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Arrhenius Equation

Calculate 2...

Question

Calculate $2 \times K_{p}$ for,
$N H_{3} ⇌ 1 / 2 N_{2} + 3 / 2 H_{2}$
(Given : $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ ; $0.5 {l i t r e}^{2} {m o l}^{- 2}$ at $400$ .) (units in atm)

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Solution

$K_{c} = \frac{1}{\sqrt{0.5}} = 1.414 m o l / L$

$Δ n = 1.5 + 0.5 - 1 = 1$

$K_{p} = K_{c} (R T)^{Δ n} = 1.41 \times (0.0821 \times 400)^{1} = 46.5 a t m$

$∴ 2 \times K = 2 \times 46.5 = 93$

Hence, the correct option is $93$

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Similar questions

Q. In a mixture of

N_{2}

H_{2}

in the ratio 1:3 at 30 atm and

300^{0} C

{N H}_{3}

at equilibrium is 17.8.
Calculate

K_{P}

{N_{2} + {3 H}_{2} ⇌ 2 N H}_{3}

.

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is the equilibrium constant for the reaction

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Q. For the reaction

N_{2} + 3 H_{2} ⇌ 2 {N H}_{3}

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

write down the expression for equilibrium constant

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400^{\circ} C

1.64 \times 10^{- 4}

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N_{2} + 3 H_{2} ⇌ 2 N H_{3}

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

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