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

Discussion Forum : Chemical Equilibrium

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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