Question

The combustion of benzene$\left(\mathrm{l}\right)$ gives ${\mathrm{CO}}_2\left(\mathrm{g}\right)$ and${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$. Given that heat of combustion of benzene at constant volume is$-3263.9\mathrm{kJ}{\mathrm{mol}}^{-1}$ at ${25}^{°}\mathrm{C}$, the heat of combustion (in $\mathrm{kJ}{\mathrm{mol}}^{-1}$) of benzene at constant pressure will be
($\mathrm{R}=8.314\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$)

• A. $-452.46$
• B. $3260$
• C. $-3267.6$
• D. $4152.6$

Answer 1

The correct option is C

$-3267.6$

Step 1: Given data

The heat of combustion of benzene at constant volume and at ${25}^{°}\mathrm{C}$$=$$-3263.9\mathrm{kJ}{\mathrm{mol}}^{-1}$

Value of $\mathrm{R}=8.314\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$

Step 2: Applying the formula

Writing a balanced equation for the combustion of benzene

$\underset{\mathrm{Benzene}}{{\mathrm{C}}_6{\mathrm{H}}_6\left(\mathrm{l}\right)}+\underset{\mathrm{Oxygen}}{\frac{150}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Carbon}\mathrm{dioxide}}{6{\mathrm{CO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Water}}{3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

Calculating the difference between moles of gaseous product

$∆\mathrm{ng}=6-\frac{15}{2}\Rightarrow -\frac{3}{2}$

Applying the formula, $∆\mathrm{H}=∆\mathrm{U}+∆\mathrm{ngRT}$ where $∆\mathrm{U}$ is heat at constant volume, R is constant, T is temperature, ng is the difference of moles

$∆\mathrm{H}=-3263.9+(-\frac{3}{2})\times 8.314\times 298\times {10}^{-3}$

$∆\mathrm{H}=-3263.9+(-3.71)$

$∆\mathrm{H}=-3267.6\mathrm{kJ}{\mathrm{mol}}^{-1}$

Therefore, the heat of combustion of benzene at constant pressure will be $∆\mathrm{H}=-3267.6\mathrm{kJ}{\mathrm{mol}}^{-1}$