Question

Consider a reaction: $A_{\left(g\right)}+B_{\left(g\right)}\rightleftharpoons C_{\left(g\right)}+D_{\left(g\right)}$

$A_{\left(g\right)},B_{\left(g\right)}$ and $C_{\left(g\right)}$ are taken in a container at 1 bar partial pressure each and adequate amount of liquid 'D' is added.

From the data given below. Calculate $a+b+c+d$

Given :

$\Delta G_f^0{A}_{\left(g\right)}=$30 kJ\ mole${}^{-1}$; $\Delta G_f^0{B}_{\left(g\right)}=$ 20 kJ mole${}^{-1}$;

$\Delta G_f^0{C}_{\left(g\right)}=$ 50 kJ mole${}^{-1}$; $\Delta G_f^0{D}_{\left(g\right)}=$ 100 kJ mole${}^{-1}$;

Vapour pressure of $D_{\left(l\right)}$ at 300 $K=\frac{1}{6}$ bar [All data at 300 K]
where
a = Equilibrium constant of the given reaction
b = Twice the partial pressure of A at equilibrium
c = Twice the partial pressure of B at equilibrium
d = Twice the partial pressure of C at equilibrium

• A. 6
• B. 5
• C. 7
• D. 8

Answer 1

The correct option is A 6

$A\left(g\right)+B\left(g\right)\rightleftharpoons C\left(g\right)+D$

$=\Delta G_{reaction}^0=\left(\Delta G_{fD}^0\right)-(\Delta G_{fA}^0+\Delta G_{fB}^0)$

$=(-50kJ+100kJ)-(30kJ+20kJ)$

$\Delta G^0=0$

$\Delta G^0=-RT/In{k}_{eq}$

$k_{eq}=1$

$D\left(I\right)\rightleftharpoons D\left(g\right)$ [Partial pressure of D gas = Vapour pressure]

$A\left(g\right)+B\left(g\right)\rightleftharpoons C\left(g\right)+D\left(g\right)$

1 bar 1 bar 1 bar 1/6 bar
(1-x) bar (1-x) bar (1+x) bar (1/6) bar

$K_{eq}=1=\frac{\frac{1}{6}\times [1+X]}{(1-X{)}^2}$

$1+x^2-2x=\frac{1}{6}+\frac{x}{6}$

$x^2-2x-\frac{x}{6}+\frac{5}{6}=0$

$x^2-\frac{13x}{6}+\frac{5}{6}=0$

$6x^2-13x+5=0$

$x=\frac{1}{2}$

$P_A=\frac{1}{2}{P}_B=\frac{1}{2}{P}_C=\frac{3}{2}$

where,

$P_A=$ partial pressure of A at equilibrium

$P_A=$ partial pressure of B at equilibrium

$P_A=$ partial pressure of C at equilibrium

Given that,

a = Equilibrium constant of the given reaction

b = Twice the partial pressure of A at equilibrium

c = Twice the partial pressure of B at equilibrium

d = Twice the partial pressure of C at equilibrium

$\therefore$ a=1 b=1 c=1 d=3

$\therefore a+b+c+d=$ 1+1+1+3 $=6$

Hence, the correct option is $A$