Chemical and Ionic Equilibrium - Result Question 5
5. In a chemical reaction, $A+2 B \stackrel{K}\rightleftharpoons 2 C+D$, 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) $\dfrac{1}{4}$
(b) 16
(c) 1
(d) 4
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Answer:
Correct Answer: 5. (d)
Solution:
- For the given chemical reaction,
[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, $K=\dfrac{[C]^2[D]}{[A][B]^2}$
Now, substituting the values in above equation, we get
$K=\dfrac{\left(a_0\right)^2 \times\left(0.5 a_0\right)}{\left(0.5 a_0\right) \times\left(0.5 a_0\right)}=4$
BETA
