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# Answer to Question #16398 in Physical Chemistry for kenneth mcdonald

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.
1
2012-10-17T08:54:45-0400
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

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18.10.12, 15:17

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kenneth mcdonald
17.10.12, 23:55

thank you excellent !

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