For the reaction C_2H_5OH_{(l)} + 3O_{2(g)} r | vedclass

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(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

$-1364.0$

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(C) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$.Here,$\Delta n_g$ is the change in the number of moles of gaseous species: $\Delta n_g = (n_{g, products}) - (n_{g, reactants}) = 2 - 3 = -1$.The temperature $T = 27\,^oC = 300 \,K$.The gas constant $R = 8.314 	imes 10^{-3} \,kJ\,K^{-1} \,mol^{-1}$.Substituting the values: $-1366.5 = \Delta U + (-1) 	imes (8.314 	imes 10^{-3}) 	imes 300$.$-1366.5 = \Delta U - 2.4942$.$\Delta U = -1366.5 + 2.4942 = -1364.0058 \,kJ\,mol^{-1} \approx -1364 \,kJ\,mol^{-1}$.

For the reaction $C_2H_5OH_{(l)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)}$,the enthalpy change $(\Delta H)$ at $27\,^oC$ is $-1366.5 \,kJ\,mol^{-1}$. What will be the value of the internal energy change $(\Delta U)$ in $kJ\,mol^{-1}$ at the same temperature?

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