Question

In the gas-phase reaction 2 A + B ⇌ 3 C + 2 D, it was found that, when 1.00 mol A, 2.00 mol B, and 1.00 mol D were mixed and allowed to come to equilibrium at 25 °C, the resulting mixture contained 0.90 mol C at a total pressure of 1.00 bar. Calculate
(i) the mole fractions of each species at equilibrium,
(ii) K, and
(iii) Δrimage.

Answer 1

Answer:

(i)The mole fractions are :

• $A=\frac{0.4}{4.3} \\=0.0930$

• $B=\frac{1.4}{4.3} \\=0.3256$

• $C=\frac{0.9}{4.3} \\=0.2093$

• $D=\frac{1.6}{4.3} \\=0.3721$

(ii)$K=0.4508$

(iii)ΔG = 1.974kJ

Explanation:

The given equation is :

$2A+B$⇄$3C+2D$

Let $\alpha$ be the number of moles dissociated per mole of $B$

Thus ,

The initial number of moles of :

• $A=1$

• $B=2$

• $C=0$

• $D=1$

$2A\\(1-2\alpha)$     +  $B\\2(1-\alpha)$  ⇄  $3C\\(3\alpha)$   + $2D\\(1+2\alpha)$

And finally the number of moles of $C[tex] is 0.9Thus ,[tex]3\alpha=0.9\\\alpha=0.3[tex]The final number of moles of:[tex]A = 1-2\alpha=1-2*0.3=0.4mol[tex] [tex]B=2(1-\alpha)=2(1-0.3)=1.4mol[tex][tex]D=1+2\alpha=1+2*0.3=1.6mol[tex]Thus , total number of moles are : 0.4+1.4+0.9+1.6=4.3(i)The mole fractions are : [tex]A=\frac{0.4}{4.3} \\=0.0930$

• $B=\frac{1.4}{4.3} \\=0.3256$

• $C=\frac{0.9}{4.3} \\=0.2093$

• $D=\frac{1.6}{4.3} \\=0.3721$

(ii)

$K=\frac{(P_C^3)(P_D^2)}{(P_A^2)(P_B)}$

Where ,

$P_A,P_B,P_C,P_D$ are the partial pressures of A,B,C,D respectively.

Total pressure = 1 bar .

∴

$P_A= 0.0930*1=0.0930$

$P_B= 0.3256*1=0.3256$

$P_C= 0.2093*1=0.2093$

$P_D= 0.3721*1=0.3721$

$K=\frac{0.2093^3*0.3721^2}{0.0930^2*0.3256} \\K=0.4508$

(iii)

Δ$G=-RTlnK$

ΔG = $-8.314*(273+25)*ln(0.4508)\\=1973.96J\\=1.974kJ$