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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
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at constant volume, the $\Delta H$ (heat of reaction at constant pressure), at constant temperature is
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Verified Answer
The correct answer is:
$\Delta H=\Delta E-R T$
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
$\begin{array}{lrl} & =2-3= & -1 \\ \text { So, } & & \Delta H=\Delta E-R T\end{array}$
where, $\Delta n=$ number of moles of gaseous products - number of moles of gaseous reactants
$\begin{array}{lrl} & =2-3= & -1 \\ \text { So, } & & \Delta H=\Delta E-R T\end{array}$
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