The value of Delta H - Delta E for the follow | vedclass

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Question

(D) The relationship between enthalpy change and internal energy change is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$,which can be r

Vedclass · Accepted Answer

$8.314 	imes 300 	imes (2)$

Vedclass · Answer

(D) The relationship between enthalpy change and internal energy change is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$,which can be rearranged as $\Delta H - \Delta E = \Delta n_g RT$. For the reaction $2NH_{3(g)} 	o N_{2(g)} + 3H_{2(g)}$,the change in the number of gaseous moles is $\Delta n_g = (1 + 3) - 2 = 2$. The temperature $T = 27 \ ^\circ C = 27 + 273 = 300 \ K$. Substituting these values,we get $\Delta H - \Delta E = 2 	imes R 	imes 300 = 8.314 	imes 300 	imes 2$. Thus,the correct option is $D$.

The value of $\Delta H - \Delta E$ for the following reaction at $27 \ ^\circ C$ will be $2NH_{3(g)} \to N_{2(g)} + 3H_{2(g)}$

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