Unit 5 A3: Energy changes in industry

1. ENTHALPY CHANGES Unit 5 A3: Energy changes in industry 1.1 Introduction to enthalpy and enthalpy changes 2 1.2 Enthalpy profile diagrams 2 1.3 Activation energy 3 1.4 Standard conditions 5 1.5 Standard enthalpy changes 5 2. CALCULATING ENTHALPY CHANGES 2.1 Determination of standard enthalpy changes 8 2.2 Calculating enthalpy changes directly from experiment 8 2.3 Comparison of experimental values with standard enthalpy values 12 2.4 Calculating enthalpy changes indirectly 12 2.5 Using bond enthalpies to calculate enthalpy changes indirectly 13 2.6 Using Hess law to calculate enthalpy changes indirectly 17 2.7 Finding the enthalpy change from enthalpy of combustion data 18 2.8 Finding the enthalpy change from enthalpy of formation data 21 Mixed questions 24 L3 Applied Science Unit 5: Physical Chemistry 1

1. ENTHALPY CHANGES 1.1 Introduction to enthalpy and enthalpy changes During a chemical reaction, bonds are broken and made. Within these bonds is chemical energy, which can be transferred to another form of energy (heat, light, sound etc) during the reaction. The amount of energy that leaves a chemical system is the exact same amount that goes into the surroundings (conservation of energy). Enthalpy, H, is the heat content (thermal energy) that is stored in a chemical reaction. We cannot directly measure the enthalpy of the reactants or the products. However, we can measure the energy absorbed or released to the surroundings during a chemical reaction. An enthalpy change, H, is the heat exchanged with the surroundings during a chemical reaction at constant pressure. This energy exchange is given units of kj mol -1. A chemical reaction will either release heat (exothermic reaction, H is negative) or absorb heat (endothermic reaction, H is positive). Enthalpy change: H = U + p V, (where U = internal energy of system, p = pressure and V = volume) Q. How can we measure the heat lost to the surroundings for a chemical reaction? How do we know if the reaction is exothermic or endothermic from an experiment? Exothermic: a reaction in which energy leaves the system to the surroundings Endothermic: a reaction in which energy enters the system from the surroundings 1.2 Enthalpy profile diagrams We can use enthalpy profile diagrams to show what happens to enthalpies during a reaction. We can determine whether the reaction is exothermic or endothermic by comparing the enthalpy of the reactants with the enthalpy of the products (looking at the enthalpy change, H). i) Exothermic reactions: Enthalpy H CH 4 (g) + 2 O 2 (g) H = - 890 kj mol -1 For exothermic reactions, heat is given out to the surroundings because the chemicals lose energy. Therefore H is negative. The enthalpy of the products is less than the enthalpy of the reactants. CO 2 (g) + 2 H 2O (g) Progress of reaction L3 Applied Science Unit 5: Physical Chemistry 2

ii) Endothermic reactions: Enthalpy H CaO (s) + CO 2 (g) H = + 178 kj mol -1 For endothermic reactions, heat is taken in from the surroundings because the reacting chemicals gain energy. Therefore H is positive. The enthalpy of the products is greater than the enthalpy of the reactants. CaCO 3 (s) Progress of reaction You must label the reactants, products, H and Ea (see later) on an enthalpy profile diagram! Note that H = Hproducts - Hreactants Q 1. A student performs two reactions. Reaction A has a positive H value and reaction B has a negative H value. Draw enthalpy profile diagrams for these reactions. Reaction A Reaction B Enthalpy H Enthalpy H Progress of reaction Progress of reaction 1.3 Activation energy We need to put in energy, the activation energy, to break the first bond and start a chemical reaction. E.g. for combustion of fuels, we supply the activation energy with a spark. Activation energy is the minimum energy required to start a reaction We can show the activation energy, Ea, on an enthalpy profile diagram: L3 Applied Science Unit 5: Physical Chemistry 3

Exothermic Endothermic Enthalpy H E a (positive) Reactants H (is negative) Enthalpy H E a (positive) Products H (is positive) Products Reactants Progress of reaction Progress of reaction The direction of the arrows is important. Activation energy is positive and the arrow points upwards. For exothermic reactions, H is negative and the arrow points down. For endothermic reactions, H is positive and the arrow points up! Q 2. Draw an enthalpy profile diagram for the following reaction: CO (g) + NO2 (g) CO2 (g) + NO (g) H = -226 kj mol -1 Ea = +134 kj mol -1 CHALLENGE: The enthalpy change for the reverse direction =... CHALLENGE: Activation energy for the reverse direction =... Q 3. Classify the reactions below as being either exothermic or endothermic. CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l) H = -890 kj mol -1... C6H12O6 (aq) + 6 O2 (g) 6 CO2 (g) + 6 H2O (l) H = -2801 kj mol -1... CaCO3 (s) CaO (s) + CO2 (g) H = +178 kj mol -1... 6 CO2 (g) + 6 H2O (l) C6H12O6 (aq) + 6 O2 (g) H = +2801 kj mol -1... L3 Applied Science Unit 5: Physical Chemistry 4

1.4 Standard conditions The enthalpy change of a reaction varies depending on the conditions that are present (e.g. temperature, pressure). Enthalpy changes for reactions are therefore measured under the same conditions. These are known as the standard conditions, which are: Pressure of 100 kpa (or 1 atmosphere) Temperature of 298 K (or 25 0 C) H Ɵ represents an enthalpy change measured under these standard conditions. Under these standard conditions, a substance will be in its standard state (the physical state of the substance at 1 atmosphere pressure and 25 0 C). Q 4. Complete the following standard states: Magnesium has the standard state Mg (s) (under standard conditions of 100 kpa and 298 K). Hydrogen has the standard state... Water has the standard state... 1.5 Standard enthalpy changes The standard enthalpy change of formation, Hf Ɵ, is the enthalpy change when one mole of a compound is made from its elements (the elements being in their standard states, reaction under standard conditions) e.g. H2 (g) + ½ O2 (g) H2O (l) Hf Ɵ = -286 kj mol -1 Note the enthalpy of formation of an element is defined as 0 kj mol -1 (If we are forming one mole of the element H2 (g) from the element H2 (g) then there is no chemical change!). The standard enthalpy change of combustion, Hc Ɵ, is the enthalpy change when one mole of a substance is completely combusted (in excess oxygen under standard conditions) e.g. C2H6 (g) + 3½ O2 (g) 2 CO2 (g) + 3 H2O (l) Hc Ɵ = -1560 kj mol -1 The standard enthalpy change of hydration, Hhyd Ɵ, is the enthalpy change when one mole of isolated gaseous ions is dissolved in water forming one mole of aqueous ions under standard conditions. e.g. K + (g) + aq K + (aq) Hhyd Ɵ = -322 kj mol -1 L3 Applied Science Unit 5: Physical Chemistry 5

Q 5. For the following reactions carried out under standard conditions, identify the enthalpy change as either Hc Ɵ, Hf Ɵ or Hhyd Ɵ. Cl - (g) + aq Cl - (aq)... CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)... 2 C (s) + 2 H2 (g) C2H4 (g)... Sr 2+ (g) + aq Sr 2+ (aq)... 2 Na (s) + C (s) + 1.5 O2 (g) Na2CO3 (s)... C5H12 (g) + 8 O2 (g) 5 CO2 (g) + 6 H2O (l)... Q 6. Write equations to show the enthalpy of formation of: Propane... Butane... Ethanol... Calcium carbonate... Q 7. Write equations to show the enthalpy of combustion of: Propane... Butane... Ethanol... Methanol... Q 8. Write equations to show the enthalpy of hydration of: Calcium... Sodium... Bromide... Sulfur... L3 Applied Science Unit 5: Physical Chemistry 6

Q 9. Past paper question (June 2012) L3 Applied Science Unit 5: Physical Chemistry 7

2. CALCULATING ENTHALPY CHANGES 2.1 Determination of standard enthalpy changes How do we determine H for a reaction? There are direct and indirect ways of doing this: Directly from experiment: Carry out an experiment and use a thermometer to determine the temperature change of the surroundings. Then use the data from the experiment to calculate H. This can be done for a reaction that has taken place either in solution or for a combustion reaction. Indirectly using either Hess s law or bond enthalpy data: H is not determined from experiment, but is calculated by using known data from a book. 2.2 Calculating enthalpy changes directly from experiment Reaction taking place in in solution Combustion reaction A simple way of calculating the enthalpy change for a reaction taking place in solution is to perform the reaction in a polystyrene cup (that acts as an insulator). For an exothermic reaction, heat is released to the surroundings (the water) and you will see a temperature increase on the thermometer. For an endothermic reaction heat is taken in from the surroundings (the water) and you see a temperature decrease on the thermometer. Substitute the temperature rise and the mass of the surroundings (water) into the equation below. Place a spirit burner containing the fuel (weighed) under a beaker of water with known volume and temperature. Burn the fuel and measure the temperature rise of the water. Weigh the spirit burner again at the end of the reaction to find the mass burned. Substitute the temperature rise and the mass of the surroundings (water) into the equation below. Q = m x c x T Heat exchanged with surroundings (J) Mass of surroundings (usually water, g) Specific heat capacity of surroundings (J g -1 K -1 ) Change in temperature of surroundings (K or 0 C) This equation gives you the heat exchanged with the surroundings in J. You must then calculate H in kj mol -1 ; see the example calculations below. L3 Applied Science Unit 5: Physical Chemistry 8

Worked example for a reaction taking place in solution: 25 cm 3 of 1.0 mol dm -3 HCl was added to 25 cm 3 of a 1.0 mol dm -3 NaOH solution in a polystyrene cup. The temperature rose by 6.9 0 C. The specific heat capacity of water is 4.18 J g -1 K -1 ; the density of water is 1.00 g cm -3. NaOH (aq) + HCl (aq) NaCl (aq) + H2O (l) Hr =? i) Calculate the energy released, in kj, during this reaction. Total mass of surroundings = 50g (water has a density of 1.00 g cm -3 ) Q = m c T = (25 + 25) x 4.18 x 6.9 = 1442.1 J = 1.4421 kj ii) Calculate the amount, in moles, of HCl that caused the temperature change. Moles = concentration x volume (in dm 3 ) = 1.0 x (25/1000) = 0.025 mol iii) Calculate the enthalpy change of the reaction. Give your answer to three significant figures. Scale the quantities to match the molar quantities in the equation. 0.025 mol released 1.4421 kj 1.0 mol would release (1.0/0.025) x 1.4421 = 57.684 kj The temperature rose, this was an exothermic reaction, the final answer should be negative! Hr = -57.7 kj mol -1 to 3 S.F. Worked example for a combustion reaction: A student burns 0.64 g of methanol in order to determine the enthalpy of combustion. The energy released was used to heat 100 cm 3 of water from to 19 0 C to 39 0 C. The specific heat capacity of water is 4.18 J g -1 K -1 ; the density of water is 1.00 g cm -3. i) Calculate the energy released during the combustion of methanol. Mass of surroundings = 100g, NOT 0.64g! Q = m c T = 100 x 4.18 x 20 = 8360 J = 8.36 kj ii) Calculate the amount, in moles, of methanol burnt. Mr of CH3OH = 32 Moles = mass / molar mass = 0.64 / 32 = 0.02 mol iii) Calculate the enthalpy of combustion of methanol. Scale up to 1 mole for enthalpy of combustion (check the definition for Hc Ɵ ) 0.02 mol released 8.36 kj 1.0 mol would release (1.0/0.02) x 8.36 = 418 kj The temperature rose, this was an exothermic reaction, the final answer should be negative! Hc = -418 kj mol -1 L3 Applied Science Unit 5: Physical Chemistry 9

Q 10. 5.5 g of BaCO3 was added to 100 cm 3 of a 2 mol dm -3 HCl solution. The temperature rose from 20 0 C to 32.7 0 C. The specific heat capacity of water is 4.18 J g -1 K -1 ; the density of water is 1.00 g cm -3. BaCO3 (s) + 2 HCl (aq) BaCl2 (aq) + H2O (l) + CO2 (g) (i) Calculate the energy released, in kj, during this reaction. Give your answer to three significant figures. (ii) Calculate the amount, in mol, of BaCO3 that caused the temperature change. Give your answer to four significant figures. (iii) Calculate the enthalpy change of the reaction. Include the sign in your answer. Give your answer to three significant figures. Q 11. A student burns 0.86 g of ethanol in order to determine the enthalpy of combustion. The energy released was used to heat 100 cm 3 of water from to 19 0 C to 37 0 C. The specific heat capacity of water is 4.18 J g -1 K -1 ; the density of water is 1.00 g cm -3. i) Calculate the energy released during the combustion of ethanol. Give your answer to four significant figures. ii) Calculate the amount, in mol, of ethanol burnt. Give your answer to three significant figures. iii) Calculate the enthalpy of combustion of ethanol. Include the sign in your answer. Give your answer to three significant figures. L3 Applied Science Unit 5: Physical Chemistry 10

Q 12. Past paper question (January 2011) Sodium ammonium thiocyanate, NH4SCN, reacts with solid barium hydroxide, B(OH)2, as shown in the equation below. 2 NH4SCN (s) + Ba(OH)2 (s) Ba(SCN)2 (s) + 2 H2O (l) + 2 NH3 (g) A research chemist carries out an experiment to determine the enthalpy change of this reaction. In the experiment, 15.22 g of NH4SCN is reacted with a slight excess of Ba(OH)2. The reaction absorbs energy, cooling the 50.0 g of water from 21.9 0 C to 10.9 0 C. i) Calculate the energy absorbed, in kj, during this reaction. The specific heat capacity of water = 4.2 J g -1 K -1. Energy =... kj [2] ii) Calculate the amount, in moles, of NH4SCN used by the research chemist. Amount =... mol [1] iii) Calculate the enthalpy change of reaction. Include the sign in your answer. Give your answer to two significant figures Hr =... kj mol -1 [3] L3 Applied Science Unit 5: Physical Chemistry 11

2.3 Comparison of experimental values with standard enthalpy values From a data book, Hc of propan-1-ol is given as -2021 kj mol -1. The value obtained from an experiment is -1881 kj mol -1. There is usually a difference between experimental values and standard enthalpy change of combustion values that you find in a text book. Why? Heat may have been lost to the surroundings There may have been incomplete combustion / an incomplete reaction Non-standard conditions may have been used A more sophisticated piece of apparatus that allows accurate measurements of energy changes is called a bomb calorimeter. This piece of apparatus ensures complete combustion and also reduces heat losses to the surroundings. Q 13. Past paper question (January 2010) Suggest two reasons why standard enthalpy changes of combustion determined experimentally are less exothermic than the calculated theoretical values....... [2] 2.4 Calculating enthalpy changes indirectly It is not always possible to measure the enthalpy change directly from experiment for the following reasons: The activation energy of the reaction is too high There is a slow reaction rate More than one reaction may be taking place (and so more than one product is formed) We can, however, calculate these enthalpy changes indirectly by using other known enthalpy values. This can be done by one of two methods that shall now be discussed in more detail: a) Using bond enthalpy data. b) Constructing enthalpy cycles and using Hess s law. L3 Applied Science Unit 5: Physical Chemistry 12

2.5 Using bond enthalpies to calculate enthalpy changes indirectly Bond enthalpy is the enthalpy change when one mole of (gaseous covalent) bonds is broken (homolytically) e.g. H H (g) 2 H (g) H = +436 kj mol -1 Bond enthalpies tell you how much energy is needed to break different bonds. Bond enthalpy values are positive as you need to put in energy to break a bond. Breaking bonds absorbs energy and is an endothermic process. Forming bonds releases energy and is an exothermic process. Energy is needed to break bonds in reactants Energy is released as new bonds are formed in products What determines whether a reaction is exothermic or endothermic overall? Q 14. Past paper question (June 2011) Many organisms use the aerobic respiration of glucose, C6H12O6, to release useful energy. The overall equation for aerobic respiration is the same as for the complete combustion of C6H12O6. i) Write the equation for the aerobic respiration of C6H12O6.... [1] ii) Explain, in terms of bond breaking and bond forming, why this reaction is exothermic....... [2] L3 Applied Science Unit 5: Physical Chemistry 13

Bond enthalpies are average values; for example, you can find C H bonds in almost every organic molecule. The C H bond strength will vary across the different environments it is found in. The bond enthalpy of a C H bond given in an exam, +415 kj mol -1, is an average value. We can use average bond enthalpy values to work out the enthalpy change of reactions involving gases using the equation (which you must learn): H = Ʃ (bond enthalpies of bonds broken) - Ʃ (bond enthalpies of bonds made) Look through the worked example below to see how this equation is used. Worked example on using bond enthalpy data to calculate enthalpy changes indirectly: Use the average bond enthalpies in the table to calculate the enthalpy change of combustion of propane. C3H8 + 5 O2 3 CO2 + 4 H2O Bond Average bond enthalpy / kj mol -1 C H +413 C C +347 O = O +498 C = O +805 O H +464 Draw displayed formulae for each molecule in the reaction so that you can easily see all of the bonds: Bonds broken bonds formed 8 x C H 8 x 413 6 x C = O 6 x 805 2 x C C 2 x 347 8 x O H 8 x 464 5 x O = O 5 x 498 6488 8542 H = Ʃ (bond enthalpies of bonds broken) - Ʃ (bond enthalpies of bonds made) = 6488 8542 = -2054 kj mol -1 L3 Applied Science Unit 5: Physical Chemistry 14

Q 15. Calculate the enthalpy of combustion of ethanol using the bond enthalpy data given below. C2H5OH (l) + 3 O2 (g) 2 CO2 (g) + 3 H2O (l) Bond Average bond enthalpy / kj mol -1 C H +413 C C +347 O = O +498 C = O +805 O H +464 C O + 358 Q 16. Calculate the enthalpy of combustion of butane using the bond enthalpy data given below. C4H10 (g) + 6.5 O2 (g) 4 CO2 (g) + 5 H2O (l) Bond Average bond enthalpy / kj mol -1 C H +413 C C +347 O = O +498 C = O +805 O H +464 L3 Applied Science Unit 5: Physical Chemistry 15

Q 17. Past paper question (January 2012) L3 Applied Science Unit 5: Physical Chemistry 16

2.6 Using Hess law to calculate enthalpy changes indirectly Hess law: for any chemical change, the enthalpy change is the same regardless of the route taken. Worked examples: Introduction to using Hess cycles; work through the examples below: A H = -200 kj mol -1 H = -120 kj mol -1 H = -80 kj mol -1 C B The enthalpy change is the same, regardless of which route you take to go from A to B (-200). From Hess law: HAB = HAC + HCB -200 = -120 + -80-200 = -200 A H = H = -150 kj mol -1 H = -100 kj mol -1 C B We do not know H AB, but we can calculate this using Hess law. From Hess law: HAB = HAC + HCB From Hess law: A H = H = -150 kj mol -1 H = -100 kj mol -1 HAB = HAC + HCB C B Going from B to C is exothermic (negative H), so going in the opposite direction, from C to B would be endothermic (positive H ) A H = H = -150 kj mol -1 H = -100 kj mol -1 C B Remember to always reverse the sign (+ / -) if going against the arrow From Hess law: HAB = HAC + HCB L3 Applied Science Unit 5: Physical Chemistry 17

We can use Hess law to calculate the: a) enthalpy change of formation from enthalpy of combustion data b) enthalpy change of reaction from enthalpy of formation data. 2.7 Finding the enthalpy change from enthalpy of combustion data Many compounds can be reacted with oxygen and the enthalpy change of combustion measured accurately. These enthalpy changes can be used to calculate the enthalpy changes for other reactions. Worked example for calculating the enthalpy change of formation when given Hc data: Use the data given below to calculate the enthalpy change of formation for methane. Substance Hc / kj mol -1 C (s) -393 H2 (g) -286 CH4 (g) -890 1. Write an equation to show the enthalpy change of formation. 2. Construct an enthalpy triangle with arrows pointing down towards combustion products if given enthalpy of combustion data. 3. Use Hess law to calculate Hf. C (s) + 2 H 2 (g) H c = -393 + (2 x -286) = -965 kj mol -1 H f CH 4 (g) H c = -890 kj mol -1 Direction of the arrows points down towards the combustion products Combustion products (CO 2 (g) and 2 H 2 O (l) ) From Hess law: Hf = -965 + 890 = -75 kj mol -1 Reverse the sign (+890) when going against an arrow in the opposite direction Note - there is also a shortcut that you can use without having to draw out a Hess cycle: Hf = Ʃ Hc (reactants) - Ʃ Hc (products) For the example above: Hf = -965 (-890) = -75 kj mol -1 L3 Applied Science Unit 5: Physical Chemistry 18

Q 18. Calculate the enthalpy change of formation of ethane using the data given below. Substance Hc / kj mol -1 C (s) -393 H2 (g) -286 C2H6 (g) -1560 Q 19. Calculate the enthalpy change of formation of ethanol using the data given below. Substance Hc / kj mol -1 C (s) -393 H2 (g) -286 C2H5OH (g) -1367 L3 Applied Science Unit 5: Physical Chemistry 19

Q 20. Past paper question (June 2009) i) Add state symbols to the equation to show each species in its standard state. 6 C (...) + 7 H2 (...) C6H14 (...) [1] ii) It is very difficult to determine the standard enthalpy change of formation of hexane directly. Suggest a reason why....... [1] iii) The standard enthalpy of formation of hexane can be determined indirectly. Calculate the standard enthalpy change of formation of hexane using the standard enthalpy changes of combustion below. Substance Hc / kj mol -1 C -394 H2-286 C6H14-4163 answer =... kj mol -1 [3] L3 Applied Science Unit 5: Physical Chemistry 20

2.8 Finding the enthalpy change from enthalpy of formation data Chemists have measured enthalpy changes of formation for many compounds accurately. These enthalpy changes can be used to calculate the enthalpy changes for other reactions. Worked example for calculating the enthalpy change of combustion when given Hf data: Use the data given below to calculate the enthalpy change of combustion for ethane. Substance Hf / kj mol -1 C2H6 (g) -84.7 CO2 (g) -394 H2O (l) -286 1. Write an equation to show the enthalpy change of combustion. 2. Construct an enthalpy triangle with arrows pointing up from the elements if given enthalpy of formation data. 3. Use Hess law to calculate Hc. Enthalpy change of formation for elements = 0 C 2 H 6 (g) + 3.5 O 2 (g) H f = -84.7 + (3.5 x 0) = -84.7 kj mol -1 Elements (2 C (s), 3 H 2 (g) and 3.5 O 2 (g) ) From Hess law: Hr = +84.7 + (-1646) = -1561.3 kj mol -1 2 CO 2 (g) + 3 H 2 O (l) Note - there is also a shortcut that you can use without having to draw out a Hess cycle: H c Hr = Ʃ Hf (products) - Ʃ Hf (reactants) For the example above: Hr = -1646 (-84.7) = -1561.3 kj mol -1 H f = (2 x -394) + (3 x -286) = -1646 kj mol -1 Direction of arrows points up from the elements Reverse the sign (+84.7) when going against an arrow in the opposite direction L3 Applied Science Unit 5: Physical Chemistry 21

Q 21. Calculate the enthalpy change of combustion of methanol using the data given below. Substance Hf / kj mol -1 CH3OH (l) -239.1 CO2 (g) -394 H2O (l) -286 Q 22. Calculate the enthalpy change of combustion of ethanol using the data given below. Substance Hf / kj mol -1 CH3CH2OH (l) -278 CO2 (g) -394 H2O (l) -286 L3 Applied Science Unit 5: Physical Chemistry 22

Q 23. Use the information below to calculate Hf for ethane, given that Hf for ethene is +52 kj mol -1. C2H4 (g) + H2 (g) C2H6 (g) Hr = -137 kj mol -1 Q 24. Past paper question (June 2012) Hess law can be used to calculate enthalpy changes of reaction. The equation for the reaction that gives the enthalpy change of formation, Hf, of N2O (g) is as follows. N2(g) + ½ O2 (g) N2O (g) i) It is not possible to measure the enthalpy change of formation of N2O directly. Suggest why it is not possible....... [1] ii) The data below can be used to calculate the enthalpy change of formation, Hf, of N2O (g). Calculate Hf for N2O (g). reaction Hr / kj mol -1 C (s) + N2O (g) CO (g) + N2 (g) -193 C (s) + ½ O2 (g) CO (g) -111 Hf =... kj mol -1 [2] L3 Applied Science Unit 5: Physical Chemistry 23

MIXED QUESTIONS: Careful! For the questions below, decide which method to use to determine the enthalpy change! Q 25. A student burns 0.27 g of hexane in order to determine the enthalpy of combustion. The energy released was used to heat 100 cm 3 of water from to 18 0 C to 38 0 C. The specific heat capacity of water is 4.18 J g -1 K -1 ; the density of water is 1.00 g cm -3. i) Calculate the energy released during the combustion of hexane. Give your answer to three significant figures. ii) Calculate the amount, in mol, of hexane burnt. Give your answer to three significant figures. iii) Calculate the enthalpy of combustion of hexane. Include the sign in your answer. Give your answer to four significant figures. Q 26. Calculate the enthalpy change for the reaction using the bond enthalpy data given in the table. N2H4 (l) + 2 F2 (g) N2 (g) + 4 HF (g) Bond Average bond enthalpy / kj mol -1 N N +163 N H +388 N = N +944 F F +158 H F +562 L3 Applied Science Unit 5: Physical Chemistry 24

Q 27. 0.327 g of Zn powder was added to 55.0 cm 3 of excess CuSO4 at 22.8 0 C. The temperature rose to 32.3 0 C. The specific heat capacity of water is 4.18 J g -1 K -1 ; the density of water is 1.00 g cm -3. Zn (s) + CuSO4 (aq) ZnSO4 (aq) + Cu (s) (i) Calculate the energy released, in kj, during this reaction. Give your answer to four significant figures. (ii) Calculate the amount, in mol, of Zn that caused the temperature change. (iii) Calculate the enthalpy change of the reaction. Include the sign in your answer. Give your answer to three significant figures. Q 28. Calculate the enthalpy change of formation of butane using the data given below. Substance Hc / kj mol -1 C (s) -393 H2 (g) -286 C4H10 (g) -2877 L3 Applied Science Unit 5: Physical Chemistry 25

Q 29. A student mixed 100.0 cm 3 of a 1.50 mol dm -3 hydrochloric acid solution with 100.0 cm 3 of a 1.50 mol dm -3 sodium hydroxide solution. Both solutions were at 19.67 0 C initially and the highest temperature reached by the reaction mixture was 34.06 0 C. The specific heat capacity of water is 4.18 J g -1 K -1 ; the density of water is 1.00 g cm -3. (i) Construct the equation for the reaction.... (ii) Calculate the enthalpy change of the reaction. Give your answer to three significant figures. L3 Applied Science Unit 5: Physical Chemistry 26

Q 30. Calculate the enthalpy change for the reaction using the bond enthalpy data given in the table. C6H14 (g) 3 C2H4 (g) + H2 (g) Bond Average bond enthalpy / kj mol -1 C H +413 C C +347 C = C +610 H H +436 Q 31. Calculate the enthalpy change of formation of benzene using the data given below. Substance Hc / kj mol -1 C (s) -393 H2 (g) -286 C6H6 (g) -3267 L3 Applied Science Unit 5: Physical Chemistry 27

Q 32. Calculate the enthalpy change for the reaction using the data given below. Substance Hf / kj mol -1 CaCO3 (s) -1207 CO2 (g) -394 CaO (s) -635 CaCO3 CaO + CO2 Q 33. Calculate the enthalpy change for the reaction using the data given below. Substance Hf / kj mol -1 CO (g) -111 CO2 (g) -394 Fe2O3 (s) -822 3 CO (g) + Fe2O3 (s) 2 Fe + 3 CO2 L3 Applied Science Unit 5: Physical Chemistry 28

Q 34. Past paper question (January 2012) You could be asked to determine the enthalpy change of reaction from an unfamiliar looking enthalpy cycle [3] L3 Applied Science Unit 5: Physical Chemistry 29