A) 8.32 kJ
B) 12.34 kJ
C) 16.25 kJ
D) 24.74 kJ
Correct Answer: C
Solution :
Combustion of benzene can be represented as follows: \[\underset{78}{\mathop{{{C}_{6}}{{H}_{6}}}}\,+{{O}_{2}}\to 6C{{O}_{2}}+{{H}_{2}}O+3250\,kJ\] 70 g of\[{{C}_{6}}{{H}_{5}}\]produce 3250 kJ heat. 1 g of will produce \[\frac{3250}{78}\,kJ\]heat. 39 g will produce \[=\frac{3250}{78}\times 0.39\,kJ\] = 16.256 kJ heat is liberated. Note: Enthalpy of combustion is the amount of heat liberated when one mole of a substance is burnt in excess of air. Hence, the correct option is [c].
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