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

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Question

(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

$-3RT$

Vedclass · Answer

(C) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$.Rearranging this equation,we get: $\Delta H - \Delta U = \Delta n_g RT$.For the given reaction: $C_3H_{8(g)} + 5O_{2(g)} \longrightarrow 3CO_{2(g)} + 4H_2O_{(l)}$.$\Delta n_g$ is the difference between the number of moles of gaseous products and gaseous reactants.$\Delta n_g = (n_{products, g}) - (n_{reactants, g}) = 3 - (1 + 5) = 3 - 6 = -3$.Substituting the value of $\Delta n_g$ into the equation: $\Delta H - \Delta U = -3RT$.

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

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