N 2+3 H 2 leftharpoons 2 NH 3 ; K c =1.2 At the start of a reaction, there are 0.249 mol N 2, 3.21 × 10-2 mol H 2 and 6.42 × 10-4 mol NH 3 in a 3.50 L reaction vessel at 375° C. Hence reaction will proceed in

Question

N 2+3 H 2 leftharpoons 2 NH 3 ; K c =1.2 At the start of a reaction, there are 0.249 mol N 2, 3.21 × 10-2 mol H 2 and 6.42 × 10-4 mol NH 3 in a 3.50 L reaction vessel at 375° C. Hence reaction will proceed in (A) Forward direction (B) Backward direction (C) At equilibrium (D) Stops

Tardigrade Admin · Accepted Answer

QC=([N H3]2/[N2][H2]3)+(((6.42/3.5))2 ×(10-4)2/((0.249)3.5)((3.21 × 10-2)3.5)3=0.61 ⇒ Q/C) < Kc ⇒ Reaction proceeds in forward direction.

Q. $N_{2} + 3 H_{2} ⇌ 2 N H_{3}; K_{c} = 1.2$
At the start of a reaction, there are $0.249 m o l N_{2}, 3.21 \times 1 0^{- 2} m o l H_{2}$ and $6.42 \times 1 0^{- 4} m o l N H_{3}$ in a $3.50 L$ reaction vessel at $37 5^{\circ} C$ . Hence reaction will proceed in

Forward direction

Backward direction

At equilibrium

Stops

$Q_{C} = \frac{[ N H _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}} + \frac{( \frac{6.42}{3.5} ) ^{2} \times ( 1 0 ^{- 4} ) ^{2}}{( \frac{0.249}{3.5} ) ( \frac{3.21 \times 1 0 ^{- 2}}{3.5} ) ^{3}} = 0.61$
$\Rightarrow Q_{C} < K_{c}$
$\Rightarrow$ Reaction proceeds in forward direction.

Q. N2​+3H2​⇌2NH3​;Kc​=1.2 At the start of a reaction, there are 0.249molN2​,3.21×10−2molH2​ and 6.42×10−4molNH3​ in a 3.50L reaction vessel at 375∘C. Hence reaction will proceed in