A sample problem for a chemistry student to solve from the subtopic Calculating the effect of changing the temperature on the equilibrium constant could be: What would be the new equilibrium constant (Kc) for the reaction 2NOCl (g) ↔ 2NO (g) + Cl2 (g) at 400 K, if the initial equilibrium constant at 300 K is 2.50 x 10^-3? Given that the enthalpy change of the reaction (∆H) is -91.84 kJ/mol and the entropy change of the reaction (∆S) is -188.8 J/mol K.

Question

A sample problem for a chemistry student to solve from the subtopic Calculating the effect of changing the temperature on the equilibrium constant could be: What would be the new equilibrium constant (Kc) for the reaction 2NOCl (g) ↔ 2NO (g) + Cl2 (g) at 400 K, if the initial equilibrium constant at 300 K is 2.50 x 10^-3? Given that the enthalpy change of the react

Answer 1

To solve this problem, we can use the Van't Hoff equation, which relates the change in temperature to the change in the equilibrium constant:ln K2/K1  = -H/R *  1/T2 - 1/T1 Where:K1 = initial equilibrium constant at T1  300 K K2 = new equilibrium constant at T2  400 K H = enthalpy change of the reaction  -91.84 kJ/mol R = gas constant  8.314 J/mol K T1 = initial temperature  300 K T2 = new temperature  400 K First, we need to convert the enthalpy change from kJ/mol to J/mol:H = -91.84 kJ/mol *  1000 J/1 kJ  = -91840 J/molNow, we can plug the values into the Van't Hoff equation:ln K2/2.50 x 10^-3  = - -91840 J/mol  /  8.314 J/mol K  *  1/400 K - 1/300 K Solve for K2:ln K2/2.50 x 10^-3  = 91840 / 8.314 *  1/400 - 1/300 ln K2/2.50 x 10^-3   11.04Now, we can find K2:K2 = 2.50 x 10^-3 * e^11.04 K2  0.075So, the new equilibrium constant  Kc  for the reaction at 400 K is approximately 0.075.