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Derivation of Kp and Kc

At 25oC, ...

Question

At $25^{o} C$ , $K_{p}$ for the reaction:
$N_{2} O_{4} (g) ⇋ 2 N O_{2} (g)$
has a value of $0.14$ atm. Calculate the value of $K_{c}$ in which the concentrations are measured in $m o l L^{- 1}$

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Solution

$K - p = K_{c} . R t^{△ n}$
$△ n =$ Moles of gaseous product-moles of gaseous reactant.
$= 2 - 1 = 1$
$∴ K_{p} = K_{c} . R T$
$∴ K_{c} = \frac{K_{p}}{R T} = \frac{0.14}{0.0821 \times 298} = 5.72 \times 10^{- 3} m o l L^{- 1}$

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300 K, K_{p}

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N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

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K_{c}

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Q. The value of

\frac{K_{p}}{K_{c}}

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are, respectively:

N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

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[A t 300 K, R T = 24.62 d m^{3} a t m m o l^{- 1}]

Q. For a reaction equilibrium,

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

, the concentrations of

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4.8 \times 10^{- 2}

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the concentrations of

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