For the reaction Ag_2O_{(s)} rightarrow 2Ag_{ | vedclass

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(D) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \De

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$\Delta H > \Delta U$

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(D) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$. For the given reaction: $Ag_2O_{(s)} \rightarrow 2Ag_{(s)} + \frac{1}{2}O_{2(g)}$. The number of moles of gaseous products is $n_p = \frac{1}{2}$ and the number of moles of gaseous reactants is $n_r = 0$. Therefore,$\Delta n_g = n_p - n_r = \frac{1}{2} - 0 = 0.5$. Since $\Delta n_g > 0$,the term $\Delta n_g RT$ is positive. Thus,$\Delta H = \Delta U + (0.5)RT$,which implies $\Delta H > \Delta U$.

For the reaction $Ag_2O_{(s)} \rightarrow 2Ag_{(s)} + \frac{1}{2}O_{2(g)}$,which of the following is true?

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