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Relation between Kp and Kc

For the react...

Question

For the reaction
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g); △ H = - 46 k J m o l^{- 1}$
Calculate the value of $K_{p} .$

Given $K_{C} = 0.5 l i t^{2} m o l^{- 2}$ and $R = 0.082 L a t m m o l^{- 1} K^{- 1}$ at $400^{o} C$

A

1.64 \times 10^{- 4}

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B

1.34 \times 10^{- 4}

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C

2.64 \times 10^{- 4}

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D

0.64 \times 10^{- 4}

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Solution

The correct option is A $1.64 \times 10^{- 4}$
$K_{P} = K_{C} (R T)^{△ n}$
$△ n = 2 - 4 = - 2$
putting the values,
$K_{P} = 0.5 (0.082 \times 673)^{- 2} = 1.64 \times 10^{- 4}$

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Similar questions

Q. For the reaction

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g); △ H = - 46 k J m o l^{- 1}

Calculate the value of

K_{p} .

K_{C} = 0.5 l i t^{2} m o l^{- 2}

R = 0.082 L a t m m o l^{- 1} K^{- 1}

400^{o} C

K_{p}

for the reaction

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

298 K

5.2 \times 10^{- 5}

. What is the value of

K_{c}

298 K

, for the reaction

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

Q. The equilibrium constant

K_{p}

for the reaction

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

1.64 \times 10^{- 4}

400^{o} C

0.144 \times 10^{- 4}

500^{o} C

. Calculate the mean heat of formation of

1

{N H}_{3}

from its elements in this temperature range.

K_{p}

for a reaction at

25^{o} C

in 10 atm is 8.568. Then

K_{c}

for the reaction at

25^{o} C

N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)

Q. Calculate the equilibrium constant

(K c)

for the formation of

N H_{3}

in the following reaction:

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

At equilibrium, the concentration of

N H_{3}, H_{2}

N_{2}

1.2 \times 10^{- 2}, 3.0 \times 10^{- 2}

1.5 \times 10^{- 2} M

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