Answer: The value of equilibrium constant for Equation 2 is [tex]K'=\sqrt[3]{\frac{1}{K}}[/tex]
Explanation:
The chemical equation whose equilibrium constant is given follows:
[tex]3I^-(aq.)+S_2O_8^{2-}(aq.)\xrightarrow{K} I_3^-(aq.)+2SO_4^{2-}(aq.)[/tex] ......(1)
The chemical equation whose equilibrium constant is to be calculated follows:
[tex]\frac{1}{3}I_3^-(aq.)+\frac{2}{3}SO_4^{2-}(aq.)\xrightarrow {K'} I^-(aq.)+\frac{1}{3}S_2O_8^{2-}(aq.)[/tex] .........(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:
[tex]K'=\sqrt[3]{\frac{1}{K}}[/tex]
Hence, the value of equilibrium constant for Equation 2 is [tex]K'=\sqrt[3]{\frac{1}{K}}[/tex]
