A) ethane
B) acetylene
C) both
D) none of these
Correct Answer: B
Solution :
Key Idea: Better fuel has higher calorific value. So, calculate calorific values first by dividing given heats of reaction by molecular masses of fuel and then compare. Given, Heat of combustion of ethane \[=-\text{ }341\text{ }kcal\] Calorific value of ethane\[({{C}_{2}}{{H}_{6}})\] \[=\frac{\Delta H\,of\text{ }combustion\text{ }of\text{ }ethane}{molecular\text{ }mass\text{ }of\text{ }ethane}\] \[=\frac{-341}{12\times 2+1\times 6}=\frac{-341}{24+6}\] \[=\frac{-340}{30}=-11.37\,kcal/g\] Heat of combustion of ethane\[=-\text{ }310.0\text{ }kcal/g\] Calorific value of acetylene\[({{C}_{2}}{{H}_{2}})\] \[=\frac{\Delta H\text{ }of\text{ }combustion\text{ }of\text{ }acetylene}{molecular\text{ }mass\text{ }of\text{ }acetylene}\] \[=\frac{-310}{12\times 2+1\times 2}\] \[=\frac{-310}{26}=-11.92\,kcal/g\] \[\therefore \]heat of combustion of acetylene is \[=-11.92\text{ }kcal/g\] \[\therefore \] acetylene is better fuel than ethane.
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