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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.

Respuesta :

Answer:

(i)The mole fractions are :

  • [tex]A=\frac{0.4}{4.3} \\=0.0930[/tex]
  • [tex]B=\frac{1.4}{4.3} \\=0.3256[/tex]
  • [tex]C=\frac{0.9}{4.3} \\=0.2093[/tex]
  • [tex]D=\frac{1.6}{4.3} \\=0.3721[/tex]

(ii)[tex]K=0.4508[/tex]

(iii)ΔG = 1.974kJ

Explanation:

The given equation is :

[tex]2A+B[/tex]⇄[tex]3C+2D[/tex]

Let [tex]\alpha[/tex] be the number of moles dissociated per mole of [tex]B[/tex]

Thus ,

The initial number of moles of :

  • [tex]A=1[/tex]
  • [tex]B=2[/tex]
  • [tex]C=0[/tex]
  • [tex]D=1[/tex]

[tex]2A\\(1-2\alpha)[/tex]     +  [tex]B\\2(1-\alpha)[/tex]  ⇄  [tex]3C\\(3\alpha)[/tex]   + [tex]2D\\(1+2\alpha)[/tex]

And finally the number of moles of [tex]C[tex] is 0.9

Thus ,

[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[/tex]
  • [tex]B=\frac{1.4}{4.3} \\=0.3256[/tex]
  • [tex]C=\frac{0.9}{4.3} \\=0.2093[/tex]
  • [tex]D=\frac{1.6}{4.3} \\=0.3721[/tex]

(ii)

[tex]K=\frac{(P_C^3)(P_D^2)}{(P_A^2)(P_B)}[/tex]

Where ,

[tex]P_A,P_B,P_C,P_D[/tex] are the partial pressures of A,B,C,D respectively.

Total pressure = 1 bar .

∴

[tex]P_A= 0.0930*1=0.0930[/tex]

[tex]P_B= 0.3256*1=0.3256[/tex]

[tex]P_C= 0.2093*1=0.2093[/tex]

[tex]P_D= 0.3721*1=0.3721[/tex]

[tex]K=\frac{0.2093^3*0.3721^2}{0.0930^2*0.3256} \\K=0.4508[/tex]

(iii)

Δ[tex]G=-RTlnK\\[/tex]

ΔG = [tex]-8.314*(273+25)*ln(0.4508)\\=1973.96J\\=1.974kJ[/tex]

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