Question

The equilibrium constant is given for one of the reactions below.

Answer 1

Answer:

12.7551

Explanation:

The given chemical equation follows:

$2HD(g)\rightarrow H_2(g)+D_2(g)$

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:

$2H_2(g)+2D_2(g)\rightarrow 4HD(g)$

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:

$K_{eq}'=(\frac{1}{0.28})^2$

Hence, the value of equilibrium constant for reverse reaction is 12.7551.