KP for the reaction N 2-3 H 2- 2 NH 3 at 400- C is 1-64 - 10-4- Calculate K c-A- 0-3 mole 2 litre-2B- 0-4 mole2litre-2C- 0-5 mole2 litre-2D- 0-6 mole2 litre-2

The correct option is C 0.5 mole2 litre 2 N_2 + 3H_2 ⇌ 2NH_3 Δ n = 2 and Kp=Kc(RT)^Δ n 1.64 × 10 4 = Kc (0.0821 × 673)^ 2 and Kc = 0.5 mole^2 litre^ 2 ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Hess' Law

KP for the re...

Question

KP for the reaction $N_{2} + 3 H_{2} ⇋ 2 N H_{3}$ at $400^{\circ} C$ is $1.64 \times 10^{- 4}$ . Calculate $K_{c}$ .

0.3 mole2 litre-2

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

0.4 mole2litre-2

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

0.5 mole2 litre-2

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

0.6 mole2 litre-2

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

Open in App

Solution

The correct option is C
0.5 mole2 litre-2

$N_{2} + 3 H_{2} ⇌ 2 N H_{3}$

$Δ n$ =-2~and $K p = K c (R T)^{Δ n}$

$1.64 \times 10 - 4 = K c (0.0821 \times 673)^{- 2}$

and $K c = 0.5 m o l e^{2} l i t r e^{- 2}$

Suggest Corrections

Similar questions

K_{p}

for the reaction

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

400^{\circ} C

1.64 \times 10^{- 4}

Calculate

K_{C}

Q. In the process

N_{2} + 3 H_{2} ⇋ 2 N H_{3}

, the initial concentration of nitrogen and hydrogen are

1

mole per litre and

3

moles per litre respectively. The equilibrium constant of the reaction is

x

. The value of

K_{c}

for

2 N H_{3} ⇋ N_{2} + 3 H_{2}

will be:

Q. The reaction

N_{2} + 3 H_{2} ⟺ 2 N H_{3}

takes place at

450^{0} C

. 1 mole of

N_{2}

and 2 mole of

H_{2}

are mixed in a 1 litre vessel and 1 mole of

N H_{3}

is formed at equilibrium. Then Kc for the above reaction is ?

Q. Consider the following reaction equilibrium:

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

. Initially, 1 mole of

N_{2}

and 3 moles of

H_{2}

are taken in a 2 litre flask.

At equilibrium state, if the number of moles of

N_{2}

is 0.6, then the total number moles of all gases present in the flask are:

K_{P}

for the reaction

N_{2} + {3 H}_{2} ⇌ {2 N H}_{3}

at 400C is

1.64 \times 10^{- 4}

. Find

K_{C}

. Also find

{△ G}^{\oplus}

using

K_{P}

and

K_{C}

values and interpret the differences.

Join BYJU'S Learning Program

KP for the reaction N2+3H2⇋2NH3 at 400∘C is 1.64×10−4. Calculate Kc.

The correct option is C 0.5 mole2 litre-2 N2+3H2⇌2NH3 Δn =-2~and Kp=Kc(RT)Δn 1.64×10−4=Kc(0.0821×673)−2 and Kc=0.5 mole2 litre−2

KP for the reaction $N_{2} + 3 H_{2} ⇋ 2 N H_{3}$ at $400^{\circ} C$ is $1.64 \times 10^{- 4}$ . Calculate $K_{c}$ .

The correct option is C
0.5 mole2 litre-2

$N_{2} + 3 H_{2} ⇌ 2 N H_{3}$

$Δ n$ =-2~and $K p = K c (R T)^{Δ n}$

$1.64 \times 10 - 4 = K c (0.0821 \times 673)^{- 2}$

and $K c = 0.5 m o l e^{2} l i t r e^{- 2}$