Question

In a chemical reaction, $A+2B\stackrel{K}{\rightleftharpoons }2C+D$, 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 $\left(K\right)$ for the aforesaid chemical reaction is:

• A. $16$
• B. $4$
• C. $1$
• D. $\frac{1}{4}$

Answer 1

The correct option is B $4$

$A+2B\rightleftharpoons 2C+D$

$t=0t=t_{eqm}$ $a_o{a}_o-x$ $1.5a_{o}1.5a_o-2x$ $02x$ $0x$

At equilibrium $\left[A\right]=\left[B\right]$

$a_0-x=1.5a_0-2x\Rightarrow x=0.5a_0$

$\left[C\right]=2\times 0.5a_o=a_o,\left[D\right]=\left[A\right]=\left[B\right]=0.5a_o$

$K_C=\frac{\left[C{]}^2\right[D]}{\left[A\right][B{]}^2}=\frac{\left(a_0{)}^2\right(0.5a_0)}{\left(0.5a_0\right)(0.5a_0{)}^2}=4$.

Hence, the correct option is B.