Question

Use the reactions below and their equilibrium constants to predict the equilibrium constant for the reaction 2A(s)⇌3D(g). A(s) ⇌ 12 B(g)+C(g), K1=0.0334 3D(g) ⇌ B(g)+2C(g), K2=2.35

Answer 1

Answer : The equilibrium constant for the reaction will be, $4.747\times 10^{-4}$

Explanation :

The following equilibrium reactions are :

(1) $A(s)\rightleftharpoons \frac{1}{2}B(g)+C(g)$           $K_1=0.0334$

(2) $3D(g)\rightleftharpoons B(g)+2C(g)$          $K_2=2.35$  

The final equilibrium reaction is :

$2A(s)\rightleftharpoons 3D(g)$         $K_{eqm}=?$

Now we have to calculate the value of $K_{eqm}$ for the final reaction.

First we have to multiply equation (1) by 2 that means we are taking square of equilibrium constant 1 and reverse the equation 2 that means we are dividing equilibrium constant 2, we get the final equilibrium reaction and the expression of final equilibrium constant is:

$K_{eqm}=\frac{(K_1)^2}{K_2}$

Now put all the given values in this expression, we get :

$K_{eqm}=\frac{(0.0334)^2}{2.35}$

$K_{eqm}=4.747\times 10^{-4}$

Therefore, the value of $K_{eqm}$ for the final reaction is, $4.747\times 10^{-4}$