For the reaction,A_{(s)} + 3B_{(g)} to 4C_{(s | vedclass

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Question

(D) 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

$\Delta H = \Delta U - 3RT$

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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 reaction $A_{(s)} + 3B_{(g)} 	o 4C_{(s)} + D_{(l)}$,we calculate the change in the number of moles of gaseous species $(\Delta n_g)$: $\Delta n_g = (n_g)_{	ext{products}} - (n_g)_{	ext{reactants}}$. Since $C$ is solid and $D$ is liquid,they are not included in $\Delta n_g$. $\Delta n_g = 0 - 3 = -3$. Substituting this into the formula: $\Delta H = \Delta U + (-3)RT = \Delta U - 3RT$.

For the reaction,$A_{(s)} + 3B_{(g)} \to 4C_{(s)} + D_{(l)}$,$\Delta H$ and $\Delta U$ are related as $\dots$

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