For the gaseous reaction,N_2O_4(g) to 2NO_2(g | vedclass

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Question

(D) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ for a gaseous reaction is given by the equation: $\D

Vedclass · Accepted Answer

$\Delta H > \Delta E$

Vedclass · Answer

(D) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ for a gaseous reaction is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$.For the reaction $N_2O_{4(g)} 	o 2NO_{2(g)}$,the change in the number of gaseous moles is $\Delta n_g = n_{p(g)} - n_{r(g)} = 2 - 1 = 1$.Since $\Delta n_g = 1$ (which is positive),the term $\Delta n_g RT$ is positive.Therefore,$\Delta H = \Delta E + RT$,which implies $\Delta H > \Delta E$.

For the gaseous reaction,$N_2O_4(g) \to 2NO_2(g)$,which of the following relations is correct?

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