds by increasing the rate products are made. If there is a decrease in concentration of reactants, the system responds by decreasing the rate products are made. An increase in concentration of reactants favours the forward reaction, and a decrease in concentration of reactants favours the reverse reaction. Adding dilute acid increases the concentration of the acid, one of the reactants, so the reaction moves in the forward direction and favours the products, therefore it will increase the rate that the added acid is used up More Cr2O7 2 (aq) would be produced and the solution would turn more orange. Adding base means that acid that reacts with the base is removed from the equilibrium (by neutralisation) and the concentration of the acid decreases. This will drive the equilibrium in the backwards direction and this favours the reactants to increase the rate of replacing the H + used up More CrO4 2 (aq), would be produced and the solution would turn more yellow Writing Excellence answers to Solids Conductivity (Ductility) questions Solids Conductivity (Ductility) QUESTION Question: Using your knowledge of structure and bonding, explain why, although both graphite and copper are good conductors of electricity, copper is suitable for electrical wires, but graphite is not. (note two properties to discuss) (you will need to fill in the chart below correctly as part of the question and use the terms in your answer) Substance Type of Type of particle Attractive forces between particles C(s) Graphite Covalent network Atom Covalent ( and weak intermolecular forces) Cu(s) copper metal Atom / cations and electrons Metallic bonds / electrostatic attraction 1. For the first (name) Graphite is a covalent network solid 2. describe the structure of this type composed of layers of C atoms covalently bonded to three other C atoms. The of using the terms above remaining valence electron is delocalised (ie free to move) between layers; 3. explain how the bonding relates to The delocalised electrons are able to carry an electrical charge the present of free moving charged particles to conduct electricity in your (property 1) 4. link to the observation Therefore graphite is able to conduct electricity (conductivity) in your question for the first 5. explain how the bonding relates to In graphite, the attractive forces holding the layers together are very weak and ductility in your (property are broken easily, so the layers easily slide over one another, 2) 6. link to the observation (forming but the attraction is not strong enough to hold the layers together and allow it wires) in your question for the first to be drawn into wires or although the layers can slide due to weak forces, if graphite was to be made into a wire the very strong covalent bonds within the layers would have to be broken. Graphite cannot form wires. 7. For the second (name) Copper is a metallic 8. describe the structure of this type composed of copper atoms packed together. Valence electrons are loosely of using the terms above held and are attracted to the nuclei of the neighbouring Cu atoms ;ie the bonding is non-directional. 9. explain how the bonding relates to These delocalised valence electrons are free moving and can carry a charge the present of free moving charged particles to conduct electricity in your (property 1) 10. link to the observation Therefore copper is able to conduct electricty (conductivity) in your question for the second 11. explain how the bonding relates In copper, the non-directional metallic bonding holds the layers together, to ductility in your allowing it to be stretched without breaking. (property 2) 12. link to the observation (forming Therefore Copper metal is malleable and can easily be drawn into wires since, wires) in your question for the as it is stretched out, second Writing Excellence answers to Equilibrium Pressure questions Writing Excellence answers to Endothermic and Exothermic questions Equilibrium Pressure QUESTION Question: The two reactions shown in the following table are both at equilibrium. Compare and contrast the effect of increasing the pressure on both reactions, with reference to the equilibrium positions. 1. State the equilibrium principle When a change is made to a system that is at equilibrium, the system responds to reduce the effect of that change. Endothermic and Exothermic QUESTION Question: Pentane, C5H12, is a liquid at room temperature. It evaporates at 36.1 C in an endothermic process. (i) Explain why the evaporation of pentane is an endothermic process. (ii) Draw, including labels, the energy diagram for the combustion of pentane, C5H12(l). Pentane combustion: C5H12(l) + 8O2(g) 5CO2(g) + 6H2O(l) ΔrH º = 3509 kj mol 1 Include in your diagram the reactants, products, and change in enthalpy. 1. define an endothermic process An Endothermic process is one where heat / energy has been absorbed and the enthalpy of the products is higher than the reactants 2. For the (name) state the type of solid that it is Pentane is a molecular solid made up of molecules held together by weak intermolecular bonds. 2. Describe the factor in your question AND Link increasing the principle to how the system responds The factor in the question above is pressure. If there is an increase in pressure, the system responds by decreasing the pressure. 3. link state change (liquid to gas) to breaking bonds requiring energy Energy is required to change pentane from a liquid to a gas. The energy / heat is used to break weak intermolecular forces / bonds / attraction between pentane molecules. (not the strong covalent bonds between atoms in the molecule) [some questions will be decreasing] 3. Generally, explain which side of the equation is favoured (relate to moles) AND the general observations at visible and particle level. This occurs by favouring the reaction, either forward or reverse direction, that produces fewer gas particles. Because there are now fewer particles hitting the sides of the container, there is less pressure. 3. link state change to endothermic process 4. draw labelled diagram including labelled axis s, reactants HR, products HP and change in enthalpy H Because energy is needed to be absorbed by the pentane to break the bonds then this process of evapouration is endothermic. 4. Specifically, in reaction one describe number of moles in both sides of the equation AND link to which direction of reaction would be favoured (and observation) In Reaction One there are two moles of gas particles on each side of the equation. Because there are the same numbers of gas particles on both sides of the reaction, then a change in pressure will have no effect as neither reaction will be favoured. 5. Specifically, in reaction two link number of moles in both sides of the equation to observation AND link to which direction of reaction would be favoured In Reaction Two however, there are four moles of gas particles on the reactant side of the equation and two moles of gas particles on the product side of the equation. Therefore, when there is an increase in pressure, the system would shift and favour the forward reaction 6. Describe how the system shift in reaction two would effect at particle level AND final observation. meaning there are now fewer gas particles overall and hence fewer gas particles hitting the sides of the container and therefore less pressure overall. 26 7

4 Writing Excellence answers to Solids Solubility questions Writing Excellence answers to Equilibrium Temperature questions Solids Solubility QUESTION Question: Justify this statement in terms of the particles, structure, and bonding of these solids. You may use diagrams in your justification. Potassium chloride is soluble in water while Silicon dioxide and copper are insoluble in water (you will need to fill in the chart below correctly as part of the question and use the terms in your answer) Substance Type of Type of particle Attractive forces between particles KCl(s) potassium chloride ionic ion Ionic bonds / electrostatic attraction SiO2(s) silicon dioxide Covalent network atoms covalent Cu(s) copper metal atom Metallic bonds / electrostatic attraction Equilibrium Temperature QUESTION Question: In a reaction, the brown gas nitrogen dioxide, NO2(g), exists in equilibrium with the colourless gas dinitrogen tetroxide, N2O4(g). The equation for this reaction is represented by: 2NO2(g) N2O4(g) brown gas colourless gas The table below shows the observations when changes were made to the system. Analyse these experimental observations. In your answer you should: link all of the observations to equilibrium principles justify whether the formation of dinitrogen tetroxide from nitrogen dioxide is endothermic or exothermic. 1. For the first (name) 2. describe the structure of this type of using the terms above 3. explain how the bonding relates to the attraction between particles in your and water particles 4. link to the observation (solubility) in your question for the first 5. For the second (name) 6. describe the structure of this type of using the terms above KCl(s) potassium chloride is an ionic solid. KCl is made up of positive K + ions, and negative Cl ions, ionically bonded in a 3D lattice. When added to water, polar water molecules form electrostatic attractions with the K + and Cl ions. The partial negative charge,, on oxygen atoms in water are attracted to the K + ions and the partial positive, +, charges on the H s in water are attracted to the Cl ions, causing KCl to dissolve in water, and therefore be soluble SiO2(s) silicon dioxide is a covalent network solid. SiO2(s) is made up of atoms covalently bonded together in a 3D lattice structure. 7. explain how the bonding relates (Covalent bonds are strong), Polar water molecules are not strong / to the attraction between particles in insufficiently attracted to the Si and O atoms, your and water particles 8. link to the observation (solubility) therefore SiO2 is insoluble in water. in your question for the second 9. For the third (name) Cu(s) copper is a metallic solid. 10. describe the structure of this type of using the terms above Cu(s) is made up of an array of atoms (or ions) held together by non-directional forces between the positive nuclei of the atoms and the delocalised / free moving valence electrons. 11. explain how the bonding relates There is no attraction between the copper atoms and the (polar) water to the attraction between particles in molecules, your and water particles 12. link to the observation (solubility) therefore Cu is insoluble in water. in your question for the third State the equilibrium principle When a change is made to a system that is at equilibrium, the system responds to reduce the effect of that change. 2. Describe the factor in your question AND Link the principle to how the system responds to cooling or heating 3. Generally, explain which side of the equation is favoured (relate to endothermic or exothermic) 4. Specifically, for your reaction with heating, link the observation to which direction of reaction would be favoured (endothermic or exothermic) 5. Describe how the system shift in heating would affect which products are made AND final observation. The factor in the question above is temperature. If there is an increase in temperature, the system responds by absorbing more (heat) energy. If there is a decrease in temperature, the system responds by releasing more (heat) energy. With Heating (increasing temperature) this occurs by favouring the reaction, either forward or reverse direction, that is endothermic. With cooling (decreasing temperature) this occurs by favouring the reaction, either forward or reverse direction, that is exothermic. With heating In this case, the colour darkened, indicating that this favoured the reverse reaction, which must be the endothermic direction. So more 2NO2(g), nitrogen dioxide, which is a brown gas, would be formed and there would be less N2O4(g) dinitrogen tetroxide, which is a colourless gas 6. Specifically, for your reaction with With cooling In this case, the colour lightened, indicating that this favoured the cooling, link the observation to which forward reaction, which must be the exothermic direction. direction of reaction would be favoured (endothermic or exothermic) 7. Describe how the system shift in cooling would affect which products are made AND final observation. So more N2O4(g) dinitrogen tetroxide, which is a colourless gas would be formed and there would be less 2NO2(g), nitrogen dioxide, which is a brown gas, Writing Excellence answers to Thermochemical Calculations questions Writing Excellence answers to Equilibrium Expression questions Thermochemical Calculations QUESTION Question: Hexane, C6H14, like pentane, will combust (burn) in sufficient oxygen to produce carbon dioxide gas and water. Pentane combustion: C5H12(l) + 8O2(g) 5CO2(g) + 6H2O(l) ΔrH º = 3509 kj mol 1 Hexane combustion: 2C6H14(l) + 19O2(g) 12CO2(g) + 14H2O(l) ΔrH º = 8316 kj mol 1 Justify which alkane pentane or hexane will produce more heat energy when 125 g of each fuel is combusted in sufficient oxygen. M(C5H12) = 72.0 g mol 1 M(C6H14) = 86.0 g mol 1 (An equation and n=m/m are required for this type of thermochemical calculation) Equilibrium Expression QUESTION Question: The following chemical equation represents a reaction that is part of the Contact Process, which produces sulfuric acid. 2SO2(g) + O2(g) 2SO3(g) ΔH = 200 kj mol 1, Kc= 4.32 at 600 C (i) Write an equilibrium constant expression for this reaction. (ii): A reaction mixture has the following concentration of gases at 600 C: [SO2(g)] = mol L 1 [O2(g)] = mol L 1 [SO3(g)] = mol L 1 Justify why this reaction mixture is not at equilibrium, using the equilibrium expression and the data provided 1. Calculate the amount of energy per mol from the equation (divide ΔrH by number mol of in equation) 1 1 mole of pentane releases 3509 kj energy 1 : Write out the equilibrium constant expression in full 2SO 2(g) + O 2(g) 2SO 3(g) 2. calculate the number of mols of the known (K) n = m/m 3. multiply amount of energy per mol (step 1) by number of mols calculated (step 2) to get energy per mass Answer with units plus 3sgf 4. Calculate the amount of energy per mol from the equation (divide ΔrH by number mol of in equation) 2 5. calculate the number of mols of the known (K) n = m/m n (pentane) = m / M n (pentane) = 125 g / 72.0 g mol 1 = 1.74 mol = 6106 kj energy released. If 2 moles of hexane release 8316 kj energy, 2 : 8316 then 1 mole of hexane releases 4158 kj energy. 2 2 n (hexane) = m / M n (hexane) = 125 g / 86.0 g mol 1 = 1.45 mol 2. Calculate the Q value by inserting all of the [ ] data given. Show working and remember order of operation and 3sgf [SO ] K [SO ] [O ] 2 3 c Q multiply amount of energy per mol (step 4) by number of mols calculated (step 5) to get energy per mass Answer with units plus 3sgf 7. compare both s with summary statement = 6029 kj energy released Pentane releases 6106 kj of energy and Hexane releases 4158 kj of energy, therefore pentane releases more energy (77.0 kj) than hexane, per 125 g of fuel. Final value will have no units 3. Write down the Kc value and compare with the Q value stating whether it is equal or not (and therefore is or is not at equilibrium) 4. Link the Q value as either being bigger (and lying to the products side as the numerator is greater) OR as being smaller (and lying to the reactants side as the numerator is smaller) Since Kc = 4.32, Q Kc, so this reaction mixture is not at equilibrium. This number is greater than the Kc value, 4.32, which indicates that the reaction lies to the products side as the larger the Kc or Q value, the greater the numerator (products). 8 25

5 Writing Excellence answers to Reaction Rate Factors Catalyst questions Writing Excellence answers to Comparing Actual and Calculated enthalpy data questions Reaction Rate Factors Catalyst QUESTION Question: A particular reaction is complete when the solution turns cloudy and the paper cross under the flask can no longer be seen. The following experiments were carried out, and the times taken for the cross to disappear recorded. Elaborate on why the reaction in Experiment 3 occurs faster than the reaction in Experiment 1. Comparing Actual and Calculated enthalpy data QUESTION Question: The accepted enthalpy change for the combustion reaction of butane gas, C4H10(g), is Δr H = 5754 kj mol 1. Explain why calculated enthalpy is different to the accepted value. In your answer, you should include at least TWO reasons. 1. state the collision theory Chemical reactions between particles of s only occur when the following conditions have been met: Particles must collide with enough energy (called activation energy EA) and with the correct orientation. If these conditions are met, the collision will be considered successful. 2. Describe the reactants in your reaction and state which factors are the same 3. Describe the reactants in your reaction and state which factor is different (the factor affecting reaction rate) 4. link the factor to the collision theory 5. link the reaction to more of the collisions being successful occurring per unit of time 6. link to more products (name products) being formed per unit of time AND link to a faster reaction rate 7. summarize the reaction with the slower reaction rate 8. Explain that both reactions will produce the same amount of product eventually as they started with the same amount of reactants In the reaction of experiment 1 and experiment 3, both are carried out under the same temperature. (we assume the concentration is also the same) The only change is the addition of a catalyst in Experiment 3 compared to experiment 1. An added catalyst means a faster rate of reaction. Particles must collide with enough energy to overcome the activation energy of the reaction. The activation energy is the energy that is required to start a reaction. When a catalyst is used, the activation energy is lowered. This is because the catalyst provides an alternative pathway for the reaction to occur in which the activation energy is lowered. Now that the activation energy has been lowered, more reactant particles will collide with sufficient energy to overcome this lowered activation energy therefore more effective collisions are occurring more frequently. Experiment 3 will produce more products initially resulting in the solution turning cloudy and the cross disappearing quicker (5s compared to 42s), resulting in a faster reaction rate Experiment 1 has no catalyst so will take longer to react (cross to disappear) as less of the collisions are effective, so will have a slower rate of reaction than experiment 3. Both reactions will eventually produce the same amount of products if the same amounts of each reactant are used. 1. state values for both calculated data (worked out from a previous question on experimental data) and accepted data Units, sign and 3sgf 2. link results from experimental data to errors in experimental design The value for calculated data worked out from a previous question on experimental data for the combustion reaction of butane gas is rh = 3370 kj mol 1 The accepted enthalpy change for the combustion reaction of butane gas, C4H10(g), is Δr H = 5754 kj mol 1. The results from this experiment are less than the accepted results, due to errors in the experimental design. The errors could include: 3. explain error number 1. Some energy is used to heat the metal can and the air surrounding the experiment / the experiment was not conducted in a closed system, therefore not the entire amount is heating the water 4. explain error number 2. Incomplete combustion of butane, which releases less energy per mol of heat, to transfer to the water 5. explain error number 3. Some butane may have escaped before being ignited and therefore not all of the fuel is combusted with the heat energy transferred 6. explain error number 4. (may need only 2 or 3 in answer) 7. make summary statement linking that not energy released is transferred to heating the water Some energy was converted to light and sound OR The butane in the gas canister was impure OR Not carried out under standard conditions etc Therefore, not all of the energy released by the combustion of butane was transferred to heating the water, and the experimental data was calculated to be less than the actual data (carried out under error free conditions) 24 9 Writing Excellence answers to Reaction rate Factors Surface Area questions Solids Summary Reaction Rate Factors Surface Area QUESTION Question: Compare and contrast the reactions of 0.5 g of magnesium ribbon, Mg(s), with 50.0 ml of mol L 1 hydrochloric acid, HCl(aq), and 0.5 g of magnesium powder, Mg(s), with 50.0 ml of mol L 1 hydrochloric acid, HCl(aq). Refer to collision theory and rates of reaction in your answer. 1. state the collision theory Chemical reactions between particles of s only occur when the following conditions have been met: Particles must collide with enough energy ( called activation energy EA) and with the correct orientation. If these conditions are met the collision will be considered successful. 2. Describe the reactants in your reaction and state which factors are the same 3. Describe the reactants in your reaction and state which factor is different (the factor affecting reaction rate) In the reaction of hydrochloric acid with Mg ribbon and Mg powder, both form the same products, magnesium chloride and hydrogen gas. Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) The concentration and amount of the hydrochloric acid is the same in both reactions as is the mass of magnesium. (we assume the temperature is also the same) However, since Mg powder has a larger surface area than Mg ribbon Name of solid Type of particle in solid Attractive force broken when solid melts Molecular molecules Weak inter molecular Metallic atoms Metallic bonding Ionic ions Electrostatic Ionic bonding Covalent Network atoms Covalent bonding Attractive force between particle weak or strong (hardness) Relative melting point solubility weak low Yes if polar No if nonpolar Electrical conductivity strong high no yes yes strong high yes Only if molten or in solution no Malleable no no strong high no no no 4. link the factor to the collision theory 5. link the reaction to more successful collisions occurring per unit of time the powder will have more Mg particles immediately available to collide than the magnesium ribbon And therefore there will be more effective collisions per second (unit of time) Shape Summary Electron regions of negative charge (lone pairs or bond groups) 6. link to more products (name products) being formed per unit of time AND link to a faster reaction rate 7. summarize the reaction with the slower reaction rate and more H2 gas will be produced initially in the magnesium powder, resulting in a faster rate of reaction. Mg ribbon will take longer to react because fewer particles are immediately available to collide, so will have a slower rate of reaction Linear no lone pairs 1 lone pair triangular planar Bent no lone pair 1 lone pair 2 lone pair Tetrahedral triangular bent pyramid 8. Explain that both reactions will produce the same amount of product eventually as they started with the same amount of reactants Both reactions will eventually produce the same volume of hydrogen gas as the same amounts of each reactant are used CO 2 BF 3 SO 2 CH 4 NH 3 H 2 O ~

6 Writing Excellence answers to Bond enthalpy questions Writing Excellence answers to Reaction Rate Factors Temperature questions Bond enthalpy QUESTION Question: Ethene gas, C2H4 (g), reacts with bromine gas, Br2(g), as shown in the equation below. Calculate the enthalpy change, rh, for the reaction between ethane and bromine gases, given the average bond enthalpies below. Show your working and include appropriate units in your answers. Reaction Rate Factors Temperature QUESTION Question: A particular reaction is complete when the solution turns cloudy and the paper cross under the flask can no longer be seen. The following experiments were carried out, and the times taken for the cross to disappear recorded. Elaborate on why the reaction in Experiment 2 occurred faster than the reaction in Experiment list types of bonds for reactants (bonds broken) and products (bonds formed) AND number of each, in a table. Watch for double or triple bonds as these are separate (Draw Lewis structures if not given) 2. write bond type for each reactant (bonds broken) and product (bonds formed). Watch for double and triple bonds as they are different. Cross off on lewis diagram as you go 3. multiply bond enthalpy by number of each bond Then Write the The number of each bond type and multiply the enthalpy 4. total reactant bond enthalpy and total product enthalpy 5. bond broken (reactants) enthalpy total minus bond formed enthalpy (products) = enthalpy change, rh 6. total enthalpy and calculate enthalpy change (sign, units and 3sgf) 7. you may have to rearrange equation if enthalpy for a bond is required rh o = Bond energies(bonds broken) Bond energies(bonds formed) Bonds broken (reactants) Bond type number enthalpy Total enthalpy Bonds formed (products) Bond type number enthalpy Total enthalpy C=C C-C C-H C-H Br-Br C-Br Total Enthalpy (bonds broken) 2463kJ Total enthalpy = = -109kJ mol -1 Not needed Total enthalpy (bonds broken) 2572kJ state the collision theory Chemical reactions between particles of s only occur when the following conditions have been met: Particles must collide with enough energy (called activation energy EA) and with the correct orientation. If these conditions are met, the collision will be considered successful. 2. Describe the reactants in your reaction and state which factors are the same 3. Describe the reactants in your reaction and state which factor is different (the factor affecting reaction rate) 4. link the factor to the collision theory (activation energy) 5. link the reaction to more of the collisions being successful occurring per unit of time 6. next link the factor to the collision theory (faster moving particles) 7. link the reaction to more successful collisions occurring per unit of time 8. link to more products (name products) being formed per unit of time AND link to a faster reaction rate 9. summarize the reaction with the slower reaction rate 10. Explain that both reactions will produce the same amount of product eventually as they started with the same amount of reactants In the reaction of experiment 1 and experiment 2, both have no catalyst added.(we assume the concentration is also the same) The only change is an increase in temperature in Experiment 2 compared to experiment 2. An increase in temperature means a faster rate of reaction. The activation energy is the energy that is required to start a reaction. When the temperature is higher, the particles have more kinetic energy. because the particles are moving with more kinetic energy, it will be more likely that when collisions occur they are more likely to be effective as a greater proportion of collisions overcome the activation energy of the reaction. When the temperature is higher, the particles have more kinetic energy; the particles are moving faster Because the particles are moving faster, there will be also more frequent collisions. Experiment 2 has more effective collisions per unit of time. (than experiment 1. Experiment 2 will produce more products initially resulting in the solution turning cloudy and the cross disappearing quicker (23s compared to 42s), resulting in a faster reaction rate Experiment 1 is at a lower temperature so will take longer to react (cross to disappear) as the particles are moving slower than in experiment 1, so will have a slower rate of reaction. Both reactions will eventually produce the same amount of products if the same amounts of each reactant are used

7 Writing Excellence answers to Identifying Unknowns questions Identifying Unknowns QUESTION Question: Question: 1c: Four separate colourless organic liquids are known to be: ethanol ethanoic acid hex-2-ene 1-aminohexane Write a procedure to identify each of these organic liquids using only the reagents listed below. acidified dichromate solution, Cr2O7 2 / H + (aq) bromine water, Br2(aq) sodium carbonate solution, Na2CO3(aq). In your answer, you should: identify the test reagents used describe any observations that would be made identify the type of reaction that occurs identify the organic product of any reaction. You do not need to include equations in your answer. Step 1. Test reagents used start with Cr2O7 2 / H + adding a bit to each sample Observations - which will turn from orange to green with ethanol No change for the other 3 samples Writing Excellence answers to Cis-Trans Isomers questions Cis-Trans Isomers QUESTION Question: Molecule D can exist as geometric (cis and trans) isomers, with both isomers having the same molecular formula. Draw the geometric (cis and trans) isomers for molecule D in the boxes below. Justify why molecule D can exist as geometric (cis and trans) isomers. Your answer should include: an explanation of the requirements for cis and trans isomers reference to the structure of molecule D. 1. Draw the cis and trans isomers Cis Trans Step 2 Step 3. Step 4. Type of reaction that occurs - oxidation Organic product of any reaction - ethanol is oxidised to ethanoic acid. Test reagents used - add sodium carbonate solution, Na2CO3(aq). to the remaining 3 samples Observations - Bubbles of gas will be produced in the ethanoic acid sample No change for the other 2 samples Type of reaction that occurs acid-base reaction Organic product of any reaction in the acid-base reaction Sodium ethanoate / ethanoate ion is formed. Test reagents used add bromine water, Br2(aq) to the remaining 2 samples Observations - the bromine water, which turns from red / brown to colourless straightaway in the hex-2-ene sample No change for the other sample (for substitution in an alkane this reaction will be seen slowly with UV light as a catalyst) Type of reaction that occurs addition reaction Organic product of any reaction in the addition reaction It will form 2,3- dibromohexane is formed Test reagents used - Hexan-1-amine will be the chemical left over that will not react with any of the given reagents. Observations (in other questions red litmus paper can be used which will turn blue for an amine) Type of reaction that occurs Organic product of any reaction If you need to select the molecule make sure that it has both: a C=C double bond and 2 different groups of each C (name of molecules not normally required as part of the question) 2. link the presence of a double C=C bond to lack of rotation 3. link the requirement of two different groups of each of the C on the double 4. link the requirements above to your specific molecule (D) 5. Explain how two geometric isomers can have the same molecular formula Name: cis 1-chlorobut-1-ene Name: trans 1-chlorobut-1-ene For cis and trans isomers to occur a carbon-carbon double bond must be present as this prevents any rotation about this bond, and the atoms or groups of atoms attached to the two carbon atoms are therefore fixed in position. They must also have two different groups attached to each carbon (involved in the double bond). This molecule has a carbon-carbon double bond. One carbon of the double bond is attached to a hydrogen atom and an ethyl group. The other is attached to a hydrogen atom and a chlorine atom. When these two requirements are met, the two haloalkenes can have the same molecular formula and the same sequence of atoms in the formula, but a different arrangement in space (a different 3D formula), hence they are cis and trans isomers Writing Excellence answers to Alkene Reactions questions Alkene Reactions QUESTION Question: Ethene, C2H4(g), reacts with aqueous potassium permanganate solution, KMnO4(aq), dilute acid, H2O / H +, and hydrogen bromide, HBr. Compare and contrast the reactions of ethene gas with each of these three reagents. In your answer, you should: describe any observations that can be made identify, with reasons, the type of reaction ethene undergoes with each reagent describe the functional group of the products formed include equations showing the structural formulae for the organic compounds for each reaction. Reaction 1 Ethene, C2H4(g) reacts with aqueous potassium permanganate solution, KMnO4(aq), Reaction 2 Ethene, C2H4(g) reacts with dilute acid, H2O / H + Reaction 3 Ethene, C2H4(g) reacts with hydrogen bromide, HBr. Observations - The purple KMnO4 turns colourless (or brown) Reaction type - This is an oxidation or addition reaction in which the double bond is broken and two OH groups attach to each C atom of the double bond. Functional group of products Ethene reacts with aqueous KMnO4 to form a diol, ethan-1,2-diol. Observations - No colour changes are observed in this reaction. (colourless to colourless) Reaction type -This is an addition reaction as once again the double bond is broken. However, in this reaction one OH group and one H atom attach to each C atom of the double bond. Functional group of products Ethene reacts with dilute acid, H2O / H +, to form ethanol. Observations - Again there is no colour change observed. (colourless to colourless) Reaction type - This reaction is an addition reaction, as the double bond is broken and two atoms are added to each C atom of the double bond. In this reaction one H and one Br atom are added. Functional group of products When ethene reacts with hydrogen bromide, bromoethane is formed. Writing Excellence answers to Polymers questions Polymers QUESTION Question: Draw TWO repeating units of the polymer formed in Reaction 5. Explain why the formation of the polymer from its monomer is classified as an addition polymerisation reaction. Compare and contrast the polymer formed in Reaction 5 to the polymer formed in Reaction 3. In your answer you should explain why the polymers formed in these two reactions are different. 1. Identify the monomer, then draw the polymer (identify C 1 and C2 in monomer either side of the double bond then draw a chain of C (4 for 2 repeating units) and add on groups of each one removing double bond) 2. explain the definition of addition polymerisation 3. molecule 1 (reaction 5) describe the 2 groups of each end of the double bonded carbons 4. molecule 2 (reaction 3) describe the 2 groups of each end of the double bonded carbons monomer reaction 5 CH 3-CH 2-CH 2-CH=CH 2 reaction 3 CH 3-CH 2-CH=CH-CH 3 polymer C1 C2 C1 C2 C1 Since the monomer for this reaction is an alkene, when polymerisation occurs, the double bond in each alkene molecule is broken, freeing up a bonding space on each of the C atoms that was part of the double bond. This allows the monomers to join together by forming covalent bonds to make the polymer. Since double bonds in the alkene are being broken and molecules added into the freed up bonding spaces to make the monomer, this is an addition reaction. Polymerisation reactions occur when many monomers are chemically joined. In Reaction 3, the polymer formed will have a carbon with one hydrogen and a methyl group, and a carbon with one hydrogen and an ethyl group, as its repeating unit, due to the double bond being on the C2 position. In Reaction 5, since the double bond is in a different position (the C1 position), the polymer formed will have as its repeating unit a carbon atom with 2 hydrogen atoms attached, and a carbon atom with one hydrogen attached and a propyl group attached. C2 Summary of the three reactions All three reactions involve the breaking of the double bond. All three reactions involve addition (adding atoms on) Two of these reactions are addition reactions and one is an oxidation reaction. Only one of the reactions gives a colour change that is easily observed. 5. Explain that molecule 1 and 2 are structural isomers but have the same molecular formula The molecular formulae of the two repeating units of both polymers are the same, but the structural formulae are different. (States repeating units are structural isomers.) 18 15

8 Writing Excellence answers to Physical Properties of Organic Compounds questions Solubility - Physical Properties of Organic Compounds QUESTION Question: Explain why two layers form in Reaction One. Hexane reacts with bromine water 1. Identify the functional group of your (hexane) and name the product formed. Hexane is an alkane, with single bonded carbons. When reacting with bromine water it will form a haloalkane (bromohexane) during a substitution reaction Writing Excellence answers to Major and Minor products questions Major and Minor Products QUESTION Question: In Reaction 1, propan-2-ol can be converted to propene. In Reaction 2, propene can be converted back to propan-2-ol. Analyse BOTH of these reactions by: describing the reagents and conditions needed for each reaction to occur identifying each type of reaction and explaining your choice explaining why Reaction 1 forms only a single organic product, but Reaction 2 forms a mixture of organic products. 2. link observation (layers forming) to the polarity of the 3. identify the polarity of the bromine water and link to the s being immiscible (forming 2 layers) Two layers form in Reaction One as hexane is non-polar and the product (bromohexane) is effectively also non-polar. The water from the bromine water is polar and therefore the non-polar organic reactant and product will not dissolve in the water; because of this, two layers form as this polar and non-polar layer do not mix. 1. Reaction 1 Propan 2-ol forms propene Reagent and conditions -To convert propan-2-ol to propene, add concentrated sulfuric acid (which is a dehydrating agent). Reaction type - It is an elimination reaction because OH and H are removed from adjacent carbon atoms and a double bond is created to form an alkene. Structural Formula Melting point - Physical Properties of Organic Compounds QUESTION Question: Identify the trends shown on the graph. Identify which alkanes will be gases at room temperature (20 C) according to the graph beside. 2. Reaction 2 Propene forms propanol Label each structure with name and whether it is major or minor Reagent type and conditions - To convert propene to propan-2-ol, add dilute (sulfuric) acid. Reaction type - This is an addition reaction because the double bond is broken forming a C-C (single) bond, allowing H and OH from water to bond to the C atoms that were double bonded together. Structural Formula 1. link the boiling point trend to number of carbons in both groups (when explaining trends on a line graph always relate one variable to the other) 2. Identify which alkanes (number of carbons) are gases at room temp. (will have boiling point below 20 C) The boiling points of both alkanes and alcohols increase as the number of C atoms increases. The boiling points of alcohols are always higher than the alkanes (with the same number of C atoms). Alkanes with 1, 2, 3, and 4 C atoms (methane, ethane, propane, and butane) will be gases at room temperature Explain why reaction one forms only one product linked to symmetry 4. State Markovnikov s rule AND Explain the reason reaction two produces two products linked to Markovnikov s rule and asymmetry, including which is major and which is minor. Product type: Major Name: Propan-2-ol Product type: Minor Name: Propan-1-ol Reaction 1 forms only one product because the carbon atom from which the H is removed (C1 or C3) does not affect the structure of the product as propan-2-ol is symmetrical. Reaction 2 produces two products because an asymmetric reagent (H-OH) adds onto an asymmetric alkene (CH3CH=CH2). There are two carbons that the H or OH can bond with (C1 and C2), so there are two possible combinations. We can predict which will be the major product by using Markovnikov s rule, which states that the carbon with the most hydrogens gains more hydrogens. This means that most of the time, C1 will get another hydrogen while C2 will get the OH in this reaction. Propan-2-ol will be the major product and propan-1-ol the minor product. Writing Excellence answers to Haloalkane reactions questions Haloalkane reactions QUESTION Question: Chloroethane, CH3CH2Cl, reacts with aqueous KOH, alcoholic KOH, and with NH3. Compare and contrast the reactions of chloroethane with the three reagents. In your answer you should include: the type of reaction occurring and the reason why it is classified as that type the type of functional group formed equations showing structural formulae for reactions occurring. Writing Excellence answers to Alcohol Reactions questions Alcohol Reactions QUESTION Question: Butan-1-ol can react separately with each of PCl5, Cr2O7 2 / H +, and concentrated H2SO4. Elaborate on the reactions of butan-1-ol with each of the three reagents. For each reaction, your answer should include: the type of reaction occurring and the reason why it is classified as that type the name of the functional group formed in each product the structural formula of the organic product. 1. Reaction 1 - Chloroethane reacts with KOH(aq) Product formed - forms an alcohol, ethanol Reaction type - in a substitution reaction; Cl is replaced by OH. Condensed CH3CH2Cl CH3CH2OH Reaction 1 Butan-1-ol reacts with PCl5 Product formed The functional group in the product is a chloro group / chloroalkane (haloalkane) 1-chlorobutane Reaction type - Reaction with PCl5 is a substitution reaction. The hydroxyl group ( OH) is replaced by a chloro group ( Cl). Condensed CH3CH2CH2CH2OH CH3CH2CH2CH2Cl Reaction 2 Chloroethane reacts with KOH(alc) Product formed - forms an alkene, ethane (plus a HCl molecule) Reaction type - in an elimination reaction; H and Cl removed / HCl formed. Condensed CH3CH2Cl CH2 = CH2 + HCl Reaction 2 Butan-1-ol reacts with Cr2O7 2 / H + Product formed - The functional group in the product is carboxylic acid. butanoic acid Reaction type Reaction with acidified dichromate is oxidation as the alcohol is oxidised to a carboxylic acid. Condensed CH3CH2CH2CH2OH CH3CH2CH2COOH Reaction 3 Chloroethane reacts with NH3(alc) Product formed - forms an amine, aminoethane Reaction type - in a substitution reaction; Cl is replaced by NH2 Condensed CH3CH2Cl CH3CH2NH2 Reaction 3 Butan-1-ol reacts with concentrated H2SO4 Product formed - The functional group in the product is a (carbon-to-carbon) double bond / alkene. But-1-ene Reaction type - Reaction with concentrated H2SO4 is an elimination reaction. A hydrogen atom and the OH group on (adjacent) carbon atoms are removed forming a (carbon-to-carbon) double bond. Condensed CH3CH2CH2CH2OH CH3CH2CH=CH