If enthalpies of formation of c2h4 ,co2, h2o be 52,-394 and -286 kj/mol ,the enthalpy of combustion of c2h4 will be
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Hey mate here's your answer
This question differs from previous ones by giving the change in enthalpy of several reactions, and asking to solve for the molar enthalpy of formation of one of the reactants.
Note also, that reference to data book values for ΔHf° of CO2(g) and H2O(l) is unnecessary, as the first two equations describe the formation of CO2(g) and H2O(l) from their respective elements. Therefore:
ΔHf°[CO2(g)] = -393 kJ mol-1
ΔHf°[H2O(l)] = -285 kJ mol-1
Since O2(g) is an element under standard conditions, its ΔHf° will be 0 kJ mol-1. We now know the ΔHf° values for all reactants and products (except methane), and ΔH°rxn, given as -890 kJ mol-1 for the reaction:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) (ΔH° = -890 kJ mol-1)
Sum of enthalpies of formation of reactants,
SDHf° (reactants) = ΔHf°[CH4(g)] + 2 x ΔHf°[O2(g)]
= ΔHf°[CH4(g)] + 2 x 0 kJ mol-1 = ΔHf°[CH4(g)]
Sum of enthalpies of formation of products,
ΣΔHf° (products) = ΔHf°[CO2(g)] + 2 x ΔHf°[H2O(l)]
= (-393 kJ mol-1) + (2 x -285 kJ mol-1)= -963 kJ mol-1.
∴ ΔH° = ΣΔHf° (products) - ΣΔHf° (reactants)
∴ -890 kJ mol-1 = -963 kJ mol-1 - ΔHf°[CH4(g)]
Solving for ΔHf°[CH4(g)]:
ΔHf°[CH4(g)] = -73 kJ mol-1
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