JEE Main On 08 April 2018 Question 26
Question: The combustion of benzene (l) gives $ C{O_2}(g) $ and $ {H_2}\text{O(l)} $ . Given that heat of combustion of benzene at constant volume is $ \text{-3263}\text{.9 kJ mo}{l^{\text{-1}}} $ at $ \text{25 }{}^\circ\text{ C} $ ; heat of combustion (in $ \text{kJ mo}{l^{\text{-1}}} $ ) of benzene at constant pressure will be: $ \text{(R=8}\text{.314 J}{K^{\text{-1}}}\text{ mo}{l^{\text{-1}}}\text{)} $ [JEE Main Online 08-04-2018]
Options:
A) 3260
B) -3267.6
C) 4152.6
D) -452.46
Show Answer
Answer:
Correct Answer: B
Solution:
$ C _6H _6(\ell )+\frac{15}{2}O _2(g)\xrightarrow{{}}6CO _2(g)+3H _2O(\ell ) $
$ \Delta n _{g}=6-7.5=-1.5 $ $ \Delta H=\Delta U+\Delta n _{g}RT $
$ \Delta H=-3263.9-\frac{1.5\times 8.314\times 298}{1000}=-3267.6kJmo{l^{-1}} $
