Question

$56$g of $N_2$ and $6$g of $H_2$ were kept at ${400}^0$C in $1$ litre vessel. The equilibrium mixture contained $27.54$g of $N{H}_3$. The approximate value for $K_c$ for the above reaction in $mo{l}^{2}li{t}^{-2}$ is?

• A. $67.98$
• B. $20$
• C. $30$
• D. $40$

Answer 1

Answer: (A)

$67.98$

initial number of mole of

$H_2$ = 6/2 = 3 moles

$N_2$ = 56/28 = 2 moles

initial number of moles: 1 3 0

$N_2+3H_2=2N{H}_3$

at equilibrium number of moles: 2-x 3-3x 2x

at equilibrium number of mole of $N{H}_3$ = 27.54/17 = 1.62

2x = 1.62

x = 0.81

at equilibrium number of mole of $N_2$ = 2 - 0.81 = 1.19

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

equilibrium concentration of

$N_2$ = 1.19/1 = 1.19M

$H_2$ = 0.57/1 = 0.57M

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

equilibrium constant $K_c=\frac{[N{H}_3{]}^2}{\left[N_2\right]\times [H_2{]}^3}$

$K_c={1.62}^2/(1.19\times {0.57}^3)$

$K_c=2.6244/0.0386$

$K_c=67.98$