Q12 / 150 · MHT CET 2008 · Chemistry

If $\Delta E$ is the heat of reaction for $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$ at constant volume, the $\Delta H$ (heat of reaction at constant pressure), at constant temperature is

  • A
    $\Delta H=\Delta E+R T$
  • Correct answer
    B
    $\Delta H=\Delta E-R T$
  • C
    $\Delta H=\Delta E-2 R T$
  • D
    $\Delta H=\Delta E+2 R T$

Solution

We know that, $\Delta H=\Delta E+\Delta n R T$ where, $\Delta n=$ number of moles of gaseous products - number of moles of gaseous reactants $\begin{array}{lrl} & =2-3= & -1 \\ \text { So, } & & \Delta H=\Delta E-R T\end{array}$

Hence, the correct answer is option: 'B'.

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