Chemical Reaction Defn: Chemical Reaction: when starting chemical species form different chemicals. Evidence to indicate that a chemical reaction has occurred: Temperature change Different coloured materials may be formed New phases may be formed (eg. from solid to gas) Defn: Chemical Reaction Equation: an equation that shows the chemicals used up and produced during a chemical reaction General form of a chemical equation is: REACTANTS! PRODUCTS Demo on Law of Conservation of Mass: Lead (II) Nitrate + Potassium Iodide! Lead (II) Iodide + Potassium Nitrate Pb(NO 3 ) 2 + KI! PbI 2 + KNO 3 Defn: Reactants start off as clear, colourless solutions Product is yellow in colour The mass before and after the reaction of the flask and its contents is the same Law of Conservation of Mass: In a chemical reaction, the total mass of the reactants is equal to the total mass of the products Law of Conservation of Atoms: In a chemical reaction, the total number of Checking your understanding: Which of the following(s) is conserved? 1. Element 2. Mass atoms before is equal to the total number of atoms after a chemical reaction. 3. Atoms 4. Volume 1
Balancing Chemical Equations: A chemical equation is balanced if there is the same number of each type of atom on each side of the equation. Example: (NH 4 ) 2 S + Pb(NO 3 ) 2! NH 4 NO 3 + PbS Guideline to balancing equations: 1. Put coefficient to balance (big numbers in front of each chemical) 2. Balance entire groups (eg. polyatomic ions: NO 3 -, NH 4 + ) first if they are found on both sides. 3. Balance metal atoms next. 4. Try to avoid balancing H, and O s until the end if possible. These atoms are often found several times throughout an equation. 5. If there are fractions in the equation (eg. ½ O 2 ), multiply the equation by the whole number (eg. 2) to eliminate the fraction. 6. Always double check your equation after balancing. Answer: (NH 4 ) 2 S + Pb(NO 3 ) 2! 2 NH 4 NO 3 + PbS Example #2: N 2 + H 2! NH 3 Answer: Balance N first N 2 + H 2! 2NH 3 Hydrogens in NH 3 now have a coefficient in front of it, check on the other side of the equation for H s and balance them too! N 2 + 3H 2! 2NH 3 Subscripts: A chemical equation can also show the states in which the reactants and products exist. The states of chemicals are shown by including the following symbols in brackets immediately after the chemical formula. (s) = solid, (l) = liquid, (g) = gas, (aq) = aqueous [ dissolved in water ] 2
Example: Solid zinc reacts with a hydrochloric acid solution to form hydrogen gas and dissolved zinc chloride. Zn (s) + 2HCl (aq)! H 2 (g) + ZnCl 2 (aq) Note: crystals, powder, and precipitate all mean the phase is a solid! Defn: Precipitate: a solid formed when two liquid or aqueous solutions react. Predicting Products The key to predicting products is to be able to identify the type of reaction that the reactants will undergo Example: 1. NO 2!? There is only one compound on the reactant side. This is a decompose tion reaction! The compound will break down into its individual elements. Therefore: 2NO 2! N 2 + 2O 2 2. Mg + ZnSO 4!? There is one element and one compound on the reactant side. This is a single replacement reaction! The metals (or non-metals) will switch places. Therefore: Mg + ZnSO 4! Zn + MgSO 4 3. Al + O 2!? There are two elements by themselves on the reactant side. This is a synthesis reaction! The elements combine to form one compound. Therefore: 4Al + 3O 2! 2Al 2 O 3 3
4. HCl + KOH!? There is an acid and a base on the reactant side. This is a neutralization reaction (a special case of double displacement reaction)! Water is one product, and the other ions combine to form a salt. Therefore: HCl + KOH! KCl + H 2 O 5. C 2 H 2 + O 2!? There is a hydrocarbon compound and oxygen on the reactant side. This is a combustion reaction! The products will be CO 2 and H 2 O Therefore: 2C 2 H 2 + 5O 2! 4CO 2 + 2H 2 O 4
Type of Reaction Definition General Formula Examples What to look for! Synthesis (S) Combination of two or more substances to form a compound A + B! AB C + O 2! CO 2 H 2 + Cl 2! 2HCl Two elements as reactants. Decomposition (D) Breaking down a molecule into simpler substances AB! A + B 2KClO 3! 2KCl + 3O 2 2Ag 2 O! 4Ag + O 2 One compound as a reactant. Single Replacement (SR) Replacing one atom in a compound by another atom AX + B! BX + A Cu + 2AgNO 3! Cu(NO 3 ) 2 + 2Ag Cl 2 + 2KI! I 2 + 2KCl Element and a compound react. Double Replacement (DR) Exchange of atoms or groups between two different compounds AX + BY! AY + BX Pb(NO 3 ) 2 + 2KI! PbI 2 + 2KNO 3 Two compounds react. Neutralization (N) Special type of double replacement. Reaction of an acid with a base HA + BOH! H 2 O + BA H 2 SO 4 + 2KOH! K 2 SO 4 + 2H 2 O HCl + NaOH! NaCl + H 2 O Acid (starts with H ) and Base (end with OH ) react Combustion (C) Reaction of hydrocarbon with oxygen to produce carbon dioxide and water C x H y + O 2! H 2 O + CO 2 CH 4 + 2O 2! CO 2 + 2H 2 O C 5 H 12 + 8O 2! 5CO 2 + 6H 2 O Hydrocarbon and O 2 5
Chemical Reactions and Energy: Molecules are held together by chemical bonds. To break a bond, energy has to be added to the bond. (eg. breaking a string requires energy from you!) To form a bond, atoms must give off their excess energy There are two types of reactions: Exothermic: gives off heat to its surroundings. (heat exits from the reactants) Eg. CaCl 2 (s) + H 2 O (l)! Ca 2+ (aq) + 2Cl (aq) [Heat pack demo] Endothermic: absorbs heat from its surroundings. (heat enters the reaction) Eg. NH 4 NO 3 (s) + H 2 O (l)! NH 4 + (aq) + NO 3 (aq) [Cold pack demo] If more energy is used to break the bonds in the reactants than is given off producing the bonds in the products, the reaction is ENDOTHERMIC. If less energy is used to break the bonds in the reactants than is given off producing the bonds in the products, the reaction is EXOTHERMIC. 7
Chemistry 11 Acid / Base Titration Notes At the beginning of the year we learned how to classify chemical reactions. One of the special double replacement reactions that we came across was called a Neutralization Reaction. Remembering back, a neutralization reaction involves a reaction between an acid and a base to give us a salt and water. Acid + Base! Salt + Water In general, acids are compounds that begin with an H in their formula. (Organic acids ends with COOH.) The following is a list of common acids that you will often see in a chemistry lab: HF Hydrofluoric Acid H 2 SO 4 Sulphuric Acid HCl Hydrochloric Acid H 2 SO 3 Sulphurous Acid HBr Hydrobromic Acid HNO 3 Nitric Acid HI Hydroiodic Acid HNO 2 Nitrous Acid H 3 PO 4 Phosphoric Acid CH 3 COOH Acetic Acid H 2 CO 3 Carbonic Acid In general, bases are compounds that end with an OH in their formula and are commonly known as hydroxides. The following is a list of common bases that you will often see in a chemistry lab: NaOH Sodium Hydroxide NH 4 OH Ammonium Hydroxide Ca(OH) 2 Calcium Hydroxide Al(OH) 3 Aluminum Hydroxide In general, a salt is defined as a substance that is neither an acid nor a base. It is an ionic compound that does not begin with an H or end with an OH. (Example: NaCl) Thus, an example of a neutralization reaction that involves HCl and NaOH is as follows: HCl + NaOH! NaCl + HOH Acid Base Salt Water 8
Titrations Chemists have developed a special process of finding an unknown concentration of a chemical in a solution based upon the neutralization reaction. This process is known as the Titration. In a titration, a solution of known concentration is reacted with a solution of an unknown concentration (but a known volume), until a desired equivalence point is reached. The equivalence point is the point at which all of the acid and base have reacted together in a solution. In other words, the moles of acid present is equal to the moles of base present in the solution! Moles of Acid = Moles of Base Thus, all that is left in the solution after the reaction is salt and water. The following is a diagram of a typical acid / base titration experiment setup: 9
We use a table and stoichiometry after we have collected the data from the experiment. to help us figure out the unknown concentration Example: Consider the reaction: H 3 PO 4 + 2KOH! K 2 HPO 4 + 2H 2 O If 0.0198 L of H 3 PO 4 with an unknown concentration is reacted with 0.0250 L of 0.500 M KOH to reach the equivalence point, what is the concentration of H 3 PO 4? Table: Molarity? H 3 PO 4 KOH 0.500 M Volume 0.0198 L 0.0250 L Road Map: Conc KOH! Moles KOH! Moles H 3 PO 4! Conc H 3 PO 4 Step 1: Calculate the moles of KOH present. Moles KOH = 0.500 M x 0.0250 L = 0.0125 mol KOH Step 2: Calculate the moles of H 3 PO 4 present. Moles H 3 PO 4 = 0.0125 mol KOH x 1 mol H 3 PO 4 = 0.00625 mol H 3 PO 4 2 mol KOH Step 3: Calculate the concentration of H 3 PO 4. Conc H 3 PO 4 = 0.00625 mol H 3 PO 4 = 0.316 M 0.0198 L In a titration experiment, usually more than one volume data is taken to make sure that the titration is done accurately. How do we approach a question like this? Let s take a look at an example. Example: Consider the reaction: Ca(OH) 2 + 2HCl! CaCl 2 + 2H 2 O A lab technician titrates a 10.0 ml sample of Ca(OH) 2 solution with 0.0156 M HCl. The following are the volume data for the titration: Trial Volume HCl used 1 15.35 ml 2 15.45 ml 3 15.80 ml 10
What is the molarity of the Ca(OH) 2 in the solution? Note: If you are given more than one trial for the volume data, choose at least two trials that have the volume within ± 0.10 ml and take an average of the two volumes. Average Volume used = (15.35 + 15.45) ml = 15.40 ml HCl 2 Table: HCl Ca(OH) 2 Molarity 0.0156 M? Volume 15.40 ml 10.0 ml Road Map: Conc HCl! Moles HCl! Moles Ca(OH) 2! Conc Ca(OH) 2 Step 1: Calculate the moles of HCl present. Moles HCl = 0.0156 M x 0.01540 L = 0.0002402 mol Step 2: Calculate the moles of Ca(OH) 2 present. Moles Ca(OH) 2 = 0.0002402 mol HCl x 1 mol Ca(OH) 2 = 0.0001201 mol 2 mol HCl Step 3: Calculate the concentration of Ca(OH) 2. Conc Ca(OH) 2 = 0.0001201 mol Ca(OH) 2 = 0.0120 M 0.0100 L 11
Excess and Limiting Reagent Study Guide Review: When 15.0g of CH 4 is reacted with an excess of Cl 2 according to: CH 4 +Cl 2 CH 3 Cl+HCl (1) What is the mass of CH 3 Cl produced? Data Table: Compound Molar Mass (g/mol) CH 4 16.0 Cl 2 71.0 CH 3 Cl 50.5 g CH 3 Cl is produced Limiting Reactant Problem: When 15.0g of CH 4 is reacted with 10.0g of Cl 2 according to the equation (1), what is the mass of CH 3 Cl produced? 15.0g CH 4! CH 4 mol! CH 3 Cl mol! CH 3 Cl g 10.0g Cl 2! Cl 2 mol! CH 3 Cl mol! CH 3 Cl g Which road map to use?? g CH 3 Cl is produced There is enough CH 4 to make g of CH 3 Cl, but only enough Cl 2 to make g of CH 3 Cl. So, Cl 2 sets a limit on the amount of CH 3 Cl produced. Therefore, Cl 2 is called the. The amount of Cl 2 is used up completely and the reaction stops. On the other hand CH 4 is called the excess because not all of it will be used and there will be some left over. The smallest amount of CH 3 Cl calculated (7.11g) is known as the of product or the. What calculation did we do to find which reactant was the LR? Calculate how much a product * could be made based on each reactant, then compared the result. The reactant that makes the least amount of product is the LR. * Caution: When a question asks for the identity of LR and there is more than one product in the reaction, pick a product, and stick to it throughout the calculation until you figure out which reactant is the LR.
Quick Practice: 2H 2 + O 2! 2H 2 O Calculate how much H 2 O can be made using 10.0g of H 2 and 50.0g of O 2 Which reactant is the limiting reagent (LR) and which is the excess reagent (ER)? What is the theoretical yield of H 2 O? Next question: When 15.0g of CH 4 is reacted with 10.0g of Cl 2 according to the equation (1), how much of the excess reactant (CH 4 ) is left? Amount of excess left = Starting mass of CH 4 Actual mass of CH 4 used = 15.0g CH 4 -? g CH 4 Based on the mass of LR given, find the mass of excess reactant actually used 10.0 g Cl 2! Cl 2 mol! CH 4 mol! CH 4 g g Cl 2 is used Amount of excess left = 15.0g CH 4 - g CH 4 = g CH 4 Quick Practice: 2H 2 + O 2! 2H 2 O We calculated how much H 2 O can be made using 10.0g of H 2 and 50.0g of O 2. H 2 is the ER and O 2 is the LR. How many gram of H 2 is left? Be able to answer: a) Which reactant is the limiting reactant? b) What is the expected mass of the product? c) How much of the excess reactant is left?
Percentage Yield Sometimes 100% of the expected amount of products can t be obtained from a reaction. Thus, we would want to know the percentage of the actual amount of product obtained when compared to the expected amount from our stoichiometry calculations. We can express this percentage as Percentage Yield of the reaction. Percentage Yield = Mass of product obtained x 100% Mass of product expected Example: When 10.0 g of CaCO 3 was used, 9.0 g of product was produced. What was the percentage yield of the reaction? Step 1: Calculate expected product Road Map: g CaCO 3! mol CaCO 3! mol CaCl 2! mol CaCO 3! g CaCO 3 Molar Mass of CaCO 3 : (40.1 g/mol) + (12.0 g/mol) + (3 x 16.0 g/mol) = 100.1 g/mol Calculate: Step 2: Calculate percentage yield % yield = mass obtained x 100% = 9.0 g x 100% = 90% mass expected 10.0 g Why is percentage yield usually less than 100%? beaker to another etc. Human error (reading measurements inaccurately How can percentage yield be more than 100%?