Question

The reaction A + 2B $\rightleftharpoons$2C + D is carried out at 298 K using only A and B initially. While mixing, the concentration of B was 1.5 times of that of A. When this system attained equilibrium, it was observed that the (new) concentrations of A and D were same. What is the value of $K_c$ (equilibrium constant) at 298 K:___

Answer 1

Let the initial concentration of A = a
Hence the concentration of B = 1.5a
$\begin{array}{ccccccc}Initialconc.& a& & 1.5a& & 0& 0\\ & A& +& 2B& \leftrightharpoons & 2C& D\\ Equb.conc.& (a-x)& & (1.5a-2x)& & 2x& x\end{array}$

$\therefore K=\frac{\left(2x{)}^2\right(x)}{(a-x)(1.5a-2x{)}^2}=\frac{4x^3}{(a-x)(1.5a-2x{)}^2}$

After the equilibrium is attained,
Since concentration of A = concentration of D
We have a - x = x or a = 2x

If we plug the above result into the $K_c$ expression , we get $K_c$ = 4