Question #16398

(a. in the equilibrium 2SO2 =2SO2 +O2 AT 1000K and A TOTAL PRESSURE of 745mm,Hg; SO3 has a mole fraction of 0.338 and SO2 as a mole fraction of 0.309. calculate the equilibrium constant for this reaction.

Expert's answer

2SO3 = 2SO2 + O2

T=1000 K

P(total) = 745 mmHg

Mole fraction of

components are

x(SO3) = 0.338

x(SO2) = 0.309

x(O2) = 1 - x(SO3) -

x(SO2) = 1 - 0.338 - 0.309 = 0.353

Partial pressures of components

are

P(SO3) = P(total) * x(SO3) = 345 mmHg * 0.338 = 251.81 mmHg

P(SO2) =

P(total) * x(SO2) = 345 mmHg * 0.309 = 230.205 mmHg

P(O2) = P(total) * x(O2)

= 345 mmHg * 0.353 = 262.985 mmHg

Equilibrium constant is

K = P^2(SO2) *

P(O2) / P^2(SO3) = 219.79

Answer: Equilibrium constant is about 220.

T=1000 K

P(total) = 745 mmHg

Mole fraction of

components are

x(SO3) = 0.338

x(SO2) = 0.309

x(O2) = 1 - x(SO3) -

x(SO2) = 1 - 0.338 - 0.309 = 0.353

Partial pressures of components

are

P(SO3) = P(total) * x(SO3) = 345 mmHg * 0.338 = 251.81 mmHg

P(SO2) =

P(total) * x(SO2) = 345 mmHg * 0.309 = 230.205 mmHg

P(O2) = P(total) * x(O2)

= 345 mmHg * 0.353 = 262.985 mmHg

Equilibrium constant is

K = P^2(SO2) *

P(O2) / P^2(SO3) = 219.79

Answer: Equilibrium constant is about 220.

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