At a temperature of 1000 K, equilibrium constant Kc for the fol

Question

At a temperature of

1000 K

, equilibrium constant

K_{c}

for the following reaction is

100 m o l^{2} L^{- 2}

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

What will be the value of

K_{p}

for the reaction below?

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

Options:

A . 0.1 atm

B . 8.21 atm

C . 831.4 atm

D . 8.314 atm

Answer: Option B
:
B

N H_{3} (g) ⇋ \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) K_{c} = K_{c}^{'}

K_{c}^{'} = (\frac{1}{100})^{\frac{1}{2}} = 0.1

K_{p} = K_{c}^{'} \times (R T)^{△ n} = 0.1 \times (0.0821 \times 1000)^{1} = 8.21 a t m

Notes: Use

R = 0.0821 a t m l m o l^{- 1} K^{- 1}

as unit of concentration has been taken as

m o l / l

while calculating

K_{c} .

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