Question

At 1000 K, the equilibrium constants for the following equilibrium are $ {K_1} $ and $ {K_2} $ .
 $ COC{l_2}(g) \rightleftharpoons CO(g) + C{l_2};{K_1} = 0.329 $
 $ 2CO(g) + {O_2}(g) \rightleftharpoons C{O_2}(g);{K_2} = 2.24 \times {10^{22}} $
For the equilibrium,
 $ 2COC{l_2}(g) + {O_2} \rightleftharpoons 2CO(g) + 2C{l_2}(g) $ , the equilibrium constant $ {K_3} $ is:
(A) $ 2.424 \times {10^{22}} $
(B) $ 2.424 \times {10^{21}} $
(C) $ 4.84 \times {10^{20}} $
(D) $ 2.424 \times {10^{20}} $

Answer 1

Hint: Chemical equilibrium is also known as dynamic equilibrium. The state in which both reactants and products are present in concentrations which have no further tendency to change with time is known as chemical equilibrium. We shall write the equilibrium constant for the 3 equations and substitute the values given to find the answer required.

Complete Step by step solution
The value of reaction quotient at chemical equilibrium is called equilibrium constant. For a given set of reaction conditions equilibrium constant is independent of initial concentration of reactants and products species. Equilibrium constant also helps us to determine the composition of the System at equilibrium. When the forward reaction equals the reverse reaction rate, equilibrium state is achieved. The equilibrium constant can be given by the multiplication of the product concentration raised to their coefficients divided by the multiplication of the reactant concentration raised to their coefficients. This relationship is known as the law of mass action.
Equilibrium constant is unit less. Equilibrium constant depends on the temperature of the reaction. This is because equilibrium is defined as a condition where temperature changes the corresponding condition changes in the reaction and equilibrium constant changes.
When the reaction is reversed, the value of new equilibrium constant for that reaction is reciprocal of the earlier equilibrium constant. When the balanced reaction is multiplied by a Factor x, the equilibrium constant is raised to $ {x^{th}} $ power. When two or more reactions are added, the new equilibrium constant is the product of the equilibrium constants of the equations added.
In order to obtain the third reaction, we will multiply the first reaction by two and add the first and the second reaction. Hence, the equilibrium constant $ {K_3} $ will be given by squaring the first equilibrium constant and multiplying it with second equilibrium constant.
For the reaction, $ 2COC{l_2}(g) + {O_2} \rightleftharpoons 2CO(g) + 2C{l_2}(g) $ ,
 $ {K_3} = {({K_1})^2} \times {K_2} $
 $ {K_3} = {(0.329)^2} \times 2.24 \times {10^{22}} $
 $ {K_3} = 0.2424 \times {10^{22}} $
 $ {K_3} = 0.2424 \times {10^{21}} $
Hence the correct answer is option B.

Note
In the chemical equilibrium, when the reaction is a reversible reaction then there is no net change in the concentration of reactants and products as at equilibrium the two opposing reactions go on equal rates, this type of equilibrium is also called dynamic equilibrium.