Dissociation of a gas takes place according to the following chemical reaction. At equilibrium, the total pressure is 1 bar at 300K. [2026]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Dissociation of a gas $A_{2}$ takes place according to the following chemical reaction. At equilibrium, the total pressure is 1 bar at 300K.

$A_{2} (g) ⇌ 2 A (g)$

The standard Gibbs energy of formation of the involved substances has been provided below:

Substance $Δ G_{f} ° / {kJ mol}^{- 1}$

$A_{2}$ - 100.00

$A$ - 50.832

The degree of dissociation of $A_{2} (g)$ is given by ${(x \times 10^{- 2})}^{1 / 2}$ where $x$ =_______. (Nearest integer).

[Given: $R = 8 {J mol}^{- 1} K^{- 1}, \log 2 = 0.3010, \log 3 = 0.48$ ]

Assume degree of dissociation is not negligible. [2026]

Ans.
(33)

$- 1.664 \times 10^{3} = - 8.3 \times 300 \ln K_{p}$

$\ln K_{p} = 0.693$

$K_{p} = 2$

$2 = \frac{4 α^{2} P_{0}}{1 - α^{2}}$

$α = \frac{1}{\sqrt{3}}$

$α = {(\frac{100}{3} \times 10^{- 2})}^{1 / 2}$

$= {(33.33 \times 10^{- 2})}^{1 / 2}$

Want Us to Email you About Special Offers & Updates?