Question

The equilibrium between gaseous isomers A, B and C can be represented as

A(g) ⇌ B(g)	$K_1$=?
B(g)⇌ C(g)	$K_2=0.4$
C(g) ⇌ A(g)	$K_3$ = 0.6

If one mole of A is taken in a closed vessel of volume 1 litre, then

• A. [A] + [B] + [C] = 1 M at any time of the reactions
• B. Concentration of C is 4.1 M at the attainment of equilibrium in all the reactions
  
• C. The value of $K_1$ is $\frac{1}{0.24}$
  
• D. Isomer [A] is least stable as per thermodynamics

Answer 1

Answer: (A), (C), (D)

The correct options are
A [A] + [B] + [C] = 1 M at any time of the reactions
C The value of $K_1$ is $\frac{1}{0.24}$

D Isomer [A] is least stable as per thermodynamics

$A\left(g\right)\rightleftharpoons B\left(g\right)$
$1-x+zx-y$

$B\left(g\right)\rightleftharpoons C\left(g\right)$
$x-yy-z$

$C\left(g\right)\rightleftharpoons A\left(g\right)$
$y-z1-x+z$

$\frac{x-y}{1-x+z}=\frac{1}{0.24}$

$0.24x-0.24y=1-x+z$

$1.24x-0.24y-z=\mathrm{1...}\left(i\right)$

$\frac{y-z}{x-y}=0.4$

$y-z=0.4x-0.4y$

$0.4x-1.4y+z=\mathrm{0...}\left(ii\right)$

$0.4x-1.4y+z=0$

$\frac{1-x+z}{y-z}=0.6$

$1-x+z=0.6y-0.6z$

$x-0.6y-1.6z=\mathrm{1...}\left(iii\right)$

on adding (i) and (ii)

$1.64x-1.64y=\mathrm{1...}\left(iv\right)$

$0.64x-2.24y+1.62=0$

$\left(ii\right)\times 1.6+\left(iii\right)gives1.64x-2.84y=\mathrm{0....}\left(v\right)$

$1.64x+1.64y=\mathrm{1....}\left(iv\right)1.2y=1$

On solving,

$y=0.833$

$x=1.43$

$z=0.59$

$\left[A\right]=0.16$

$[B=0.6$

$\left[C\right]=0.24$

A) Hence,
[A] + [B] + [C] = 1 M at any time of the reactions. since the reaction does not involve any association or dissociation. the total number of moles is equal to the initial number of moles at any time.

B) $\left[C\right]=0.24$

C) $K_1=\frac{1}{K_2{K}_3}=\frac{1}{0.6\times 0.4}=\frac{1}{0.24}=4.16$

D) $K_1=4.16$ and it is larger than $K_2$ and $K_3$. Hence $\left[A\right]$ is the least stable.