Question

The equilibrium constant of the reaction $ {A_2}(g) + {B_2}(g) \rightleftharpoons 2AB(g) $ at $ {100^ \circ }C $ is $ 16 $ . Initially equal moles of $ {A_2} $ and $ {B_2} $ are taken in $ 2L $ container. Then find the mole $ \% $ of $ {A_2} $ in the equilibrium mixture.

Answer 1

Hint :In this we will use the equilibrium constant formula to calculate the value of $ a $ . It is formulated as amounts of products divided by amounts of reactants and each amount is raised to the power of its coefficient in the balanced chemical equation.

Complete Step By Step Answer:
We have the given chemical reaction $ {A_2}(g) + {B_2}(g) \rightleftharpoons 2AB(g) $
The equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium, a state approached by a dynamic chemical system after sufficient time has elapsed at which its composition has no measurable tendency towards further change.
Given the initial composition of a system, known equilibrium constant values can be used to determine the composition of the system at equilibrium. However, reaction parameters like temperature, solvent, and ionic strength may all influence the value of the equilibrium constant.
So, the equilibrium constant for the given reaction is:
 $ = K = \dfrac{{{{[AB]}^2}}}{{[{A_2}][{B_2}]}} $
 $ = 16 = \dfrac{{{{(2a)}^2}}}{{(1 - a)(1 - a)}} $
 $ = 16 = \dfrac{{4{a^2}}}{{{{(1 - a)}^2}}} $
 $ = 4 = \dfrac{{2a}}{{(1 - a)}} $
 $ = 4 - 4a = 2a $
 $ = 4 = 6a $
 $ a = \dfrac{2}{3} $
So, we know that the $ \% $ mole of $ {A_2} $ at equilibrium is:
 $ \dfrac{{no.\;of\;molesat\;{A_2}}}{{total\;moles\;at\;equilibrium}} \times 100 $
Number of moles of $ {A_2} $ is $ \dfrac{1}{3} $ and total moles at equilibrium is $ 2 $ .
So, $ \dfrac{1}{3} \times \dfrac{1}{2} \times 100 $
 $ \dfrac{{100}}{6} = 16.67\% $
So, the $ \% $ mole of $ {A_2} $ at equilibrium is $ 16.67\% $ .

Note :
When the concentration values are measured on the mole fraction scale all concentrations and activity coefficients are dimensionless quantities. The quotient is constant under the conditions in which the equilibrium constant value is determined. These conditions are usually achieved by keeping the reaction temperature constant and by using a medium of relatively high ionic strength as the solvent.