Question

3 moles of $N_2$ and 9 moles of $H_2$ are taken in an evacuated vessel. The reaction starts at $t=0$ and an equillibrium is attained at $t=t_1$. The amount of ammonia is found to be 34 gm at $t=2t_1$. Mark the correct option(s):

• A. $W\left(N_2\right)+W\left(H_2\right)+W\left(N{H}_3\right)=118gmatt=t_1$
• B. $W\left(N_2\right)+W\left(H_2\right)+W\left(N{H}_3\right)=102gmatt=2t_1$
• C. $\frac{W\left(N_2\right)}{W\left(H_2\right)}=\frac{14}{3}att=\frac{t_1}{3}$
• D. $\frac{W\left(N_2\right)}{W\left(H_2\right)}=\frac{14}{3}att=\frac{t_1}{2}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $W\left(N_2\right)+W\left(H_2\right)+W\left(N{H}_3\right)=102gmatt=2t_1$
C $\frac{W\left(N_2\right)}{W\left(H_2\right)}=\frac{14}{3}att=\frac{t_1}{3}$
D $\frac{W\left(N_2\right)}{W\left(H_2\right)}=\frac{14}{3}att=\frac{t_1}{2}$

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$
Initially
3 9
At equillibrium
3-x 9-3x

At $t=t_1$ the reaction attains equillibrium therefore the amount of $N{H}_3$ remains constant after $t_1$ time
2x = 2
x = 1
Moles of $N_2=2;H_2=6;N{H}_3=2$
So,
$W\left(N_2\right)+W\left(H_2\right)+W\left(N{H}_3\right)=102gmatt=2t_1$

Molar ratio of $N_2$ and $H_2$ remains same at $\frac{t_1}{3}$ and $\frac{t_1}{2}$
$\frac{W\left(N_2\right)}{W\left(H_2\right)}=\frac{(3-x)28}{(9-3x)2}=\frac{14}{3}$