Question

$28g$ of $N_2$ and $6g$ of $H_2$ were kept at ${400}^{o}C$ in $1L$ vessel. The equilibirium mixture contained $27.54g$ of $N{H}_3$. The approximate value for $K_c$ for the above reaction in $M^2$ is ________.

• A. $10$
• B. $75$
• C. $30$
• D. $40$

Answer 1

The correct option is B $75$

Initial number of mole:

Moles of $H_2=\frac{mass}{molarmass}=\frac{6}{2}=3moles$

Moles of $N_2=\frac{28}{28}=1moles$

initial number of moles: 1 3 0

$N_2+3H_2\iff 2N{H}_3$

At equilibrium number of mole: 1-x 3-3x 2x

At equilibrium number of mole of $N{H}_3=\frac{27.54}{17}=1.62mol$

$2x=1.62$

$x=0.81$

At equilibrium number of mole of $N_2=1-0.81=0.19mol$

At equilibrium number of mole of $H_2=3(1-0.81)=0.57mol$

Equilibrium concentration of

$N_2=0.19/1=0.19M$

$H_2=0.57/1=0.57M$

$N{H}_3=1.62/1=1.62M$

Equilibrium constant, $K_c=\frac{[NH3{]}^2}{\left[N2\right]\times [H2{]}^3}$

$K_c=\frac{{1.62}^2}{(0.19\times {0.57}^3)}$

$Kc=74.6\approx 75$