Question

The reaction A+B gives C+D proceed to right hand side upto 99.9%. the equilibrium constant K of reaction will be?

Answer 1

Answer: The equilibrium constant will be very high for the reaction.

Explanation:

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{eq}$ is written as:

$K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

There are 3 conditions:

• When $K_{eq}>1$; the reaction is product favored.
• When $K_{eq}<1$; the reaction is reactant favored.
• When $K_{eq}=1$; the reaction is in equilibrium.

It is given that reaction is proceeding 99.9 % to the right side. This means that it is product favored.

Hence, the equilibrium constant will be very high for the reaction.