Question

When $3.00mole$ of $A$ and $1.00mole$ of $B$ are mixed in a $1.00litre$ vessel, the following reaction takes place:

$A\left(g\right)+B\left(g\right)\leftrightharpoons 2C\left(g\right)$.

The equilibrium mixture contains $0.5mole$ of $C$. What is the value of equilibrium constant for the reaction?

• A. $0.12$
• B. $6$
• C. $1.5$
• D. $3$

Answer 1

Answer: (A)

$0.12$

$A\left(g\right)+B\left(g\right)\leftrightharpoons 2C\left(g\right)$

3 1 0

3-x 1-x 2x

Given that, $2x=0.5$

$x=\frac{1}{4}$

Now,

$K_c=\frac{[C{]}^2}{\left[A\right]\left[B\right]}$

$[c{]}^2={\left(\frac{1}{4}\right)}^2=\frac{1}{16}$

$\left[A\right]=3-\frac{1}{4}=\frac{11}{4}$

$\left[B\right]=1-\frac{1}{4}=\frac{3}{4}$

$K_c=\frac{\frac{1}{16}}{\frac{11}{4}\times \frac{3}{4}}$

$Kc=0.12$

The equilibrium constant for the reaction is 0.12.

Hence, option $A$ is correct.