Question

If the equilibrium constant for the conversion of SO2 to SO3 is 1.458e^-5 exp(1.181e4/T), the SO2-to-SO3 ratio exceeds 1.0 at 1600K and 10atm if the starting SO2/O2 ratio is 1/10. (True of False and summarize how you reached your conclusions.)

Answer 1

The given Equilibrium constant

K = 1.458*10^-5 * exp(1.181*10^4/T)

At T = 1600 K

K = 1.458*10^-5 * exp(1.181*10^4/1600)

K = 0.02341

The given ratio

SO2/O2 = 1/10

The balanced reaction with ICE TABLE

2SO2 + O2 = 2SO3

I 1 10

C -2x - x +2x

E 1-2x 10-x 2x

Total moles = 1 - 2x + 10 - x + 2x = 11-x

Mol fraction at equilibrium

ySO3 = 2x/(11-x)

yO2 = (10-x)/(11-x)

ySO2 = (1-2x)/(11-x)

Equilibrium constant expression of the reaction

K = y²SO3*P² / yO2*P * y²SO2*P²

0.02341 = (2x/11-x)² / (10-x/11-x)*P * (1-2x/11-x)²

0.02341*10 = 2x*(11-x)/(10-x)*(1-2x)

0.234 = (22x - 2x²)/(10-21x+2x²)

22x - 2x² = 2.34 - 4.914x + 0.468x²

2.468x² - 26.914x + 2.34 = 0

x = 0.0876

Ratio

SO2/SO3 = (1-2x) / 2x

= (1 - 2*0.0876)/(2*0.0876)

= 4.71

SO2/SO3 exceeds 1

True.