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Derivation of Kp and Kc

A mixture of ...

Question

A mixture of 0.373 atm of NO(g) and 0.31 atm of $C l_{2} (g)$ is prepared at $500 K$ . The reaction $2 N O (g) + C l_{2} (g) ⇌ 2 N O C l (g)$ takes place. The total pressure at equilibrium is 0.544 atm. Determine $K_{p}$ for the reaction.

50.08 a t m^{- 1}

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10.08 a t m^{- 1}

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50.07 a t m^{- 1}

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None of the above

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Solution

The correct option is C $50.07 a t m^{- 1}$
The initial pressure are
$P_{N O} = 0.373 a t m$
$P C l_{2} = 0.310 a t m$
$P_{N O C l} = 0 a t m$

The equilibrium pressure are
$P_{N O} = 0.373 - 2 x a t m$
$P_{C l_{2}} = 0.310 - x a t m$
$P_{N O C l} = 2 x a t m$

Total equilibrium pressure $= (0.373 - 2 x) + (0.310 - x) + (2 x) a t m .$ But it is equal to 0.544 atm.
$= (0.373) + (0.310 - x) = 0.544$
$x = 0.139 a t m$

The equilibrium pressure constant
$K_{p} = \frac{P_{N O C l}^{2}}{P_{N O} \times P_{C l_{2}}}$
$K_{p} = \frac{(2 x)^{2}}{(0.373 - 2 x)^{2} \times (0.310 - x)}$
$K_{p} = \frac{(2 (0.139))^{2}}{(0.373 - 2 (0.139))^{2} \times (0.310 - 0.139)}$

$K_{p} = 50.07 a t m^{- 1}$

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Similar questions

C l_{2} (g)

500 K

2 N O (g) + C l_{2} (g) ⇌ 2 N O C l (g)

K_{p}

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N O (g)

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{C l}_{2} (g)

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2 N O (g) + {C l}_{2} ⇌ 2 N O C l

0.544 a t m

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P C l_{5}

P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)

K_{p} = 1.78 a t m

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P C l_{5} ⇌ P C l_{3} (g) + C l_{2} (g) .

523 K, K_{p} = 1.78 a t m .

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{S O}_{2} {C l}_{2} (g) ⇌ {S O}_{2} (g) + {C l}_{2} (g)

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