C_2H_5OH(l) + 3O_2(g) rightarrow 2CO_2(g) + 3 | vedclass

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

Vedclass · Accepted Answer

$-1364.0 	ext{ kJ mol}^{-1}$

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 products and reactants.For the reaction: $C_2H_5OH(l) + 3O_2(g) ightarrow 2CO_2(g) + 3H_2O(l)$.$\Delta n_g = (n_{g, 	ext{products}}) - (n_{g, 	ext{reactants}}) = 2 - 3 = -1$.Given: $\Delta H = -1366.5 	ext{ kJ mol}^{-1}$, $T = 27 + 273 = 300 	ext{ K}$, $R = 8.314 	imes 10^{-3} 	ext{ kJ K}^{-1} 	ext{ mol}^{-1}$.Substituting the values: $-1366.5 = \Delta U + (-1) 	imes (8.314 	imes 10^{-3}) 	imes 300$.$-1366.5 = \Delta U - 2.4942$.$\Delta U = -1366.5 + 2.4942 = -1364.0058 	ext{ kJ mol}^{-1} \approx -1364.0 	ext{ kJ mol}^{-1}$.

$C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$. The enthalpy change $(\Delta H)$ for the above reaction at $27^{\circ}C$ is $-1366.5 \text{ kJ mol}^{-1}$. Calculate the internal energy change $(\Delta U)$ for the same reaction at this temperature.

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