At 300 K,the enthalpy change for the reactio | vedclass

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Question

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by: $\Delta H = \Delta U + \Delta n_g RT$.

Vedclass · Accepted Answer

$-2800.00 \ kJ \ mol^{-1}$

Vedclass · Answer

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by: $\Delta H = \Delta U + \Delta n_g RT$. Here,$\Delta n_g$ is the change in the number of moles of gaseous species: $\Delta n_g = (n_{products, g}) - (n_{reactants, g})$. For the reaction $C_3H_8(g) + 5O_2(g) ightarrow 3CO_2(g) + 4H_2O(l)$,the gaseous moles are: $\Delta n_g = 3 - (1 + 5) = 3 - 6 = -3$. Given: $\Delta H = -2800 \ kJ \ mol^{-1}$,$T = 300 \ K$,$R = 8.314 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$. Substituting the values: $\Delta U = \Delta H - \Delta n_g RT = -2800 - (-3 	imes 8.314 	imes 10^{-3} 	imes 300)$. $\Delta U = -2800 + 7.48 = -2792.52 \ kJ \ mol^{-1}$. Note: If the reaction was assumed to have $\Delta n_g = 0$,the answer would be $-2800 \ kJ \ mol^{-1}$ (Option $B$).

At $300 \ K$,the enthalpy change for the reaction $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)$ is $-2800 \ kJ \ mol^{-1}$. Calculate the internal energy change $(\Delta U)$ for the reaction at the same temperature.

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