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Standard XII

Chemistry

Relation between Kp and Kc

At a temperat...

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?

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

0.1 atm

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8.21 atm

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831.4 atm

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8.314 atm

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Solution

The correct option is B 8.21 atm
$N H_{3} (g) ⇋ \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) K_{c} = K_{c}^{'}$

$K_{c}^{'} = (\frac{1}{100})^{\frac{1}{2}} = 0.1$
$K_{p} = K_{c}^{'} \times (R T)^{△ n} = 0.1 \times (0.0821 \times 1000)^{1} = 8.21 a t m$

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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Similar questions

Q. 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)

Q. The equilibrium constant

K_{c}

for the reaction

N_{2 (g)} + 3 H_{2} ⇌ {2 N H}_{3 (g)}

0.286

500^{o} C

. Calculate

K_{c}

and

K_{p}

for the reaction

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

Q. Assertion :For reaction

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

Unit of

K_{C} = L^{2} {m o l}^{- 2}

Reason: For the reaction

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

, Equilibrium constant

K_{C} = \frac{{[{N H}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}

Q. For the reaction;

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

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

. Find the value of

K_{p}

for each of the following reactions at the same temperature:

(i)

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

;

(ii)

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

;

(iii)

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

Q. One mole of

N_{2} (g)

is mixed with 2 moles of

H_{2} (g)

in a 4 litre vessel. If

50

% of

N_{2} (g)

is converted to

H_{2} (g)

by the following reaction :

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

What will be the value of

K_{c}

for the following equilibrium?

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

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At a temperature of 1000 K, equilibrium constant Kc for the following reaction is 100 mol2L−2. N2(g)+3H2(g) ⇋ 2NH3(g) What will be the value of Kp for the reaction below? NH3(g) ⇋ 12 N2(g)+32 H2(g)

The correct option is B 8.21 atmNH3(g) ⇋ 12 N2(g)+32 H2(g)Kc=K′c K′c=(1100)12=0.1 Kp=K′c×(RT)△ n=0.1×(0.0821×1000)1=8.21 atm Notes: Use R=0.0821 atm l mol−1K−1 as unit of concentration has been taken as mol/l while calculating Kc.

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?

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