5.0 Important Relationship Involving Equilibrium Constant

  Chemical Equilibrium

• If the reaction is reversed, the value of equilibrium constant is inversed. $$If\ \ \ \ A + B \rightleftharpoons C + D\ \ \ \ \ ,\ \ {K_c} = 10$$$$Then\ for\ \ \ \ C + D \rightleftharpoons A + B\ \ \ \ \ ,\ \ {K_c} = {10^{ - 1}}$$

• If the reaction is multiplied by a factor $n$ then the value of equilibrium constant becomes $${K_c}^\prime = {\left( {{K_c}} \right)^n}$$ For example, $$A + B \rightleftharpoons 2P\ \ \ ;\ \ \ {K_c} = {K_c}$$$$\frac{1}{2}A + \frac{1}{2}B \rightleftharpoons P \ \ \ ;\ \ \ {K_c}^\prime = {\left( {{K_c}} \right)^{\frac{1}{2}}}$$$$2A + 2B \rightleftharpoons 4P \ \ \ ;\ \ \ {K_c}^\prime = {\left( {{K_c}} \right)^2}$$

• When we combine individual equation, we have to multiply their equilibrium constant for net reaction.$$A \rightleftharpoons B\ \ \ \ \ ;\ \ \ \ \ \ {K_1}$$$$B \rightleftharpoons C\ \ \ \ \ ;\ \ \ \ \ \ {K_2}$$Then, $$A \rightleftharpoons C\ \ \ \ \ \ ;\ \ \ \ \ \ {K_1} \times {K_2}$$

• The value of the equilibrium constant for a particular reaction is always constant depending only upon the temperature of the reaction and is independent of the concentrations of the reactants with which we start or the direction from which the equilibrium is approached.

• The value of the equilibrium constant is not affected by the addition of a catalyst to the reaction. This is because it increases the speed of the forward and backward reactions to the same extent.

Example 4. ${N_2} + {O_2} \rightleftharpoons 2NO$ has equilibrium constant $2.5 \times {10^{^{ - 5}}}$, calculate equilibrium constant for the reaction $$NO \rightleftharpoons \frac{1}{2}{N_2} + \frac{1}{2}{O_2}$$

Solution: The given reaction is reversed and then multiplied by $\frac{1}{2}$, therefore relation between ${{K_c}}$ and ${K_c}^\prime $ will be given by $${K_c}^\prime = \sqrt {\frac{1}{{{K_c}}}} = \sqrt {\frac{1}{{2.5 \times {{10}^{^{ - 5}}}}}} = \frac{1}{{5 \times {{10}^{^{ - 3}}}}} = 2 \times {10^{^2}}$$