Question

The ion pair MgSO4 is formed in solution via the reaction below: In a system where the initial state is [Mg"2-0.10 M, [S04-21,-0.10 M and [MgSOJ,-0.00 M, the equilibrium constant was found to be 1.015. Calculate the ratio of the rate constants for the forward (k) and reverse (kb) reactions (kf/ ko). Calculate the state of the system at equilibrium.

Answer 1

Initial concentration of Mg^2 +, [Mg^2 +]_0 = 0.1 M Initial concentration of SP_4^2, [SO^2 -_4]_0 0.1 M Initial concentration of Mg SO_4, [Mg SO_4]_0 = 0 M Mg^2 + + SO^2 -_4 Mg SO_4 At t = 0, 0.1 0.1 0 Initial state At t = 0 0.1 - x 0.1 - x x Equilibrium State where t is time when equilibrium has been achieved. Equilibrium Constant, k = 1.015 At Equilibrium, concentration Mg^2 +, [Mg^2 +] = 0.1 - x Concentration SO^2 -_4, [Mg SO_4] = x Therefore From definition of equilibrium constant, k = [Mg SO_4]/[Mg^2 +] [SO^2 -_4] 1.015 = x/(0.1 - x) (0.1 - x) Therefore 1.015 {0.01 + x^2 - 0.2 x} = x Therefore 1.015 x^2 - 0.203 x + 0.01015 = 0 Solving above equation gives: x = 1.17672 & 0.00849817 Since x can't be greater than 0.1 otherwise equilibrium concentration of Mg^2 + & SO^2 -_4 shall be negative.