For the reaction C_{(s)} + frac{1}{2}O_{2(g)} | vedclass

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

Vedclass · Accepted Answer

$\Delta H > \Delta U$

Vedclass · Answer

(C) 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 reaction $C_{(s)} + \frac{1}{2}O_{2(g)} 	o CO_{(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = n_{g(products)} - n_{g(reactants)}$. Here,$n_{g(products)} = 1$ (for $CO_{(g)}$) and $n_{g(reactants)} = \frac{1}{2}$ (for $O_{2(g)}$). Therefore,$\Delta n_g = 1 - 0.5 = 0.5$. Since $\Delta n_g > 0$,the term $\Delta n_g RT$ is positive. Thus,$\Delta H = \Delta U + 0.5RT$,which implies $\Delta H > \Delta U$.

For the reaction $C_{(s)} + \frac{1}{2}O_{2(g)} \to CO_{(g)}$ at constant $T$ and $P$,which of the following statements is correct?

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