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3 years ago

Consider the reaction:

Chemistry

1 answer:

goldfiish [28.3K]3 years ago

5 0

Answer:

K = Ka/Kb

Explanation:

P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?

P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka

PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb

K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)

Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)

Kb = [PCl₅]/ ([PCl₃] [Cl₂])

Since [PCl₅] = [PCl₅]

From the Ka equation,

[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)

From the Kb equation

[PCl₅] = Kb ([PCl₃] [Cl₂])

Equating them

Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])

(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)

(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)

Comparing this with the equation for the overall equilibrium constant

K = Ka/Kb

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3 years ago

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Answer:

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Explanation:

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The partial pressure of each gas is:

$P_{NO_{2}}=\frac{n_{NO_{2}*R*T_{2}}}{V}$

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so total pressure is:

$P_{total}=(n_{NO_{2}}+n_{NO})*\frac{R*T_{2}}{V}$

replacing the moles we get:

$P_{total}=(2*\frac{P_{1}V}{RT_{1}})*\frac{R*T_{2}}{V}$

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replacing this value in the equation we get:

$P_{total}=2*\frac{P_{1}V}{RT_{1}}*\frac{R*3T_{1}}{V}$

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