The difference between the heats of reaction at constant pressure and constant volume for the reaction, at a constant temperature $\mathrm{T}$ is $2 \mathrm{C}_6 \mathrm{H}_6(\mathrm{l})+15 \mathrm{O}_2(\mathrm{~g}) \rightarrow 12 \mathrm{CO}_2(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$

JEE CHEMISTRY MCQ: THERMODYNAMICS

1 Medium

1 of 1

Application +4 | -1

The difference between the heats of reaction at constant pressure and constant volume for the reaction, at a constant temperature

\mathrm{T}

2 \mathrm{C}_6 \mathrm{H}_6(\mathrm{l})+15 \mathrm{O}_2(\mathrm{~g}) \rightarrow 12 \mathrm{CO}_2(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l})

View answer and explanation

Correct option: A

-3 R T

Option A (-3 RT) is correct because the difference between heats of reaction at constant pressure and volume is given by

\Delta H - \Delta U = \Delta n_g RT

, where

\Delta n_g

is the change in the number of gaseous moles.

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}