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Question: The equilibrium constant of the reaction \( {A_2}(g) + {B_2}(g) \rightleftharpoons 2AB(g) \) at \( {...

The equilibrium constant of the reaction A2(g)+B2(g)⇌2AB(g){A_2}(g) + {B_2}(g) \rightleftharpoons 2AB(g) at 100∘C{100^ \circ }C is 1616 . Initially equal moles of A2{A_2} and B2{B_2} are taken in 2L2L container. Then find the mole %\% of A2{A_2} in the equilibrium mixture.

Explanation

Solution

Hint : In this we will use the equilibrium constant formula to calculate the value of aa . 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 A2(g)+B2(g)⇌2AB(g){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=[AB]2[A2][B2]= K = \dfrac{{{{[AB]}^2}}}{{[{A_2}][{B_2}]}}
=16=(2a)2(1−a)(1−a)= 16 = \dfrac{{{{(2a)}^2}}}{{(1 - a)(1 - a)}}
=16=4a2(1−a)2= 16 = \dfrac{{4{a^2}}}{{{{(1 - a)}^2}}}
=4=2a(1−a)= 4 = \dfrac{{2a}}{{(1 - a)}}
=4−4a=2a= 4 - 4a = 2a
=4=6a= 4 = 6a
a=23a = \dfrac{2}{3}
So, we know that the %\% mole of A2{A_2} at equilibrium is:
no.  of  molesat  A2total  moles  at  equilibrium×100\dfrac{{no.\;of\;molesat\;{A_2}}}{{total\;moles\;at\;equilibrium}} \times 100
Number of moles of A2{A_2} is 13\dfrac{1}{3} and total moles at equilibrium is 22 .
So, 13×12×100\dfrac{1}{3} \times \dfrac{1}{2} \times 100
1006=16.67%\dfrac{{100}}{6} = 16.67\%
So, the %\% mole of A2{A_2} at equilibrium is 16.67%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.