SATHEE: Chemical and Ionic Equilibrium

Chemical and Ionic Equilibrium - Result Question 6

####6. Two solids dissociate as follows:

$\begin{aligned} A (s) ⇌ B (g) + C (g); K_{p_{1}} = x a t m^{2} & D (s) ⇌ C (g) + E (g); K_{p_{2}} = y a t m^{2} \end{aligned}$

The total pressure when both the solids dissociate simultaneously is

(2019 Main, 12 Jan I)

(a) $\sqrt{x + y}$ atm

(b) $x^{2} + y^{2}$ atm

(d) $2 (\sqrt{x + y}) a t m$

Show Answer

Answer:

Correct Answer: 6. (c)

Solution:

The equilibrium reaction for the dissociation of two solids is given as:

At equilibrium

$A (s) ⇌ B (g) + C (g)$

$K_{p_{1}} = x = p_{B} \cdot p_{C} = p_{1} (p_{1} + p_{2})$

Similarly, $D (s) ⇌ C (g) + E (g)$

At equilibrium

$I m_{p_{2}} = y = p_{C} \cdot p_{E} = (p_{1} + p_{2} \cdot p_{2}) p_{2}$

On adding Eq. (i) and (ii), we get.

$K_{p_{1}} + K_{p_{2}} = x + y = p_{1} (p_{1} + p_{2}) + p_{2} (p_{1} + p_{2})$

$= {(p_{1} + p_{2})}^{2}$

or $\sqrt{x + y} = p_{1} + p_{2}$

Now, total pressure is given as

$\begin{aligned} p_{T} & = p_{B} + p_{C} + p_{E} & = p_{1} + (p_{1} + p_{2}) + p_{2} & = 2 (p_{1} + p_{2}) \end{aligned}$

On substituting the value of $p_{1} + p_{2}$ from Eq. (iii) to Eq. (iv), we get

$p_{T} = 2 \sqrt{x + y}$

Chemical and Ionic Equilibrium - Result Question 6

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Chemical and Ionic Equilibrium - Result Question 6

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu