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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 $400^{o} C$

1.64 \times 10^{- 4}

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1.34 \times 10^{- 4}

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2.64 \times 10^{- 4}

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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

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

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

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

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

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

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

K_{c}

N_{2} + 3 H_{2} ⇌ 2 {N H}_{3}

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

K_{P}

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

K_{c}

N_{2 (g)} + 3 H_{2} ⇌ {2 N H}_{3 (g)}

0.286

500^{o} C

K_{c}

K_{p}

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

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)

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