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(B) The relationship between enthalpy change at constant pressure $(\Delta H)$ and constant volume $(\Delta U)$ is given by $\Delta H = \Delta U + \De

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$1000 \ atm^3 \ M^{-3}$

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(B) The relationship between enthalpy change at constant pressure $(\Delta H)$ and constant volume $(\Delta U)$ is given by $\Delta H = \Delta U + \Delta n_g RT$. Given $\Delta H - \Delta U = 1.8 \ Kcal/mol = 1800 \ cal/mol$ at $T = 300 \ K$. Substituting the values: $1800 = \Delta n_g 	imes 2 	imes 300$,which gives $\Delta n_g = 3$. The relationship between $K_P$ and $K_C$ is $K_P = K_C(RT)^{\Delta n_g}$. Thus,$\frac{K_P}{K_C} = (RT)^{\Delta n_g}$. Given $T = \frac{1}{0.00821} \ K$ and $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1} = 0.00821 \ L \ atm \ K^{-1} \ mmol^{-1}$. Using $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,we have $RT = 0.0821 	imes \frac{1}{0.00821} = 10$. Therefore,$\frac{K_P}{K_C} = (10)^3 = 1000 \ atm^3 \ M^{-3}$.

For a gaseous reversible reaction,the enthalpy of reaction at constant pressure is $1.8 \ Kcal/mol$ greater than that at constant volume at $300 \ K$. The value of $\left( \frac{K_P}{K_C} \right)$ for the reaction at $T = \left( \frac{1}{0.00821} \right) \ K$ is:

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