SATHEE: Chemical and Ionic Equilibrium

Chemical and Ionic Equilibrium - Result Question 55

####55. The equilibrium constant $K_{p}$ for this reaction at $298 K$ , in terms of $β_{equilibrium}$ is

(2016 Adv.)

(a) $\frac{8 β^{2}_{equilibrium}}{2 - β_{equilibrium}}$

(b) $\frac{8 β^{2}_{equilibrium}}{4 - β_{equilibrium}^{2}}$

(d) $\frac{4 β^{2}_{equilibrium}}{4 - β^{2}_{equilibrium}}$

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Solution:

$\begin{aligned} X_{2} (g) ⇌ 2 X (g) & At \begin{array}{lll} t = 0 & 1 & 0 \end{array} & At equilibrium (1 - \frac{x}{2}) x (where, x = β_{e q}) & Total moles = (1 + \frac{x}{2}) and Mole fraction, X_{2} (g) = \frac{(1 - \frac{x}{2})}{(1 + \frac{x}{2})} & X (g) = (\frac{x}{1 + \frac{x}{2}}) and p = 2 bar & Partial pressure, p_{X 2} = (\frac{1 - \frac{x}{2}}{1 + \frac{x}{2}}) \cdot p and p_{X} = \frac{p \cdot x}{(1 + \frac{x}{2})} & \begin{matrix} ∴ K_{p} = p_{X}^{2} / p_{X_{2}} = \frac{{[p x / (1 + \frac{x}{2})]}^{2}}{p \frac{(1 - x / 2)}{(1 + \frac{x}{2})}} = \frac{4 p x^{2}}{(4 - x^{2})} = \frac{8 β_{e q}^{2}}{(4 - β_{e q}^{2})} \end{matrix} \end{aligned}$

Chemical and Ionic Equilibrium - Result Question 55

Solution:

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Chemical and Ionic Equilibrium - Result Question 55

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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