SATHEE: Chemical and Ionic Equilibrium

Chemical and Ionic Equilibrium - Result Question 54

####53. One mole of $N_{2}$ and 3 moles of $P C l_{5}$ are placed in a $100 L$ vessel heated to $227^{\circ} C$ . The equilibrium pressure is 2.05 atm. Assuming ideal behaviour, calculate the degree of dissociation for $P C l_{5}$ and $K_{p}$ for the reaction,

$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$

$(1984, 6 M)$

Topic 2 Ionic Equilibrium

Objective Questions I (Only one correct option)

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Solution:

Total moles of gases at equilibrium $= \frac{p V}{R T} = \frac{2.05 \times 100}{0.082 \times 500} = 5.0$

Out of this 5 moles, 1.0 mole is for $N_{2} (g)$ and remaining 4 moles for $P C l_{5}$ and its dissociation products.

$\begin{matrix} P C l_{5} ⇌ \underset{x}{P C l_{3}} + \underset{x}{C l_{2}} 3 + x = 4 \Rightarrow x = 1 \end{matrix}$

Degree of dissociation $= \frac{1}{3} = 0.33$

$\begin{array}{lllc} N_{2} + & 3 H_{2} & ⇌ 2 N H_{3} Initial : & 1.0 & 3.0 & 0 Equilibrium & 1 - 0.25 & 3 - 0.75 & 0.05 & = 0.75 & = 2.25 [N_{2}] = & \frac{0.75}{4}, [H_{2}] = & \frac{2.25}{4}, [N H_{3}] = \frac{0.50}{4} \end{array}$

$\begin{matrix} K_{c} = \frac{{[N H_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}} = \frac{(0.50)^{2}}{(0.75) (2.25)^{3}} \times 16 = 0.468 L^{2} m o l^{- 2} Also for : \frac{1}{2} N_{2} + \frac{3}{2} H_{2} ⇌ N H_{3} K_{c}^{'} = \sqrt{K_{c}} = 0.68 \end{matrix}$

Chemical and Ionic Equilibrium - Result Question 54

Topic 2 Ionic Equilibrium

Objective Questions I (Only one correct option)

Solution:

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Chemical and Ionic Equilibrium - Result Question 54

Topic 2 Ionic Equilibrium

Objective Questions I (Only one correct option)

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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