SATHEE: Equilibrium - Result Question 9

Equilibrium - Result Question 9

####10. If the concentration of $O H^{-}$ ions in the reaction $F e (O H)_{3} (s) \to F e^{3 +} (a q) + 3 O H^{-} (a q)$ is decreased by $\frac{1}{4}$ times, then equilibrium concentration of $F e^{3 +}$ will increase by : [2008]

(a) 8 times

(b) 16 times

(d) 4 times

Show Answer

Solution:

$K = \frac{[F e^{3 +}] [O H^{-}]^{3}}{[F e (O H)_{3}]}$

$. = (F e^{3 +}) (O H^{-})^{3} [∴ solid] = 1] .$

If $(O H^{-})$ is decreased by $\frac{1}{4}$ times then for given reaction, equilibrium constant to remain constant, we have to increase the concentration of $[F e^{3 +}]$ by a factor of $4^{3}$ i.e. $4 \times 4 \times 4 = 64$ . Thus, option (c) is correct answer.

To maintain the equilibrium constant if concentration of one of the product decreases by $\frac{1}{n}$ times then the concentration of another product is increased by the factor of $(n)^{x}$ , where $x$ is the stoichiometric coeficient.

Equilibrium - Result Question 9

Solution:

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Equilibrium - Result Question 9

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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