For a reaction involving ionic reactants described by the mechanism given by

Question

A + B <===> (A-B)* ===> P
log k= log k1Kc + 1.0332 ZaZb I^1/2 where I is the ionic strength of the solution. Given log k1Kc = -2.861
S2O8^2- (aq) + 2I^- (aq) =======> 2SO4^2- (aq) + l2 (aq)
Determine k( L.mol/s) for a reaction mixture containing C (KI) = 0.010 M and C (Na2S2O8) = 1.5 x 10^-4 M.

Answer 1

To determine k for the reaction, we first need to find the value of Kc, which is the equilibrium constant.

The given equation is:
S2O8^2- (aq) + 2I^- (aq) =======> 2SO4^2- (aq) + l2 (aq)

From the equation, we can see that the stoichiometric coefficients of S2O8^2- and I^- are both 1, while the stoichiometric coefficients of SO4^2- and I2 are both 2. Therefore, we can write the equilibrium expression as:

Kc = ([SO4^2-]^2 [I2]) / ([S2O8^2-] [I^-]^2)

Now, we can substitute the given concentrations into the equilibrium expression:

Kc = ([SO4^2-]^2 [I2]) / ([S2O8^2-] [I^-]^2)
= (2x)^2 (x) / ((1.5x10^-4) (0.010)^2)
= 4x^3 / 1.5x10^-6

Simplifying, we get:
Kc = 2.6667x10^12

Next, we can substitute the value of log k1Kc (which is -2.861) and the given equation for log k into the given equation:

log k = log k1Kc + 1.0332 ZaZb I^1/2

log k = -2.861 + 1.0332(1)(2) (0.5I^1/2)
log k = -2.861 + 2.0664 (0.5I^1/2)

Now, we need to calculate the value of I, the ionic strength of the solution. In this case, I can be calculated as the sum of the product of the square of the concentration of each ionic species multiplied by its charge:

I = (C(S2O8^2-)(2)^2) + (C(I^-)(1)^2) + (C(SO4^2-)(2)^2) + (C(I2)(2)^2)
= (1.5x10^-4)(2)^2 + (0.010)(1)^2 + (0)(2)^2 + (0)(2)^2
= 6x10^-4 + 0.01 + 0 + 0
= 6.01x10^-3

Substituting the value of I into the equation for log k:

log k = -2.861 + 2.0664 (0.5(6.01x10^-3)^1/2)
log k = -2.861 + 2.0664 (0.5(2.447x10^-2)^1/2)
log k = -2.861 + 2.0664 (0.5(4.947x10^-1)^1/2)
log k = -2.861 + 2.0664 (0.5(0.7016)^1/2)
log k = -2.861 + 2.0664 (0.5(0.8370)^1/2)
log k = -2.861 + 2.0664 (0.5(0.9150))
log k = -2.861 + 2.0664 (0.5(0.4575))
log k = -2.861 + 2.0664 (0.2288)
log k = -2.861 + 0.4713
log k = -2.3897

Finally, we can find k by taking the antilog of -2.3897:

k = 10^(-2.3897)
k = 0.004215 L.mol/s

Therefore, the value of k for the reaction mixture is 0.004215 L.mol/s.