Question

write the equilibrium expression for the following reaction.based upon the value of k,would there be more or less product in the equilibrium mixture?why?


CaO(s)+CH4(g)+2H2O(g) ⇄ CaCO3(s)+4H2(g) kp=2344

Answer 1

Answer: $K_p=\frac{[H_2]^4}{[CH_4]^1[H_2O]^2}$

As value of K is greater than 1, the reaction is forward driven.

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. For equilibrium constant in terms of pressure , only gaseous reactants and products are considered.

For the given chemical reaction:

$CaO(s)+CH_4(g)+2H_2O(g)\rightleftharpoons CaCO_3(s)+4H_2(g)$

The expression for $K_p$ is written as:

$K_p=\frac{[H_2]^4}{[CH_4]^1[H_2O]^2}$  

As > 1, the reaction will be favoured in the forward direction and thus more products will be there.