SATHEE: Chemical and Ionic Equilibrium

Chemical and Ionic Equilibrium - Result Question 18

####18. Consider the following equilibrium in a closed container

$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$

At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statements hold true regarding the equilibrium constant $(K_{p})$ and degree of dissociation $(α)$ ?

$(2002, 3 M)$

(a) Neither $K_{p}$ nor $α$ changes

(b) Both $K_{p}$ and $α$ change

(d) $K_{p}$ does not change but $α$ changes

Show Answer

Solution:

$\begin{aligned} N_{2} O_{4} (g) ⇌ \underset{2}{2} ⇌ \underset{2}{2} N O_{2} (g) \begin{matrix} Total 1 + α \end{matrix} p_{i} : \frac{1 - α}{1 + α} p & \frac{2 α}{1 + α} p K_{p} = \frac{4 α^{2}}{1 - α^{2}} p \end{aligned}$

At constant temperature, halving the volume will change both $p$ and $α$ but $K_{p}$ remains constant.

Chemical and Ionic Equilibrium - Result Question 18

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Chemical and Ionic Equilibrium - Result Question 18

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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