Exam

JEE Main

Year

2013

Date

9 April

Mode

Online

Subject

Chemistry

Chapter

Chemical Equilibrium

JEE Main 2013 · 9 April · Online · Chemistry · Chemical Equilibrium

In reaction A+2 ~B leftharpoons 2 C+D , initial concentration of B...

In reaction $\mathrm{A}+2 \mathrm{~B} \rightleftharpoons 2 \mathrm{C}+\mathrm{D}$, initial concentration of $\mathrm{B}$ was $1.5$ times of $\mathrm{[A]}$, but at equilibrium the concentrations of $\mathrm{A}$ and $\mathrm{B}$ became equal. The equilibrium constant for the reaction is :

Step-by-step solution

Hence $\mathrm{K}_{\mathrm{C}}=\frac{(2 x)^2 \times x}{(a-x)(1.5 a-2 x)^2}$ Given, at equilibrium $$\begin{array}{ll}\therefore & (a-x)=(1.5 a-2 x) \\ \therefore & a=2 x\end{array}$$ On solving $K_C=4$