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 come to 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

The balanced reaction with ICE TABLE

2A + B = 3C + 2D

I 1.5 2 - 1

C - 2x - x +3x +2x

E (1.5-2x) (2-x) (3x) (1+2x)

At equilibrium

Moles of C = 3x = 1.20

x = 0.40

Moles of A = 1.5 - 2x = 1.5 - 2*0.40 = 0.70

Moles of B = 2 - 0.4 = 1.60

Moles of D = 1 + 2*0.40 = 1.80

Total Moles at equilibrium = 0.70 + 1.60 + 1.20 + 1.80

= 5.30 mol

Part a

Mol fraction of A = moles of A / total moles

xA = 0.70/5.30 = 0.132

xB = 1.60/5.30 = 0.302

xC = 1.20/5.30 = 0.226

xD = 1.80/5.30 = 0.340

Part b

Kx = [xD]²[xC]³ / [xB] [xA]²

Kx = (0.340*0.340*0.226*0.226*0.226) / (0.302*0.132*0.132)

= 0.25

Part c

Kp = [pD]²[pC]³ / [pB] [pA]²

Kp = [xD*P]²[xC*P]³ / [xB*P] [xA*P]²

Kp = [0.340*2]²[0.226*2]³ / [0.302*2] [0.132*2]²

Kp = 0.0427 / 0.042

Kp = 1.02

Part d

Go = - RT ln Kp

= - 8.314 J/mol·K x (25+273)K x ln (1.02)

= - 49.06 J/mol