The heat change for the following reaction at | vedclass

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(C) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by: $\Delta H = \Delta E + \Delta n_g RT$,w

Vedclass · Accepted Answer

Less than $7.3 \ kcal$

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(C) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by: $\Delta H = \Delta E + \Delta n_g RT$,where $\Delta n_g$ is the change in the number of moles of gaseous species.For the reaction: $A_2B_{(s)} 	o 2A_{(s)} + 1/2B_{2(g)}$,the value of $\Delta n_g = (1/2) - 0 = 0.5 \ mol$.Given: $\Delta H = +7.3 \ kcal$,$T = 298 \ K$,and $R \approx 0.002 \ kcal \ K^{-1} \ mol^{-1}$.Substituting the values: $\Delta E = \Delta H - \Delta n_g RT = 7.3 - (0.5 	imes 0.002 	imes 298) = 7.3 - 0.298 = 7.002 \ kcal$.Since $7.002 \ kcal < 7.3 \ kcal$,the heat change at constant volume $(\Delta E)$ is less than $7.3 \ kcal$.

The heat change for the following reaction at $298 \ K$ and at constant pressure is $+7.3 \ kcal$: $A_2B_{(s)} \to 2A_{(s)} + 1/2B_{2(g)}$,$\Delta H = +7.3 \ kcal$. The heat change at constant volume would be:

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