A mixture of 2 mol of carbon monoxide and 1 | vedclass

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(B) The chemical equation for the reaction is: $2CO(g) + O_2(g) \rightarrow 2CO_2(g)$.Calculate the change in the number of moles of gaseous species,$

Vedclass · Accepted Answer

$\Delta H < \Delta E$

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(B) The chemical equation for the reaction is: $2CO(g) + O_2(g) \rightarrow 2CO_2(g)$.Calculate the change in the number of moles of gaseous species,$\Delta n_g = n_p - n_r = 2 - (2 + 1) = 2 - 3 = -1$.The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by: $\Delta H = \Delta E + \Delta n_g RT$.Substituting $\Delta n_g = -1$,we get: $\Delta H = \Delta E - RT$.Since $R$ and $T$ are positive values,$\Delta H = \Delta E - RT$ implies that $\Delta H < \Delta E$.

$A$ mixture of $2 \ mol$ of carbon monoxide and $1 \ mol$ of oxygen in a closed vessel is ignited to convert the carbon monoxide to carbon dioxide. If $\Delta H$ is the enthalpy change and $\Delta E$ is the change in internal energy,then:

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Identify the reaction for which $\Delta H \neq \Delta E$ :

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