Question

At a certain temperature, the equilibrium constant $K_c$ is 0.25 for the reaction,
$A_2\left(g\right)+B_2\left(g\right)\rightleftharpoons C_2\left(g\right)+D_2\left(g\right)$
If we take 1 mole of each of the four gases in a 10 litre container, what would be the equilibrium concentration of $A_2\left(g\right)$ ?

• A. 0.331 M
• B. 0.033 M
• C. 0.133 M
• D. 1.33 M

Answer 1

The correct option is C 0.133 M

$Concentration=\frac{Moles}{Volume}$
Volume= 10 L
Each species has 1 mole.
So concentration of each species :
$\text{Concentration}=\frac{1}{10}M=0.1M$
$Q_C=\frac{\left[C_2\right]\left[D_2\right]}{\left[A_2\right]\left[B_2\right]}=\frac{0.1\times 0.1}{0.1\times 0.1}=1$

$\text{Given that}{K}_c=0.25$

$\because Q_c>K_c$ so reaction will proceed in backward direction:
$\begin{array}{cccccccc}& A_2\left(g\right)& +& B_2\left(g\right)& \rightleftharpoons & C_2\left(g\right)& +& D_2\left(g\right)\\ Conc.eqm& \frac{1+x}{10}& & \frac{1+x}{10}& & \frac{1-x}{10}& & \frac{1-x}{10}\end{array}$
$0.25=\frac{{\left(\frac{1-x}{10}\right)}^2}{{\left(\frac{1+x}{10}\right)}^2}0.5=\frac{(1-x)}{(1+x)}1.5x=0.5\Rightarrow x=0.333$

$\left[A_2\right(g\left)\right]=\frac{1+x}{10}=\frac{1.333}{10}=0.133$


Theory:

Application of Reaction Quotient(Q) :

Case 1 $ReactionQuotient\left(Q\right)>Equilibriumconstant{K}_{eq};$
For a given reaction $R⟶P$
When $Q>K_{eq}$ the product concentration $\left[P\right]$ is very high as compared to concentration of reactants $\left[R\right]$. Since concentration of products is more so there will be more number of collisions among product molecules and product starts converting back to reactants until $Q$ becomes equal to $K_{eq}$ i.e. If $Q>K_{eq}$ then it favours the backward reaction.

Case 2 $ReactionQuotient\left(Q\right)<Equilibriumconstant{K}_{eq}:$
For a given reaction $R⟶P$
When $Q<K_{eq}$ the reactant concentration $\left[R\right]$ is very high as compared to concentration of products $\left[P\right]$. Since concentration of reactants is more so there will be more number of collisions among reactant molecules and reactant starts converting back to products until $Q$ becomes equal to $K_{eq}$ i.e. If $Q<K_{eq}$ then it favours the forward reaction.
Note :- Equilibrium is achieved when the rate of backward reaction is equal to the rate of forward reaction and not when the concentration of reactants and products becomes equal.