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

Introduction

If the equili...

Question

If the equilibrium constant for the reaction:
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)$
at $750 K$ is $49$ , then the equilibrium constant for the reaction,
${N H}_{3} (g) ⇌ \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g)$
at the same temperature is:

1 / 49

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

49

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

7

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

49^{2}

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

1 / 7

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

Open in App

Solution

The correct option is E $1 / 7$
If the equilibrium constant for the reaction:
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)$ .....(1)
at $750 K$ is $49$ , then the equilibrium constant for the reaction,
${N H}_{3} (g) ⇌ \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g)$ .....(2)
at the same temperature is $1 / 7$ .
When reaction (1) is reversed and divided by 2, the reaction (2) is obtained.
$K_{2} = \sqrt{\frac{1}{K_{1}}}$
$K_{2} = \sqrt{\frac{1}{49}}$
$K_{2} = \frac{1}{7}$

Suggest Corrections

Similar questions

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

\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) ⇌ {N H}_{3} (g)

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

K

\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) ⇌ {N H}_{3} (g)

\sqrt{K} .

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

400 K, K_{p} = 41 a t m^{- 2}

K_{p}

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

\frac{1}{2} N_{2 (g)} + \frac{3}{2} H_{2 (g)} ⇌ N H_{3 (g)}

2 N_{2 (g)} + 6 H_{2 (g)} ⇌ 4 N H_{3 (g)}

H_{2} + I_{2} ⇌ 2 H I

700 K

49

H I ⇌ \frac{1}{2} H_{2} + \frac{1}{2} I_{2}

527^{0}

N H_{3} (g) ⇌ \frac{1}{2} N_{2} (g)

\frac{3}{2} H_{2} (g)

K_{c} = 4

K_{c}

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

Join BYJU'S Learning Program