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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

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$-6.2$

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(C) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$.Therefore,the difference is $\Delta H - \Delta U = \Delta n_g RT$.For the reaction: $C_2H_{6(g)} + 3.5O_{2(g)} ightarrow 2CO_{2(g)} + 3H_2O_{(l)}$,the number of moles of gaseous products is $2$ and gaseous reactants is $1 + 3.5 = 4.5$.$\Delta n_g = n_{g(products)} - n_{g(reactants)} = 2 - 4.5 = -2.5$.Temperature $T = 25 + 273 = 298 \ K$.$\Delta H - \Delta U = -2.5 	imes 8.314 	imes 298 = -6193.93 \ J$.Converting to $kJ$: $-6193.93 \ J = -6.19 \ kJ \approx -6.2 \ kJ$.

Calculate the difference between $\Delta H$ and $\Delta U$ for the following reaction at $25^{\circ} C$:
$C_2H_{6(g)} + 3.5O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)}$
(Given: $R = 8.314 \ J \ K^{-1} \ mol^{-1}$) (in $kJ$)

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Calculate the difference between $\Delta H$ and $\Delta U$ for the following reaction at $25^{\circ} C$:$C_2H_{6(g)} + 3.5O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)}$(Given: $R = 8.314 \ J \ K^{-1} \ mol^{-1}$) (in $kJ$)