Butane cap c sub 4 cap h sub 10 g , ΔH_f=−125.7 , combusts in the presence of oxygen to form CO sub 2 times g , ΔH_f=−393.5kJ/mol , and cap h sub 2 cap o g , ΔH_f=−241.82 in the reaction: 2 cap c sub 4 cap h sub 10 g plus 13 cap o sub 2 g rightarrow 8 CO sub 2 g plus 10 cap h sub 2 cap o g What is the enthalpy of combustion, per mole, of butane? Use ΔH_rxn=∑(ΔH_f_1products)−∑(ΔH_f_1reactants)
Answer
A
2 comma 657 point 4 kJ over mol
B
5314 point 8 kJ over mol
C
negative 2 comma 657 point 4 kJ over mol
D
negative 5314 point 8 kJ over mol
Explanation
Use the formation-enthalpy equation with the coefficients in the balanced reaction. The products total 8(−393.5) + 10(−241.82) = −5566.2 kJ, and the reactants total 2(−125.7) + 13(0) = −251.4 kJ. Thus, ΔH for the reaction as written is −5566.2 − (−251.4) = −5314.8 kJ for 2 mol of butane. Per mole, this is −5314.8 ÷ 2 = −2657.4 kJ/mol, choice C.
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