A reaction mixture containing H 2- N 2 and NH 3 has partial pressures 2 atm - 1 atm and 3 atm respectively at 725 K- If the value of K p for the reaction- N 2 g -3 H 2 g - 2 NH 3 g is 4-28 - 10-5 atm -2 at 725 K- in which direction the net reaction will go-A- No net reactionB- Direction of reaction cannot be predictedC- ForwardD- Backward

The correct option is D BackwardThe given reaction is, N_2(g) + 3H_2(g)⇌ 2NH_3(g) K_p=4.28 × 10^ 5 atm^ 2 Q_P = (P_NH_3)^2P_N_2× (P_H_2)^3 = (3)^2(1)(2)^3 Q_p ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Characteristics of Equilibrium Constant

A reaction mi...

Question

A reaction mixture containing $H_{2}, N_{2}$ and $N H_{3}$ has partial pressures 2 atm, 1 atm and 3 atm respectively at 725 K. If the value of $K_{p}$ for the reaction, $N_{2 (g)} + 3 H_{2 (g)} ⇌ 2 N H_{3 (g)}$ is $4.28 \times 10^{- 5} a t m^{- 2}$ at 725 K, in which direction the net reaction will go?

Forward

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Backward

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No net reaction

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Direction of reaction cannot be predicted

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B Backward
The given reaction is, $N_{2 (g)} + 3 H_{2 (g)} ⇌ 2 N H_{3 (g)}$
$K_{p} = 4.28 \times 10^{- 5} a t m^{- 2}$ $Q_{P} = \frac{(P_{N H_{3}})^{2}}{P_{N_{2}} \times (P_{H_{2}})^{3}} = \frac{(3)^{2}}{(1) (2)^{3}}$
$Q_{p} = \frac{9}{8} a t m^{- 2} = 1.125 a t m^{- 2}$ .
Since value of $Q_{P}$ is larger than $K_{P}$ $(4.28 10^{- 5} a t m^{- 2})$ , it indicates net reaction will proceed in backward direction.

Suggest Corrections

Similar questions

Q. A reaction mixture containing

H_{2}, N_{2}

and

N H_{3}

has partial pressures

2

atm,

1

atm and

3

atm respectively at

725 K

. If the value of

K_{p}

for the reaction,

N_{2 (g)} + 3 H_{2 (g)} ⇌ 2 N H_{3 (g)}

4.28 \times 10^{- 5} a t m^{- 2}

725 K

, in which direction the net reaction will go?

Q. A reaction mixture containing

H_{2}, N_{2}

and

N H_{3}

has partial pressure

2 a t m, 1 a t m

and

3 a t m

respectively at

725 K

. If the value of

K_{P}

for reaction,

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

4.28 \times 10^{- 5} a t m^{- 2}

725 K

, in which direction the net reaction will go :

Q. In a reaction mixture containing

H_{2}, N_{2}

and

{N H}_{3}

at partial pressure of

1

atm,

2

atm and

3

atm respectively, the value of

K_{p}

725 K

4.28 \times 10^{- 5} {a t m}^{- 2}

. In which direction the net reaction will go?

N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)

Q. A reaction mixture containing

H_{2}, N_{2}

and

N H_{3}

has partial pressures 2 atm,1 atm, and 3 atm respectively at 725K.

If the value of

K_{p}

for the reaction,

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

4.28 \times 10^{- 5} {a t m}^{- 2}

at 725K,in which direction the net reaction will go :

Q. A reaction mixture containing

H_{2}, N_{2}

and

N H_{3}

has partial pressure 2 atm, 1atm and 3 atm respectively at 725 K. If the value of K for the reaction.

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

4.28 \times 10^{- 5}

atm

^{- 2}

at 725 K, in which direction the net reaction will go:

Join BYJU'S Learning Program

Explore more

Characteristics of Equilibrium Constant

Standard XII Chemistry

Join BYJU'S Learning Program

A reaction mixture containing H2,N2 and NH3 has partial pressures 2 atm, 1 atm and 3 atm respectively at 725 K. If the value of Kp for the reaction, N2(g)+3H2(g)⇌2NH3(g) is 4.28×10−5 atm−2 at 725 K, in which direction the net reaction will go?

The correct option is B BackwardThe given reaction is, N2(g)+3H2(g)⇌2NH3(g) Kp=4.28×10−5 atm−2 QP=(PNH3)2PN2×(PH2)3=(3)2(1)(2)3 Qp=98atm−2=1.125 atm−2. Since value of QP is larger than KP (4.28 10−5atm−2), it indicates net reaction will proceed in backward direction.