Question

A reaction is, $A+B\to C+D$, initially we start with equal concentrations of $A$ and $B$. At equilibrium, we find the moles of $C$ is two times of $A$. What is the equilibrium constant of the reaction?

• A. $2$
• B. $4$
• C. $1/2$
• D. $1/4$

Answer 1

The correct option is B $4$

$A+B\rightleftharpoons C+D$

At equilibrium $\left[C\right]=2\left[A\right]$

Let us consider $x=0$ be the no. of moles of $\left[C\right]$

At equilibrium $\left[C\right]=2\left[A\right]$

$x=2(1-x)$

$3x=2\Rightarrow x=2/3$

$A+B\rightleftharpoons C+D$

$\begin{array}{ccccc}Initial& 1& 1& 0& 0\\ Atequilibrium& 1-x& 1-x& x& x\\ Sofinalno& 1-2/3& 1-2/3& 2/3& 2/3\end{array}$

$K_c=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}=\frac{2/3}{1/3}.\frac{2/3}{1/3}=\frac{2}{3}\times \frac{2}{3}\times 3\times 3$

$=4$

Ans $\left(B\right)4$