What is the value of the equilibrium constant for the following reaction at 400 K ? 2NOCl(g) hArr 2NO(g) + Cl_(2)(g) Delta H^(@) = 77.5 kJ mol^(-1), R = 8.3124 J mol^(-1) K^(-1), Delta S = 135 J K^(-1) mol^(-1).

What is the value of the equilibrium constant for the following reacti

1.	What is the value of the equilibrium constant for the following reaction at 400 K ? 2NOCl(g) hArr 2NO(g) + Cl_(2)(g) Delta H^(@) = 77.5 kJ mol^(-1), R = 8.3124 J mol^(-1) K^(-1), Delta S = 135 J K^(-1) mol^(-1).
Answer» Solution :`Delta G = Delta H - T Delta S` `Delta G = 77.5 XX 1000 J - 400 K + 135 J K^(-1) = 77500 J - 54000 J = 23500 J` `Delta G = -2.303 RT LOG K` `23500J = -2.303 xx 8.314 xx 400 K log K` `log K = (-23500)/(19.147 xx 400 K) = (-235)/(76.588) = -3.068` K = ANTILOG (-3.608) `"Antilog" (0.932 - 4)= "Antilog" (0.93) xx 10^(-4) = 8.55 xx 10^(-4)`.