For the formation of NH 3 from N 2 and H 2 at 500 K, the concentration of N 2, H 2 and NH 3 at equilibrium are 1.5 × 10-2 M , 3.0 × 10-2 M and 1.2 × 10-2 M, respectively. The equilibrium constant for the reverse reaction is

Question

For the formation of NH 3 from N 2 and H 2 at 500 K, the concentration of N 2, H 2 and NH 3 at equilibrium are 1.5 × 10-2 M , 3.0 × 10-2 M and 1.2 × 10-2 M, respectively. The equilibrium constant for the reverse reaction is (A) 3.56 × 102 (B) 2.81 × 10-3 (C) 3.56 × 10-2 (D) 2.81 × 103

Tardigrade Admin · Accepted Answer

Given: Concentration of [ N 2] =1.5 × 10-2 M [ H 2] =3.0 × 10-2 M [ NH 3] =1.2 × 10-2 M The reaction for the formation of NH 3 is N 2+3 H 2 leftharpoons 2 NH 3 Thus, the reverse reaction will be 2 NH 3 leftharpoons N 2+3 H 2 and equilibrium constant (Kc) will be KC=([ N 2] ⋅[ H 2]3/[ NH 3]2)=([1.5 × 10-2][3.0 × 10-2]3/[1.2 × 10-2]2) =((1.5 × 10-2)(27.0 × 10-6)/1.44 × 10-4)=(40.5/1.44) × 10-4 =28.125 × 10-4 or KC=2.81 × 10-3

Q. For the formation of $N H_{3}$ from $N_{2}$ and $H_{2}$ at $500 K$ , the concentration of $N_{2}, H_{2}$ and $N H_{3}$ at equilibrium are $1.5 \times 1 0^{- 2} M, 3.0 \times 1 0^{- 2} M$ and $1.2 \times 1 0^{- 2} M$ , respectively. The equilibrium constant for the reverse reaction is

$3.56 \times 1 0^{2}$

$2.81 \times 1 0^{- 3}$

$3.56 \times 1 0^{- 2}$

$2.81 \times 1 0^{3}$

Q. For the formation of NH3​ from N2​ and H2​ at 500K, the concentration of N2​,H2​ and NH3​ at equilibrium are 1.5×10−2M,3.0×10−2M and 1.2×10−2M, respectively. The equilibrium constant for the reverse reaction is

3.56×102

2.81×10−3

3.56×10−2

2.81×103

Q. For the formation of $N H_{3}$ from $N_{2}$ and $H_{2}$ at $500 K$ , the concentration of $N_{2}, H_{2}$ and $N H_{3}$ at equilibrium are $1.5 \times 1 0^{- 2} M, 3.0 \times 1 0^{- 2} M$ and $1.2 \times 1 0^{- 2} M$ , respectively. The equilibrium constant for the reverse reaction is

$3.56 \times 1 0^{2}$

$2.81 \times 1 0^{- 3}$

$3.56 \times 1 0^{- 2}$

$2.81 \times 1 0^{3}$