JEE Main 12 Jan 2019 Morning Question 10
Question: In a chemical reaction, $ A+2B2C+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 [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) 1
B) $ \frac{1}{4} $
C) 4
D) 16
Show Answer
Answer:
Correct Answer: C
Solution:
-
$ \begin{pmatrix} {} & A & + & 2B & & 2C+ & D \\ Initial & a & {} & 1.5a & {} & 0 & 0 \\ At.eq. & a-x & {} & 1.5a-2x & {} & 2x & x \\ \end{pmatrix} $
As given, at equilibrium, $ a-x=1.5a-2x\Rightarrow x=0.5a $ $
$ [C]=2x=0.5a\times 2=a, $
$ [D]=x=0.5a $
$ [B]=1.5a-2\times 0.5a=0.5a $
$ [A]=a-x=a-0.5a=0.5a $
$ k=\frac{{{[C]}^{2}}[D]}{[A]{{[B]}^{2}}}=\frac{a^{2}\times 0.5a}{{{(0.5a)}^{2}}\times 0.5a}=\frac{a^{2}}{0.25a^{2}}=4 $
