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

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None of these

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(D) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by: $\Delta H = \Delta E + \Delta n_g RT$.Here,$\Delta n_g$ is the change in the number of moles of gaseous species.For the reaction $A_2B_{(s)} 	o 2A_{(s)} + 1/2 \ B_{2(g)}$,$\Delta n_g = (1/2) - 0 = 0.5 \ mol$.Given $\Delta H = + 7.3 \ kcal$,$R = 0.001987 \ kcal \ K^{-1} \ mol^{-1} \approx 0.002 \ kcal \ K^{-1} \ mol^{-1}$,and $T = 298 \ K$.$\Delta E = \Delta H - \Delta n_g RT = 7.3 - (0.5 	imes 0.002 	imes 298) = 7.3 - 0.298 = 7.002 \ kcal$.Since $7.002 \ kcal$ is not listed in the options,the correct choice is $(D)$.

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

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