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Derivation of Kp and Kc

The value of ...

Question

The value of $\frac{K_{p}}{K_{c}}$ for the following reactions at $300 K$ are, respectively:

$N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)$

$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$

$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$
$[A t 300 K, R T = 24.62 d m^{3} a t m m o l^{- 1}]$

1, 24.61 d m^{3} a t m m o l^{- 1}, 606.0 d m^{6} a t m^{2} m o l^{- 2}

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1, 4.1 \times 10^{- 2} d m^{- 3} s m^{- 1} m o l^{- 1}, 606.0 d m^{6} a t m^{2} m o l^{- 2}

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606.0 d m^{6} a t n^{6} m o l^{- 2}, 1.65 \times 10^{- 3} d m^{3} a t m^{- 2} m o l^{- 1}

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1, 24.62 d m^{3} a t m m o l^{- 1}, 1.65 \times 10^{- 3} d m^{- 6} a t m^{- 2} m o l^{2}

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Solution

The correct option is B $1, 24.62 d m^{3} a t m m o l^{- 1}, 1.65 \times 10^{- 3} d m^{- 6} a t m^{- 2} m o l^{2}$
$N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)$
$\frac{k_{p}}{k_{c}} = (R T)^{Δ n_{g}} = (R T)^{o} = 1$
$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$
$\frac{k_{p}}{k_{c}} = (R T)^{1} = 24.62 d m^{3} a t m m o l^{- 1}$
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$
$\frac{k_{p}}{k_{c}} = (R T)^{- 2} = \frac{1}{(R T)^{2}} = 1.65 \times 10^{- 3} d m^{- 6} a t m^{- 2} m o l^{2}$

Hence, the correct option is D.

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