Using information from the text Appendix, calculate ?H for the reaction. 2C2H6(g

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The following equation shows the combustion of ethane. 2C2H6 + 7O2 ---> 4CO2 +6H2O Use Hess's law to calculate the enthalpy of combustion for ethane, C2h6, from these data. C2H4(g) +3O2(g) --> 2CO2(g) + 2H2O(g) / Hc= -1323 kJ/mol C2H4(g) + H2(g) --> C2H6(g) / Hr = -137 kJ/mol H2(g) + 1/2 O2(g) --> H2O(g) / Hf degrees = -242 kJ/mol THESE ABOVE VALUES (or somewhat close to)WOULD BE IN APPENDIX SOLUTION The idea is to rearrange the three equations (using each equation only once) until we get the desired equation. First off, notice that the desired equation has 2C2H6 on the left. Equation (2) has C2H6 on the right, so let's reverse it AND multiply by 2 to get 2C2H6. (2) reversed x 2 : 2C2H6(g) => 2C2H4(g) + 2H2(g) delta H = +274 kJ (let's call this eqn. A) In doing the above we created 2C2H4 on the right that do not appear in the desired equation. So we need to get 2C2H4 on the left to cancel out the 2C2H4 on the right. Equation (1) has a C2H4 on the left, so let's double it: (1) x 2: 2C2H4(g) + 6O2(g) => 4CO2(g) + 4H2O(g) delta H = -2646 kJ (Let's cal this eqn B) If we add together equations A + B, we get 2C2H6(g) + 6O2(g) => 4CO2(g) + 4H2O(g) + 2H2(g) delta H = 274 - 2646 = -2372 kJ We still need another O2 on the left side (we have 6O2 and need 7). Take equation 3 (the one we haven't used yet) and double it: (3) x 2: 2H2(g) + O2(g) => 2H2O(g) ............delta H = -484 kJ (call this eqn. C) If we add C to A+B, we get 2C2H6(g) + 7O2(g) + 2H2(g) => 4CO2(g) + 6H2O(g) + 2H2(g) delta H = -2372 - 484 = -2856 kJ Note how the 2H2 cancels from both sides and we get what we wanted!

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