Question

If the equilibrium constant for the reaction 3I- (aq) + S2O82- (aq)  I3- (aq) + 2SO42- (aq) is K, what is the equilibrium constant for the reaction (1/3)I3- (aq) + (2/3)SO42- (aq)  I- (aq) + (1/3)S2O82- (aq) ?

Answer 1

Answer: The value of equilibrium constant for Equation 2 is
`K'=\sqrt[3]{(1)/(K)}`

Step-by-step explanation:

The chemical equation whose equilibrium constant is given follows:

`3I^-(aq.)+S_2O_8^(2-)(aq.)\xrightarrow{K} I_3^-(aq.)+2SO_4^(2-)(aq.)` ......(1)

The chemical equation whose equilibrium constant is to be calculated follows:

`(1)/(3)I_3^-(aq.)+(2)/(3)SO_4^(2-)(aq.)\xrightarrow {K'} I^-(aq.)+(1)/(3)S_2O_8^(2-)(aq.)` .........(2)

As, the Equation 2 is the result of the reverse of one-third of Equation 1. So, the equilibrium constant for the Equation 2 will be the cube root of inverse of equilibrium constant of Equation 1.

The value of equilibrium constant for Equation 2 is:

`K'=\sqrt[3]{(1)/(K)}`

Hence, the value of equilibrium constant for Equation 2 is
`K'=\sqrt[3]{(1)/(K)}`