Answer:
The value of equilibrium constant for given reaction will be 28.18.
Explanation:
[tex]H_2 (g) + I_2(g)\rightleftharpoons 2 HI (g)[/tex]
The value of equilibrium constant = K = 794
The expression of an equilibrium will be given by ;
[tex]K=\frac{[HI]^2}{[H_2][I_2]}[/tex]
[tex]\frac{1}{2}H_2 (g) + \frac{1}{2}I_2(g)\rightleftharpoons HI(g)[/tex]
The value of equilibrium constant for above reaction= K' = ?
The expression of an equilibrium will be given by ;
[tex]K'=\frac{[HI]}{[H_2]^{\frac{1}{2}}[I_2]^{\frac{1}{2}}}[/tex]
Squaring both sides:
[tex]K'^2=\frac{[HI]^2}{[H_2][I_2]}[/tex]
[tex]K'^2=K=794[/tex]
[tex]K'=\sqrt{794}=28.18[/tex]
The value of equilibrium constant for given reaction will be 28.18.
