It's hard to know exactly what is meant by "without 'detailed' calculations" estimate... We don't know how detailed the author wants it nor do we know what is considered an estimate.
I think the idea here is that with such a large Kc the equilibrium is far to the right which means that for all practical purposes all of the SO2 an O2 react to form 0.134 mol SO3. The concn then will be about 0.134/0.275L = about 0.487 or about 0.5M
Without detailed equilibrium calculations, estimate the equilibrium concentration of SO3 when a mixture of 0.134 mol of SO2 and 0.067 mol of O2 in a 275 mL flask at 300oC combine to form SO3.
2 SO2(g) + O2(g) = 2 SO3(g)
Kc = 6.3x10^9
I tried doing an ICE chart, and also tried working through an equilibrium expression to solve for the concentration of SO3, but both methods got me wrong answers
2 answers
At 627ºC, sulfur dioxide and oxygen gases combine to form sulfur trioxide gas. At equilibrium, the concentrations for sulfur dioxide, oxygen, and sulfur trioxide gases are .0060M, .0054M, and .0032M, respectively.
a. Write a balanced equation for the formation of one mole of sulfur trioxide.
b. Calculate K for the reaction @ 627ºC. (Note that gases need to be in atm.)
a. Write a balanced equation for the formation of one mole of sulfur trioxide.
b. Calculate K for the reaction @ 627ºC. (Note that gases need to be in atm.)
0 answers