Question

The equilibrium constants at $298K$ for a reaction $A+B\rightleftharpoons C+D$ is 100. If the initial concentration of all the four species were $1M$ each then equilibrium concentration of $D\left({\text{in mol L}}^{-1}\right)$ will be:

• A. 1.182
• B. 0.182
• C. 0.818
• D. 1.818

Answer 1

The correct option is D 1.818

the given reaction is
$A+B\rightleftharpoons C+DK=100$
and the concentration of different species are $\left[A\right]=\left[B\right]=\left[C\right]=\left[D\right]=1M$
So, reaction quotient $Q_c=\frac{\left[C\right]\times \left[D\right]}{\left[A\right]\times \left[B\right]}=1$
$Q=\frac{1\times 1}{1\times 1}=1$
$\because Q<K$ so reaction moves in forward direction
$\begin{array}{cccccccc}& A& +& B& \rightleftharpoons & C& +& D\\ \text{Initial}& 1& & 1& & 1& & 1\\ \text{At equ.}& 1-x& & 1-x& & 1+x& & 1+x\end{array}$

Comparing the equilibrium constant,
$K_{eqb}=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}=\frac{(1+x{)}^2}{(1-x{)}^2}=100\Rightarrow \frac{1+x}{1-x}=10$
Solving this we get,
$x=\frac{9}{11}$
Concentration of D,
$\therefore \left[D\right]=1+x=1+\frac{9}{11}=1.818\text{M}$
hence the concentration of D is $1.818mol{L}^{-1}$