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Arrhenius Equation

K c for the r...

Question

$K_{c}$ for the reaction ; $N_{2} + 3 H_{2} ⇌ 2 {N H}_{3}$ is 0.5 ${m o l}^{- 2} L^{2}$ at 400 K. Find $K_{P}$ .
[Given that $R = 0.082 L - a t m K^{- 1} {m o l}^{- 1}$ ]

K_{P} = 4.648 \times 10^{- 4} {(a t m)}^{- 2}

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K_{P} = 46.36 \times 10^{- 5} {(a t m)}^{- 2}

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K_{P} = 8.12 \times 10^{- 5} {(a t m)}^{- 2}

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K_{P} = 2.190 \times 10^{- 3} {(a t m)}^{- 2}

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Solution

The correct options are
A $K_{P} = 4.648 \times 10^{- 4} {(a t m)}^{- 2}$
B $K_{P} = 46.36 \times 10^{- 5} {(a t m)}^{- 2}$
Given that;

$N_{2} + 3 H_{2} ⇌ 2 {N H}_{3}$

$K_{c} = 0.5 M^{- 2}$

$Δ n_{g} = 2 - (1 + 3) = - 2$

$K_{P} = K_{C} {(R T)}^{△ n_{g}}$
$∴ K_{P} = 0.5 \times {(0.082 \times 400)}^{- 2}$
$K_{P} = 4.648 \times 10^{- 4} {(a t m)}^{- 2}$

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