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Chemistry Question on Equilibrium

1.1 mol of A is mixed with 2.2 mol of B and the mixture is kept in one litre flask till the equilibrium is reached. At equilibrium, 0.2 mol of C is formed. If the equilibrium reaction is A+2B→2C+D,A + 2B \rightarrow 2C + D, the value of equilibrium constant is

A

0.0002

B

0.0004

C

0.001

D

0.003

Answer

0.001

Explanation

Solution

A+2B⇌2C+D\begin{matrix}A&+&2B&{\rightleftharpoons}&2C&+&D\end{matrix}
Initialcone1.12.200\begin{matrix}Initial cone&1.1&2.2&0&0\end{matrix}
Econe(1.1−0.1)(2.2−0.2)0.20.1\begin{matrix}E cone&\left(1.1-0.1\right)&\left(2.2-0.2\right)&0.2&0.1\end{matrix}
[∵α=0.1]\left[\because \alpha=0.1\right]
K=(0.2)2(0.1)(1.0)(2.0)2=0.0044=0.001K=\frac{\left(0.2\right)^{2}\left(0.1\right)}{\left(1.0\right)\left(2.0\right)^{2}}=\frac{0.004}{4}=0.001
=1×10−3=1\times10^{-3}