Question

(a) Find ∆G◦ and K at 298.15 K for the gas-phase reaction
2 SO2(g) + O2(g) 2 SO3(g)

(b) If a stoichiometric mixture of SO2 and O2 is allowed to come to equilibrium at 298.15 K and 1.000bar, find the partial pressure of SO2.

Answer 1

a)

Kp = P-SO3^2 / (PSO2)^2 (PO2)

dG = Grxn = Gprod - Greact =2*(-370.4) - (-300.13*2 + 0) = -140.54 kJ/mol

so..

dG = -RT*ln(Kc)

dG = -8.314*298*ln(Kc) = -140.54*10^3

Kc = exp((-140.54*10^3)/(-8.314*298)) = 4.3182*10^24

Kp = Kc*(RT)^dn

dn = 2-(2+1) = -1

Kp = Kc*(RT)^-1

Kp = (4.3182*10^24)(0.082*298)^-1

Kp = 1.76*10^23

so..

b)

stoichiometric mix of SO2 = O2

to equilbirium and P = 1 bar

find.. SO2:

Kp = 1.76*10^23

P = 1 bar, T = 298.15K

PV = nRT

n/V = P/(RT)

M = (1)/(0.0831*298) = 0.0403815

total molarity = 0.0403815

since stoichiometric:

SO2 = 2x

O2 = x

so

0.0403815/3 = 0.0134605

[SO2] = 2*0.0134605 = 0.026921 M

[O2] = 0.0134605

in equilbirium

[SO2] = 0.026921 -2x

[O2] = 0.0134605-x

[SO3] = 2x

4.3182*10^24 = (2X)^2 / (0.026921 -2x)^2 ( 0.0134605-x)

since x >> 1

then

4.3182*10^24 = (2X)^2 / (0.026921 -2x)^2

sqrt( 4.3182*10^24 ) = 2x / (0.026921 -2x)

x = 0.026921 M

[SO2] = 0 (approx)

Partial Pressure

P = M*RT

P = 0 atm (approx)