Question

In the gas-phase reaction 2A+B ⇄ 3C + 2D, it was found that when 1.50 mole A, 2.00 mole B and 1.00 mole D were mixed and allowed to cometo equilibrium at 25oC, the resulting mixture contained 1.20 mol C at a total pressure of 2.00 bar. Calculate (a) the mole fractions of each species at equalibrium x_A=_______, x_B=_______, x_c=_______, x_D=________ (b) Kx =_________ (c)Kp=_______ (d) ∆_rG^o=_________

Please enter all answers in (a) with 3 decimals. for example, 0.4467 is written as 0.447. for (b),(c) and (d) please enter answers with 2 decimals. for example, -25.445 is written as -25.45. In (d) use unit kJ/mole

Answer 1

Let x be the no. of moles of B reacting

By stoichiometry we can write

2A+B ⇄ 3C + 2D

(A_o –2 x) + (B_o – x) ⇄ (C_o + 3x) + (D_o + 2x)

Resulting mixture contains 1.2 moles of C

(C_o + 3x) = 1.2

Initially C is not present in the mixture, therefore C_o = 0

And x = 1.2 / 3 = 0.4

Ao = 1.5 mol

Bo = 2 mol

Do = 1 mol

Total no. of moles at equilibrium = (A_o –2 x) + (B_o – x) + (C_o + 3x) + (D_o + 2x)

= (1.5 – 2*0.4) + (2 – 0.4) + (0 + 3*0.4) + (1 + 2*0.4)

= 0.7 + 1.6 + 1.2 + 1.8 = 5.3 moles

X_A = 0.7 / 5.3 = 0.132

X_B = 1.6 / 5.3 = 0.302

X_C = 1.2 / 5.3 = 0.226

X_D = 1.8 / 5.3 = 0.340

b)

Kx = X_C³ * X_D² / (X_A² * X_B)

= 0.226³ * 0.340² / (0.132² * 0.302)

= 0.25

c)

Total pressure, Pt = 2 bar

Partial pressure of A, P_A = Pt * X_A = 2 * 0.132 = 0.264 bar

Partial pressure of B, P_B = Pt * X_B = 2 * 0.302 = 0.604 bar

Partial pressure of C, P_C = Pt * X_C = 2 * 0.226 = 0.452 bar

Partial pressure of D, P_D = Pt * X_D = 2 * 0.340 = 0.68 bar

Kp = P_C³ * P_D² / (P_A² * P_B)

= 0.452³ * 0.680² / (0.264² * 0.604)

= 1.02

d)

For gas phase

R = 8.314 J/mol-K

T = 25 C = 298 K

ΔG_r^o = - RT ln (Kp)

= - 8.314 J/mol-K *298 K * ln(1.02)

= -41.54 J/mol

= -41.54 x 10^-3 kJ/mol