Question

56 g of nitrogen and 8 g hydrogen gas are heated in a closed vessel. At equilibrium 34 g of ammonia are present. The equilibrium number of moles of nitrogen, hydrogen and ammonia are respectively.

**

• A. ** 1,2,2
• B. 2,2,1
• C. 1,1,2
• D. 2,1,2

Answer 1

Answer: (C)

1,1,2

Answer 2

$\mathrm { N } _ { 2 } + 3 \mathrm { H } _ { 2 }$ ⇌ $2 \mathrm { NH } _ { 3 }$ ….. (i)

at $t = 0$ 56 gm

=

at equilibrium $2 - 1$ $4 - 3$ 34 gm

=1 mole = 1 mole = 2 mole

According to eq. (i) 2 mole of ammonia are present & to produce 2 mole of $\mathrm { NH } _ { 3 }$, we need 1 mole of $N _ { 2 }$ and 3 mole of hence $2 - 1 = 1$ mole of $N _ { 2 }$ and $4 - 3 = 1$ mole of are present at equilibrium in vessel.