THERMODYNAMICS (Advanced questions) Link to Data Page Your feedback on these self-help problems is appreciated. Click here to send an e-mail. For on-line help with this topic, see the chemcal module "Chemical Energy and Calorimetry" Shortcut to Questions Q: 1 2 3 4 5 6 7 8 9 10 1 Calcium reacts completely with excess water to produce calcium hydroxide and hydrogen. (1.) Write a balanced equation for the reaction. (2.) Calculate the enthalpy change, ΔH°, when 0.300 mole of calcium and 14.0 mole of water are mixed at 298 K. 2 Hydrogen gas and oxygen gas react according to the following equations: 2H2(g) + O2(g) → 2H2O(g) (ΔH° = -484 kJ mol-1) 2H2(g) + O2(g) → 2H2O(1) (ΔH° = -570 kJ mol-1) (1.) How much more energy is stored in one mole of hydrogen molecules and 0.5 mole of oxygen molecules than is stored in one mole of gaseous water molecules? (2.) What is the enthalpy change for the formation of one mole of liquid water from hydrogen gas and oxygen gas? 3 Suitable experiments have given the following results: Reaction equation 1: H2(g) + ½O2(g) → H2O(l) (ΔH° = -285 kJ mol-1) Reaction equation 2: Pb(s) + ½O2(g) → PbO(s) (ΔH° = -219 kJ mol-1) Reaction equation 3: H2O(l) → H2O(g) (ΔH° = +43 kJ mol-1) Calculate ΔH° for the reaction PbO(s) + H2(g) → Pb(s) + H2O(g) 4 A reaction involved in the production of Fe from iron ore is Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) (ΔH° = -18 kJ mol-1) (1.) What mass of CO must react to release 54 kJ of heat? (2.) What volume of CO at 298 K and 1.00 atmosphere is needed to produce 1.0 kg of Fe? 5 For the photosynthesis of fructose according to the equation 6CO2(g) + 6H2O(l) → C6H12O6(s) + 6O2(g) (ΔH° = 2828 kJ mol-1) (1.) Calculate the standard heat of formation of fructose. (2.) How many kilojoule of heat energy are required by a plant to produce 50.0 g of fructose? 6 Hydrogen gas (2.00 mole) reacts with excess chlorine and the heat liberated is used with 100% efficiency to convert a mixture of excess N2 and excess O2 into N2O. What mass of N2O is formed if all reactants and products are in the gaseous state? 7 At room temperature, nitrogen trichloride, NCl3, is an unstable yellow oil. When exposed to strong light it explodes, producing only N2 and Cl2, and releasing 230 kilojoule of heat per mole of NCl3. (1.) Write an equation for the reaction. (2.) Calculate ΔH° for the reaction as written. (3.) Calculate the change in enthalpy accompanying the complete decomposition of a specimen of NCl3 having a mass of 24.0 g. (4.) If this specimen were completely decomposed, what volume of gas would be produced if measurements were made at 100 ° C and 4.00 atmosphere? 8 Using values given in the table of Average Bond Enthalpies, calculate the heats of hydrogenation of (1.) N2 to the entirely single bonded product hydrazine, N2H4 (g), and (2.) C2H2 to the entirely singly bonded product ethane, C2H6 (g). 9 Normal octane, C8H18, has an unbranched chain of single bonded carbon atoms. The density of octane at 298 K is 0.704 g cm-3 and Hv is 42 kJ mol-1. Calculate from average bond enthalpies the heat evolved in the combustion of 4.00 litre of liquid octane in excess oxygen at 298 K and 1.00 atmosphere pressure. 10 Calculate the bond energy of the oxygen-oxygen bond in the hydrogen peroxide molecule given the following enthalpy changes: H2(g) + O2(g) → HOOH(g) (ΔH° = -136 kJ mol-1) H2(g) + ½O2(g) → H2O(g) (ΔH° = -242 kJ mol-1) H2(g) → 2H(g) (ΔH° = +436 kJ mol-1) O2(g) → 2O(g) (ΔH° = +498 kJ mol-1) THERMODYNAMICS (Advanced answers) 1 (1.) Balanced equation: Ca(s) + 2H2O(l) → Ca(OH)2(s) + H2(g) (2.) Sum of enthalpies of formation of reactants, Σ ΔHf° (reactants) = ΔHf°[Ca(s)] + ΔHf°[H2O(l)] = (0 kJ mol-1) + (2 x -285 kJ mol-1) = -570 kJ mol-1. Sum of enthalpies of formation of products, ΣΔHf° (products) = ΔHf°[Ca(OH)2(s)] + ΔHf°[H2(g)] = (-986 kJ mol-1) + (0 kJ mol-1) = -986 kJ mol-1. ∴ ΔH° = ΣΔHf° (products) - ΣΔHf° (reactants) = (-986 kJ mol-1) - (-570 kJ mol-1) = -416 kJ mol-1. This is for the reaction of 1 mol Ca. The enthalpy change for 0.3 mol Ca = -416 kJ mol-1 x 0.3 mol = -125 kJ. 2 (1.) The balanced reaction for the production of 1 mole of H2O(g) can be obtained by dividing the stoichiometric coefficients and likewise, the enthalpy for the reaction, by 2 : H2(g) + ½O2(g) → H2O(g) (ΔH° = (-484 / 2) kJ mol-1 = -242 kJ mol-1 The negativeΔH° value indicates the reaction is exothermic, and represents the difference in energy contained in the reactant bonds (1 mol H2(g) and 0.5 mol O2(g)), and the energy in the product bonds (1 mol H2O(g)). Therefore there is 242 kJ mol-1 more energy stored in the reacting 1 mol H2, 0.5 mol O2, than in the 1 mol H2O produced. (2.) From the equation representing the formation of H2O(l): 2H2(g) + O2(g) → 2H2O(l) (ΔH° = -570 kJ mol-1) This represents the enthalpy change for the production of 2 mol H2O(l). The production of 1 mol H2O(l) can be found by dividing the equation (and the enthalpy of reaction) by 2. ThereforeΔH° for production of 1 mol H2O(l) = (-570 / 2) kJ mol-1 = -285 kJ mol-1. 3 This question can be approached in two slightly different ways. The first uses Hess' law to combine all the given data into a final equation. The second method approaches the problem in smaller parts. Both answers are valid, but the second takes significantly longer. First method: Combine reaction equation 1 with reaction equation 3 and the reverse of reaction equation2. The overall reaction resuting is the desired equation, thus: H2(g) + ½O2(g) + PbO(s) + H2O(l) → H2O(l) + Pb(s) + ½O2(g) + H2O(g) Cancelling common terms: PbO(s) + H2(g)→Pb(s) + H2O(g) The enthalpy change for the combined reaction equation is obtained from the sum of the individual reaction equation enthalpies, reversing the sign of that for reaction equation 2 because the equation was reversed. ΔH° = (-285) + (+43) + (-(-219)) kJ mol-1 = -23 kJ mol-1 Second method: To calculate ΔH° for the final reaction requires the values of ΔHf°[H2O(g)] and ΔHf°[PbO(s)]. Note that ΔHf° for Pb(s) and H2(g) are both zero as they are elements in their standard states. The ΔHf° value of PbO(s) is given by the second equation: ΔHf°[PbO(s)] = -219 kJ mol-1 The value of ΔHf°[H2O(g)] can be found by combining the data given for the first and the third reaction equations: From the first equation,ΔHf°[H2O(l)] = -285 kJ mol-1 From the third equation, H2O(l) → H2O(g) (ΔH° = +43 kJ mol-1) ΔH° = ΣΔHf° (products) - ΣΔHf° (reactants) +43 kJ mol-1 = ΔHf°[H2O(g)] - (-285 kJ mol-1) ∴ ΔHf°[H2O(g)] = +43 kJ mol-1 - 285 kJ mol-1 = -242 kJ mol-1 Now the final equation can be considered using the ΔHf° values obtained: PbO(s) + H2(g) → Pb(s) + H2O(g) Sum of enthalpies of formation of reactants, ΣΔHf° (reactants) = ΔHf°[PbO(s)] + ΔHf°[H2(g)] = (-219 kJ mol-1) + (0 kJ mol-1) = -219 kJ mol-1. Sum of enthalpies of formation of products, ΣΔHf° (products) = ΔHf°[Pb(s)] + ΔHf°[H2O(g)] = (0 kJ mol-1) + (-242 kJ mol-1) = -242 kJ mol-1. ∴ ΔH° = ΣΔHf° (products) - ΣΔHf° (reactants) = (-242 kJ mol-1) - (-219 kJ mol-1) = -23 kJ mol-1. 4 (1.) Balanced equation: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) (ΔH = -18 kJ mol-1) From the equation, 3.0 mol CO reacts to produce 18 kJ heat. Therefore moles of CO required to produce 54 kJ heat = 3.0 mol x (54 kJ / 18 kJ) = 3.0 mol x 3.0 = 9.0 mol. Mass CO = moles of CO x molar mass CO = 9.0 x 28.01 = 2.5 x 102 g. (2.) Moles of Fe = mass Fe / molar mass Fe = 1.0 kg / 55.85 = 18 mol. From the equation, each mol Fe produced requires 1.5 mol CO, therefore moles of CO = 1.5 x moles of Fe = 1.5 x 18 = 27 mol. Volume CO (298 K, 1 atm) = no. mol CO x molar volume = 27 mol x 24.5 L mol-1 = 6.6 x 102 L. 5 (1.) 6CO2(g) + 6H2O(l) → C6H12O6(s) + 6O2(g) (ΔH° = 2828 kJ mol-1). From the balanced equation and given ΔH° value we can solve for ΔHf°[C6H12O6(s)]. Sum of enthalpies of formation of reactants, ΣΔHf° (reactants) = 6 x ΔHf°[CO2(g)] + 6 x ΔHf°[H2O(l)] = (6 x -393 kJ mol-1) + (6 x -285 kJ mol-1) = -4068 kJ mol-1. Sum of enthalpies of formation of products, ΣΔHf° (products) = ΔHf°[C6H12O6(s)] + 6 x ΔHf°[O2(g)] = ΔHf°[C6H12O6(s)] + (0 kJ mol-1) = ΔHf°[C6H12O6(s)]. ΔH° = ΣΔHf° (products) - ΣΔHf° (reactants) ΣΔHf° (products) = ΔH° + ΣΔHf° (reactants) ∴ ΔHf°[C6H12O6(s)] = 2828 kJ mol-1 + (-4068 kJ mol-1) = -1240 kJ mol-1. (2.) Moles of C6H12O6 = mass C6H12O6 / molar mass C6H12O6 = 50.0 g / 180.16 g mol-1 = 0.278 mol. The production of 1 mol C6H12O6 requires 2828 kJ energy (given), therefore the energy required for the production of 0.278 mol C6H12O6 = 0.278 mol x 2828 kJ mol-1 = 786 kJ. 6 Balanced equation: H2(g) + Cl2(g) → 2HCl(g) Sum of enthalpies of formation of reactants, ΣΔHf° (reactants) = ΔHf°[H2(g)] + ΔHf°[Cl2(g)] = (0 kJ mol-1) + (0 kJ mol-1) = 0 kJ mol-1. Sum of enthalpies of formation of products, ΣΔHf° (products) = 2.00 x ΔHf°[HCl(g)] = (2.00 x -92 kJ mol-1) = -184 kJ mol-1. ∴ ΔH° = ΣΔHf° (products) - ΣΔHf° (reactants) = (-184 kJ mol-1) - (0 kJ mol-1) = -184 kJ mol-1. Since chlorine is in excess in the first reaction, hydrogen is the limiting reactant and will determine the amount of heat produced: Heat produced = no. mol H2 x ΔH° = 2.00 mol x -184 kJ mol-1 = 368 kJ. This energy will be the limiting factor in determining how much N2 and O2 will react. ΔH° for the reaction N2(g) + ½O2(g) → N2O(g) needs to be found: Sum of enthalpies of formation of reactants, ΣΔHf° (reactants) = ΔHf°[N2(g)] + 0.5 x ΔHf°[O2(g)] = (0 kJ mol-1) + (0 kJ mol-1) = 0 kJ mol-1. Sum of enthalpies of formation of products, ΣΔHf° (products) = ΔHf°[N2O(g)] = +82 kJ mol-1. ∴ ΔH° = ΣΔHf° (products) - ΣΔHf° (reactants) = (+82 kJ mol-1) - (0 kJ mol-1) = +82 kJ mol-1. Therefore each mole of N2O requries 82 kJ energy (assuming 100% efficiency). Moles of N2O produced by 368 kJ = 368 kJ / 82 kJ mol-1 = 4.49 mol. Mass N2O = no. mol N2O x molar mass N2O = 4.49 mol x 44.02 g mol-1 = 198 g. 7 (1.) NCl3(l) → 0.5N2(g) + 1.5Cl2(g) (2.) 230 kJ heat is released per mole NCl3 (given), so ΔH° for reaction as written in part (1.) will be -230 kJ mol-1 (note energy release implies the reaction is exothermic, hence the negative ΔH° value) The equation in (1.) can be written a number of different ways, and the ΔH° value assigned to the reaction will also vary (for instance, another valid equation would be 2NCl3(l) → N2(g) + 3Cl3(g), in which case the ΔH° value becomes 2 x -230 kJ mol-1 = -460 kJ mol-1, for the reaction as written.) (3.) Moles of NCl3 = mass NCl3 / molar mass NCl3 = 24.0 g / 120.36 g mol-1 = 0.200 mol. ΔH° = moles of NCl3 x ΔH° per mol NCl3 = 0.200 mol x -230 kJ mol-1 = -46.0 kJ. (4.) The volume of the gases produced is found using the ideal gas equation PV = nRT. In this case, where pressure is specified in atmospheres, R = 0.0821 litre atmosphere kelvin-1 mole-1 is the more convenient form to use. This gives V (litres) = nRT / P, where n = moles of gas produced, R = 0.0821 L atm K-1 mol-1, T = temperature in Kelvin, P = pressure in atmospheres. Note that the question refers to the volume of the gases produced, and using the ideal gas equations the fact that the gas is a mixture of two different gases is not important - the only important factor is the number of moles present. From the balanced equation, 1 mol NCl3(l) gives 0.5 mol N2(g) and 1.5 mol Cl2(g), ie 2 moles of gaseous products formed per mol NCl3. ∴ 0.200 mol NCl3(l) produces 0.400 mol gas. ∴ V = (0.400 mol x 0.0821 L atm K-1 mol-1 x 373 K) / 4.00 atm = 3.06 L. 8 The heat of reaction is found from the relationship ΔHf° = ΔH°reactant bonds broken + ΔH°product bonds formed. All reactants and products are in the gaseous phase, so bond enthalpies can be used without having to allow for any phase changes. (1.) For the reaction N2(g) + 2H2(g) → N2H4(g): Bonds broken (ΔH positive) ΔH°reactant bonds broken = ΔH°[N≡ N] + 2 x ΔH°[H-H] = 945 kJ mol-1 + (2 x 436 kJ mol-1) = +1817 kJ mol-1. Bonds formed (ΔH negative) ΔH°product bonds formed = ΔH°[N-N] + 4 x ΔH°[N-H] = -163 kJ mol-1 + (4 x -391 kJ mol-1) = -1727 kJ mol-1. Overall reaction ΔH° = ΔH°reactant bonds broken + ΔH°product bonds formed = +1817 kJ mol-1 + (-1727 kJ mol-1) = +90 kJ mol-1.   (2.) For the reaction C2H4(g) + H2(g) → C2H6(g): Bonds broken (ΔH positive) ΔH°reactant bonds broken = ΔH°[C=C] + 4 x ΔH°[C-H] + ΔH°[H-H] = 614 kJ mol-1 + (4 x 413 kJ mol-1) + 436 kJ mol-1 = 2702 kJ mol-1.   Bonds formed (ΔH negative) ΔH°product bonds formed = ΔH°[C-C] + 6 x ΔH°[C-H] = -348 kJ mol-1 + (6 x -413 kJ mol-1) = -2826 kJ mol-1. Overall reaction ΔH° = ΔH°reactant bonds broken + ΔH°product bonds formed = 2702 kJ mol-1 + (-2826 kJ mol-1) = -124 kJ mol-1. 9 For the reaction C8H18(g) + 12½O2(g) → 8CO2(g) + 9H2O(g) ΔH°reactant bonds broken = 7 x ΔH° [C-C] + 18 x ΔH° [C-H] + 12.5 x ΔH° [O=O] = (7 x 348 kJ mol-1) + (18 x 413 kJ mol-1) + (12.5 x 498 kJ mol-1) = 16095 kJ mol-1. ΔH°product bonds formed = 16 x ΔH° [C=O] + 18 x ΔH0[O-H] = (16 x -745 kJ mol-1) + (18 x -463 kJ mol-1) = -20254 kJ mol-1. ∴ ΔH° = ΔH°reactant bonds broken + ΔH°product bonds formed = 16095 kJ mol-1 + (-20254 kJ mol-1) = -4159 kJ mol-1. This ΔH° value needs to be corrected by the ΔHv° value provided for octane, which is added to allow for the energy absorbed by the liquid octane to allow it to vaporise and burn, hence reducing the amount of heat released by the reaction: Corrected ΔH° = -4159 kJ mol-1 + 42 kJ mol-1 = -4117 kJ mol-1. Mass C8H18 = Volume C8H18 x density C8H18 = 4.00 x 103 cm3 x 0.704 g L-1 = 2816 g. No. mol C8H18 = mass C8H18 / molar mass C8H18 = 2816 g / 114.22 g mol-1 = 24.7 mol. The heat released by the reaction was found to be -4117 kJ mol-1, so the heat released by 4.00 L C8H18 = no. mol C8H18 x heat released per mole = 24.7 mol x -4117 kJ mol-1 = 1.02 x 105 kJ mol-1. 10 For the reaction H2(g) + O2(g) → H2O2(g) (ΔH° = -136 kJ mol-1) ΔH°reactant bonds broken = ΔH°[H2(g) → 2H(g)] + ΔH°[O2(g) → 2O(g)] = +436 kJ mol-1 + 498 kJ mol-1) = +934 kJ mol-1. ΔH°product bonds formed = ΔH°[O-O] + 2 x ΔH°[O-H] - Equation A Note that both ΔH°[O-O] and ΔH°[O-H] are unknown. In order to solve for ΔH°[O-O], we need to obtain ΔH°[O-H] from equation 2 in the question: For the reaction H2(g) + ½O2(g) → H2O(g) (ΔH = -242 kJ mol-1) ΔH°reactant bonds broken = ΔH°[H2(g) → 2H(g)] + 0.5 x ΔH°[O2(g) → 2O(g)] = +436 kJ mol-1 + (0.5 x 498 kJ mol-1) = +685 kJ mol-1. ΔH°product bonds formed = 2 x ΔH°[O-H] ΔH° = -242 kJ mol-1 Rearranging ΔH° = ΔH°reactant bonds broken + ΔH°product bonds formed: ΔH°product bonds formed = ΔH° - ΔH°reactant bonds broken ∴ 2 x ΔH°[O-H] = -242 kJ mol-1 - (+685 kJ mol-1) = 927 kJ mol-1. ∴ ΔH°[O-H] = 927 kJ mol-1 / 2 = 464 kJ mol-1. Now that we have a value for ΔH°[O-H], we can return to equation A and solve for ΔH°[O-O]: ΔH°product bonds formed = ΔH°[O-O] + 2 x ΔH°[O-H] - Equation A ΔH° = -136 kJ mol-1 Rearranging ΔH° = ΔH°reactant bonds broken + ΔH°product bonds formed: ΔH°product bonds formed = ΔH° - ΔH°reactant bonds broken ∴ ΔH°[O-O] + (2 x 464 kJ mol-1) = -136 kJ mol-1 + 934 kJ mol-1 ΔH°[O-O] + 928 kJ mol-1 = 798 kJ mol-1 ΔH°[O-O] = -130 kJ mol-1    