The combustion of benzene $(l)$ gives $C{O_2}\left( g \right)$ and ${H_2}O\left( l \right)$. Given that heat of combustion of benzene at constant volume is $-3263.9\,kJ\,mo{l^{ - 1}}$ at $ 25^\circ C$, heat of combustion (in $kJ\,\,mo{l^{ - 1}}{\rm{)}}$ of benzene at constant pressure will be $\left( {R = 8.314\,J{K^{ - 1}}\,mo{l^{ - 1}}} \right)$

JEE CHEMISTRY MCQ: THERMODYNAMICS

1 Hard

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Application +4 | -1

The combustion of benzene

(l)

gives

C{O_2}\left( g \right)

and

{H_2}O\left( l \right)

. Given that heat of combustion of benzene at constant volume is

-3263.9\,kJ\,mo{l^{ - 1}}

25^\circ C

, heat of combustion (in

kJ\,\,mo{l^{ - 1}}\mathrm{)}

of benzene at constant pressure will be

\left( {R = 8.314\,J{K^{ - 1}}\,mo{l^{ - 1}}} \right)

View answer and explanation

Correct option: D

−3267 .6

Option D (−3267 .6) is correct because the heat of combustion at constant pressure is calculated using the formula

\Delta {H_p} = \Delta U + \Delta {n_g}RT

Practice All Chapter MCQs →

\ \mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}(\mathrm{g}), \Delta \mathrm{H}_1=\mathrm{a}

\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}_2(\mathrm{~g}), \Delta \mathrm{H}_2=\mathrm{b}

\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})-2 \mathrm{CO}(\mathrm{g}) ; \Delta \mathrm{H}_3=\mathrm{c}

0.18 \mathrm{~g}

4 \mathrm{~K}.

\left(\Delta_{\mathrm{C}} \text{H}^{\circ}=-2.8\times 10^6 \mathrm{~J} / \mathrm{mol} \text{ glucose}\right)

{CsI}

{CsCl}

{CsCl}

{Cs} I

\begin{array}{cccccccccccccccccccc} {}&{\Delta H(KJ/mol)} \\ {1/2A \to B}&{ + 150} \\ {3B \to 2C + D}&{ - 125} \\ {E + A \to 2D}&{ + 350} \end{array}

{B}+{D} \rightarrow {E}+2 {C}, \Delta {H}

\Delta H

\Delta u

\mathrm{C}{\mathrm{H}_4}\left( \mathrm{g} \right) + 2{\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right) + {\mathrm{H}_2}\mathrm{O}\left( l \right);\Delta H = - 890.3\mathrm{kJ}