SATHEE: Equilibrium Question 762

Equilibrium Question 762

Question: If $K_{1}$ and $K_{2}$ are respective equilibrium constants for the two reactions $X e F_{6} (g) + H_{2} O (g) \to X e O F_{4} (g) + 2 H F (g)$

$X e O_{4} (g) + X e F_{6} (g) \to X e O F_{4} (g) + X e O_{3} F_{2} (g)$ the equilibrium constant for the reaction $X e O_{4} (g) + 2 H F (g) \to X e O_{3} F_{2} (g) + H_{2} O (g)$ will be

Options:

A) $\frac{K_{1}}{K_{2}^{2}}$

B) $K_{1} . K_{2}$

C) $\frac{K_{1}}{K_{2}}$

D) $\frac{K_{2}}{K_{1}}$

Show Answer

Answer:

Correct Answer: D

Solution:

$K = K_{2} \times \frac{1}{K_{1}} = \frac{K_{2}}{K_{1}} .$

Equilibrium Question 762

Question: If $K_{1}$ and $K_{2}$ are respective equilibrium constants for the two reactions $X e F_{6} (g) + H_{2} O (g) \to X e O F_{4} (g) + 2 H F (g)$

Options:

Answer:

Solution:

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Equilibrium Question 762

Question: If K1 and K2 are respective equilibrium constants for the two reactions XeF6(g)+H2O(g)→XeOF4(g)+2HF(g)

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: If $K_{1}$ and $K_{2}$ are respective equilibrium constants for the two reactions $X e F_{6} (g) + H_{2} O (g) \to X e O F_{4} (g) + 2 H F (g)$