Calculate Delta H for the following reaction | vedclass

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(A) The relationship between $\Delta H$ and $\Delta U$ is given by: $\Delta H = \Delta U + \Delta n_g RT$For the reaction $2C(s) + 3H_2(g) \rightarrow

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$-85.00$

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(A) The relationship between $\Delta H$ and $\Delta U$ is given by: $\Delta H = \Delta U + \Delta n_g RT$For the reaction $2C(s) + 3H_2(g) ightarrow C_2H_6(g)$:Calculate the change in the number of gaseous moles, $\Delta n_g = n_p(g) - n_r(g) = 1 - 3 = -2 	ext{ mol}$.Given $\Delta U = -80 	ext{ kJ} = -80000 	ext{ J}$, $T = 300 	ext{ K}$, and $R = 8.314 	ext{ J K}^{-1} 	ext{mol}^{-1}$.$\Delta H = -80000 	ext{ J} + (-2 	ext{ mol} 	imes 8.314 	ext{ J K}^{-1} 	ext{mol}^{-1} 	imes 300 	ext{ K})$$\Delta H = -80000 	ext{ J} - 4988.4 	ext{ J} = -84988.4 	ext{ J} \approx -85.00 	ext{ kJ}$.

Calculate $\Delta H$ for the following reaction at $300 \text{ K}$: $2C(s) + 3H_2(g) \rightarrow C_2H_6(g)$ if $\Delta U$ for the reaction is $-80 \text{ kJ}$ $(R = 8.314 \text{ J K}^{-1} \text{mol}^{-1})$ (in $\text{ kJ}$)

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