Question

$1.1$ mole of A is mixed with $1.2$ mol of B and the mixture is kept in a $1$ L flask till the equilibrium $A+2B\rightleftharpoons 2C+D$ is reached. At equilibrium $0.1$ mol of D is formed. The $K_c$ of the reaction?

• A. $0.002$
• B. $0.004$
• C. $0.001$
• D. $0.003$

Answer 1

The correct option is B $0.004$

For the equillibrium reaction:

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

volume of flask=1L

Initial moles of A=$1.1$ mol

initial concentration of A=$[A{]}_i=1.1M$

initial mole of B =$1.2$ mol

$[B{]}_i=1.2M$

At equilibrium the cocentrations of all species are:

$[A{]}_{eq}=1.1-x,[B{]}_{eq}=1.2-2x,[C{]}_{eq}=2x,[D{]}_{eq}=x$

Given that $[D{]}_{eq}=0.1mol\times 1L=0.1M$

thus x=0.1 M

or $[A{]}_{eq}=1.1-0.1=1,[B{]}_{eq}=1.2-2\times 0.1=1,[C{]}_{eq}=2\times 0.1=0.2M,[D{]}_{eq}=0.1M$

$K_c=\frac{[D{]}_{eq}\times ([C{]}_{eq}{)}^2}{[A{]}_{eq}\times ([B{]}_{eq}{)}^2}$

upon substitution we get:

$K_c=\frac{0.1\times (0.2{)}^2}{1\times (1{)}^2}$

$K_c=0.004$