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Law of Mass Action

Consider the ...

Question

Consider the two reactions:
$X e F_{6} (g) + H_{2} O (g) ⇌ X e O F_{4} (g) + 2 H F (g); K_{1}$
$X e O_{4} (g) + X e F_{6} (g) ⇌ X e O F_{4} (g) + X e O_{3} F_{2} (g); K_{2}$
Then, the equilibrium constant for the following reaction will be:
$X e O_{4} (g) + + 2 H F (g) ⇌ X e O_{3} F_{2} (g) + H_{2} O (g)$ is given by

A

K_{1} / K_{2}^{2}

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B

K_{1} / K_{2}

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C

K_{1}^{2} / K_{2}

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D

K_{2} / K_{1}

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Solution

The correct option is D $K_{2} / K_{1}$
Consider the two reactions:
$X e F_{6} (g) + H_{2} O (g) ⇌ X e O F_{4} (g) + 2 H F (g); K_{1}$ ....(1)
$X e O_{4} (g) + X e F_{6} (g) ⇌ X e O F_{4} (g) + X e O_{3} F_{2} (g); K_{2}$ ....(2)
Then, the equilibrium constant for the following reaction will be:
$X e O_{4} (g) + + 2 H F (g) ⇌ X e O_{3} F_{2} (g) + H_{2} O (g)$ ....(3) is given by
$K_{2} / K_{1}$
When balanced equations (1) is subtracted from equation (2), equation (3) is obtained.
When balanced equations are subtracted, the equilibrium constant for the combined process is equal to the ratio of the equilibrium constants for each step.

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Similar questions

K_{1}

K_{2}

are the respective equilibrium constant for the reactions:
(i)

X e F_{6 (g)} + H_{2} O_{(g)} ⇌ X e O F_{4 (g)} + 2 H F_{(g)}; K_{1}

X e O_{4 (g)} + X e F_{6 (g)} ⇌ X e O F_{4 (g)} + X e O_{1} F_{2}; K_{2}

The equilibrium constant for the reaction

X e O_{4 (g)} + 2 H F_{(g)} ⇌ X e O_{3} F_{2 (g)} + H_{2} O_{(g)}

K_{1}

K_{2}

are the respective equilibrium constants for the two reactions

X e F_{6} (g) + H_{2} O (g) ⇌ X e O F_{4} (g) + 2 H F (g)

X e O_{4} (g) + X e F_{6} (g) ⇌ X e O F_{4} (g) + X e O_{3} F_{2} (g)

the equilibrium constant of the following reaction is:

X e O_{4} (g) + 2 H F (g) ⇌ X e O_{3} F_{2} (g) + H_{2} O (g)

K_{1}

K_{2}

are the respective equilibrium constants for the two reactions:

X e F_{6} (g) + H_{2} O (g) \Leftrightarrow X e O F_{4} (g) + 2 H F (g)

X e O_{4} (g) + X e F_{6} (g) \Leftrightarrow X e O F_{4} (g) + X e O_{3} F_{2} (g)

The equilibrium constant of the reaction

X e O_{4} (g) + 2 H F (g) \Leftrightarrow X e O_{3} F_{2} (g) + H_{2} O (g)

K_{1}

K_{2}

are the respective equilibrium constants for the two reactions:
(i)

X e F_{6} (g) + H_{2} O (g) ⇌ X e O F_{4} (g) + 2 H F (g)

X e O_{4} (g) + X e F_{6} (g) ⇌ X e O F_{4} (g) + X e O_{3} F_{2} (g)

then the equilibrium constant for the reaction:

X e O_{4} (g) + 2 H F (g) ⇌ X e O_{3} F_{2} (g) + H_{2} O (g)

Equilibrium constants $K_{1}$ and $K_{2}$ for the following equilibria are

$N O (g) + 0.5 O_{2} (g) \to K_{1} N O_{2} (g)$ and
$2 N O_{2} (g) \to K_{2} 2 N O (g) + O_{2} (g)$

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