Answer:
12.7551
Explanation:
The given chemical equation follows:
[tex]2HD(g)\rightarrow H_2(g)+D_2(g)[/tex]
The equilibrium constant for the above equation is 0.28.
We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:
[tex]2H_2(g)+2D_2(g)\rightarrow 4HD(g)[/tex]
The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.
If the equation is multiplied by a factor of '2', the equilibrium constant of the reverse reaction will be the square of the equilibrium constant of initial reaction.
The value of equilibrium constant for reverse reaction is:
[tex]K_{eq}'=(\frac{1}{0.28})^2[/tex]
Hence, the value of equilibrium constant for reverse reaction is 12.7551.
