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

(i)The mole fractions are :

• 
  `A=(0.4)/(4.3) \\=0.0930`

• 
  `B=(1.4)/(4.3) \\=0.3256`

• 
  `C=(0.9)/(4.3) \\=0.2093`

• 
  `D=(1.6)/(4.3) \\=0.3721`

(ii)
`K=0.4508`

(iii)ΔG = 1.974kJ

Step-by-step 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 0.9</p><p>Thus ,</p><p>[tex]3\alpha=0.9\\\alpha=0.3[tex]</p><p><em><strong>The final number of moles of:</strong></em></p><ul><li><em><strong>[tex]A = 1-2\alpha=1-2*0.3=0.4mol[tex] </strong></em></li></ul><ul><li><em><strong>[tex]B=2(1-\alpha)=2(1-0.3)=1.4mol[tex]</strong></em></li></ul><ul><li><em><strong>[tex]D=1+2\alpha=1+2*0.3=1.6mol[tex]</strong></em></li></ul><p>Thus , total number of moles are : 0.4+1.4+0.9+1.6=4.3</p><p><strong>(i)The mole fractions are : </strong></p><ul><li><strong>[tex]A=(0.4)/(4.3) \\=0.0930`

• 
  `B=(1.4)/(4.3) \\=0.3256`

• 
  `C=(0.9)/(4.3) \\=0.2093`

• 
  `D=(1.6)/(4.3) \\=0.3721`

(ii)

`K=((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=(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`