Answer to Question #16398 in Physical Chemistry for kenneth mcdonald
2012-10-15T08:14:42-04:00
(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 Answer: Equilibrium constant is about 220.
Need a fast expert's response?
Submit order
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS !
Comments
You're welcome. We are glad to be helpful. If you really liked our service please press like-button beside answer field. Thank you!
thank you excellent !
Leave a comment