Chemical and Ionic Equilibrium 1 Question 5
5. In a chemical reaction, $A+2 B \rightleftharpoons \stackrel{K}{\rightleftharpoons} 2 C+D$, the initial concentration of $B$ was 1.5 times of the concentration of $A$, but the equilibrium concentrations of $A$ and $B$ were found to be equal. The equilibrium constant $(K)$ for the aforesaid chemical reaction is
(2019 Main, 12 Jan I)
(a) $\frac{1}{4}$
(b) 16
(c) 1
(d) 4
Show Answer
Solution:
- For the given chemical reaction,
$$ \begin{aligned} & \begin{aligned} A+2 B & \rightleftharpoons 2 C+D \ \text { At } t=0 & \rightleftharpoons 25 a \end{aligned} \ & \begin{array}{lcccc} \text { At, } t=0 & a_{0} & 1.5 a_{0} & 0 & 0 \ t=t_{\mathrm{eq}} & a_{0}-x & 1.5 a_{0}-2 x & 2 x & x \end{array} \end{aligned} $$
$[x=$ degree of dissociation $]$
Given, at equilibrium.
$$ \begin{gathered} {[A]=[B]} \ a_{0}-x=1.5 a_{0}-2 x \ x=0.5 a_{0} \ \therefore[A]=a_{0}-x=a_{0}-0.5 a_{0}=0.5 a_{0} \ {[B]=1.5 a_{0}-2 x=1.5 a_{0}-2 \times 0.5 a_{0}=0.5 a_{0}} \ {[C]=2 x=2 \times 0.5 a_{0}=a_{0}} \ {[D]=x=0.5 a_{0}} \end{gathered} $$
Now, $\quad K=\frac{[C]^{2}[D]}{[A][B]^{2}}$
Now, substituting the values in above equation, we get
$$ K=\frac{\left(a_{0}\right)^{2} \times\left(0.5 a_{0}\right)}{\left(0.5 a_{0}\right) \times\left(0.5 a_{0}\right)}=4 $$
