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(C) The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta E + \Delta n_g RT$.Rearranging this,we get $\De

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$N_{2(g)} + O_{2(g)} 	o 2NO_{(g)}$

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(C) The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta E + \Delta n_g RT$.Rearranging this,we get $\Delta E - \Delta H = - \Delta n_g RT$.For $(\Delta E - \Delta H)$ to be maximum,the value of $-\Delta n_g$ must be maximum,which means $\Delta n_g$ must be the most negative (minimum).Calculating $\Delta n_g$ (moles of gaseous products - moles of gaseous reactants) for each reaction:$A: \Delta n_g = (1 + 1) - 1 = 1$$B: \Delta n_g = (1 + 1) - 0 = 2$$C: \Delta n_g = 2 - (1 + 1) = 0$$D: \Delta n_g = (0 + 1) - 0 = 1$Since $\Delta E - \Delta H = - \Delta n_g RT$,the expression is maximum when $\Delta n_g$ is minimum. However,the question asks for the maximum value. Re-evaluating: if $\Delta n_g$ is negative,the value is positive. Among the given options,$C$ has the lowest $\Delta n_g$ $(0)$,making the value $0$. For $A, B, D$,$\Delta n_g$ is positive,making $(\Delta E - \Delta H)$ negative. Thus,the maximum value is $0$ for option $C$.

At a definite temperature,for which of the following reactions will the value of $(\Delta E - \Delta H)$ be maximum?

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