Chemical and Ionic Equilibrium - Result Question 61
####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
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Answer:
Correct Answer: 5. (a)
Solution:
- For the given chemical reaction,
$$ \begin{aligned} & \begin{aligned} A+2 B & \rightleftharpoons 2 C+D \ \text { At } t=0 & \rightleftharpoons 25 a \end{aligned} \ & \begin{array}{lcccc} \text { At, } t=0 & a _0 & 1.5 a _0 & 0 & 0 \ t=t _{eq} & a _0-x & 1.5 a _0-2 x & 2 x & x \end{array} \end{aligned} $$
$[x=$ 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
$$ K=\frac{\left(a _0\right)^{2} \times\left(0.5 a _0\right)}{\left(0.5 a _0\right) \times\left(0.5 a _0\right)}=4 $$
