Consider the equilibrium `H _(2) + I _(2) Harr 2HI.` Calculate the equilibrium constant of the reverse reaction when the equlibrium concentration of `H _(2), I _(2) and HI` are `1.14xx10^(-2),0.12 xx10^(-2) and 2.50 xx10^(-2) mol L ^(-1),` respectively

Options

A. 46.4
B. 0.021
C. 18.42
D. 0.054

Answer

B

Explanation

`H_2 + I_2 iff 2HI`
Reverse of above reaction is
`2HI iff H_2 + I_2 `
Equilibrium constant `(K) = ( [H_2][I_2])/([HI]^2)`
Given,
Concentration of `H_2 = 1.14 xx 10^(-2)` mol/L
Concentration of `I_2 = 0.12 xx 10^(-2)` mol/L
concentration of `HI = 2.52 xx 10^(-2)` mol/L
`K = ([1.14 xx 10^(-2) ] [ 0.12 xx 10^(-2) ] )/([2.52 xx 10^(-2) ]^2 )`
K = 0.021