For the reaction,C_3H_{8(g)} + 5O_{2(g)} to 3 | vedclass

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Question

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

Vedclass · Accepted Answer

$-3RT$

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(C) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$,which implies $\Delta H - \Delta E = \Delta n_g RT$. For the reaction $C_3H_{8(g)} + 5O_{2(g)} 	o 3CO_{2(g)} + 4H_2O_{(l)}$,the change in the number of gaseous moles $(\Delta n_g)$ is calculated as: $\Delta n_g = (	ext{moles of gaseous products}) - (	ext{moles of gaseous reactants})$. $\Delta n_g = 3 - (1 + 5) = 3 - 6 = -3$. Substituting this value into the equation: $\Delta H - \Delta E = -3RT$.

For the reaction,$C_3H_{8(g)} + 5O_{2(g)} \to 3CO_{2(g)} + 4H_2O_{(l)}$ at constant temperature,$\Delta H - \Delta E$ is

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