Question

For the reaction $A + B \rightleftharpoons 3C$ at ${25^0}C$
A $3L$ vessel contains $1,2$ and $4$ moles of A , B and C respectively . Predict the direction of reaction if
a) ${K_c}$ for the reaction is $10$
b) ${K_c}$ for the reaction is $15$
c) ${K_c}$ for the reaction is $10.66$
A. a) Backward direction b) Forward direction c) At equilibrium
B. a) Forward direction b) Backward direction c) Forward direction
C. a) Backward direction b) Forward direction c) Back direction
D. a) Backward direction b) Backward direction c) At equilibrium

Answer 1

Hint: We know that equilibrium constant (${K_c}$) can be defined as the ratio of concentration of product to the ratio of concentration of reactant .each concentration of reactant and product raised to the power of their stoichiometric ratios. We are also familiar with the reaction quotient (${Q_c}$) which measures the relative amounts of products and reactants present during the reaction at specific points. The reaction quotient is useful in determination of the direction of reaction. There is a important relation between ${K_c}$ and ${Q_c}$ which is as follows :
Complete answer:
1 . If ${Q_c} > {K_c}$ then the direction of the reaction will be backward means the reverse reaction will be favoured.
2 . If ${Q_c} < {K_c}$ then the direction of the reaction will be forward . 3 . If ${Q_c} = {K_c}$ it shows that the reaction is at equilibrium means both the forward and backward reactions continue to occur.
Now we will write reaction quotient for the given reaction $A + B \rightleftharpoons 3C$ the expression for ${Q_c}$ is written as : ${Q_c} = \dfrac{{{{[C]}^3}}}{{{{[A]}^1}{{[B]}^1}}}$, here $[A],[B]$ and $[C]$ is the active mass of A,B and C respectively. Active mass can be calculated by taking the ratio of number of moles to volume of solution in L . Then here,
The reaction quotient =${Q_c} = \dfrac{{{{[(\dfrac{4}{3})]}^3}}}{{[\left( {\dfrac{1}{3}} \right)][\left( {\dfrac{2}{3}} \right)]}}$
                                                   $ \Rightarrow {Q_c} = 10.66$
a) ${K_c}$ for the reaction is $10$ that means ${Q_c} > {K_c}$ then the direction of the reaction will be backward.
b) ${K_c}$ for the reaction is $15$ that means ${Q_c} < {K_c}$ then the direction of the reaction will be forward.
c) ${K_c}$ for the reaction is $10.66$ that means ${Q_c} = {K_c}$ it shows that the reaction is at equilibrium.

So option A is correct that is a) Backward direction b) Forward direction c) At equilibrium.

Note: We have approached this problem with the help of relation between ${K_c}$ and ${Q_c}$. Then we calculated the reaction quotient and compared its value with the given values of equilibrium constant which gives the direction of the reaction.