One mole of A(g) is allowed to… | JEE Advanced 2018 P2

One mole of A(g) is allowed to dissociate as A(g)rightleftharpoons 2B(g) at 400 K and constant total pressure. The equilibrium constant is Kp=0.50…

One or more options correct

One mole of

\mathrm{A(g)}

is allowed to dissociate as

\mathrm{A(g)\rightleftharpoons 2B(g)}

400\ \mathrm{K}

and constant total pressure. The equilibrium constant is

K_p=0.50

(standard pressure

1\ \mathrm{bar}

, treat

\mathrm{bar}=\mathrm{atm}

). Take

R=8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}

. The correct statement(s) is/are

A. At a total pressure of $1$ bar the degree of dissociation of $\mathrm{A}$ is $1/3$
B. When the total pressure is raised to $4$ bar the degree of dissociation becomes exactly half its value at $1$ bar
C. The standard Gibbs energy change at $400\ \mathrm{K}$ is approximately $+2.3\ \mathrm{kJ\,mol^{-1}}$
D. At $1$ bar the average molar mass of the equilibrium mixture is $2/3$ of the molar mass of $\mathrm{A}$

Show answer and solution

Answer: A, C

For

\mathrm{A\rightleftharpoons 2B}

with degree of dissociation

\alpha

at total pressure

P

K_p=\dfrac{4\alpha^2P}{1-\alpha^2}

. At

P=1

4\alpha^2/(1-\alpha^2)=0.5

gives

\alpha^2=1/9

, so

\alpha=1/3

: (A) is correct. At

P=4

16\alpha^2/(1-\alpha^2)=0.5

gives

\alpha^2=0.5/16.5

\alpha=0.174

, which is

0.52

1/3

, not exactly half (the

1-\alpha^2

term spoils the simple

P^{-1/2}

scaling): (B) is wrong.

\Delta G^\circ=-RT\ln K=-8.314\times400\times\ln0.5=+2305\ \mathrm{J\,mol^{-1}}\approx+2.3\ \mathrm{kJ\,mol^{-1}}

: (C) is correct. Moles at equilibrium are

1+\alpha=4/3

for the same mass, so the mean molar mass is

M_A/(1+\alpha)=\tfrac34M_A

, not

\tfrac23M_A

: (D) is wrong. Answer: A, C.

Original question written in the JEE Advanced 2018 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake