In the gas-phase reaction 2A+B 3C + 2D, it was found that when 1.50 mole A, 2.00
ID: 588792 • Letter: I
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 xA=_______, xB=_______, xc=_______, xD=________ (b) Kx =_________ (c)Kp=_______ (d) rGo=_________
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
Explanation / Answer
Let x be the no. of moles of B reacting
By stoichiometry we can write
2A+B 3C + 2D
(Ao –2 x) + (Bo – x) (Co + 3x) + (Do + 2x)
Resulting mixture contains 1.2 moles of C
(Co + 3x) = 1.2
Initially C is not present in the mixture, therefore Co = 0
And x = 1.2 / 3 = 0.4
Ao = 1.5 mol
Bo = 2 mol
Do = 1 mol
Total no. of moles at equilibrium = (Ao –2 x) + (Bo – x) + (Co + 3x) + (Do + 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
XA = 0.7 / 5.3 = 0.132
XB = 1.6 / 5.3 = 0.302
XC = 1.2 / 5.3 = 0.226
XD = 1.8 / 5.3 = 0.340
b)
Kx = XC3 * XD2 / (XA2 * XB)
= 0.2263 * 0.3402 / (0.1322 * 0.302)
= 0.25
c)
Total pressure, Pt = 2 bar
Partial pressure of A, PA = Pt * XA = 2 * 0.132 = 0.264 bar
Partial pressure of B, PB = Pt * XB = 2 * 0.302 = 0.604 bar
Partial pressure of C, PC = Pt * XC = 2 * 0.226 = 0.452 bar
Partial pressure of D, PD = Pt * XD = 2 * 0.340 = 0.68 bar
Kp = PC3 * PD2 / (PA2 * PB)
= 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
Gro = - RT ln (Kp)
= - 8.314 J/mol-K *298 K * ln(1.02)
= -41.54 J/mol
= -41.54 x 10-3 kJ/mol
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