For the system 2SO2(g) + O2(g) <--> 2SO3 (g), change in enthalpy is negative for the production of SO3. At a particular temperature, 8.00 moles of sulfur dioxide and 10.00 moles of sulfur trioxide are introduced into a 2.00 L container. The system is allowed to reach equilibrium. The value of the equilibrium constant for this reaction is 3.40.
a) Calculate the concentrations of all species at equilibrium.
b) Predict the effect of the following changes on the value of the equilibrium constant and on the number of moles of sulfur trioxide present in the mixture at equilibrium. (assume that in each cause all other factors remain constant).
i) decreasing the volume of the system
ii) adding oxygen to the equilibrium mixture
iii) raising the temperature of the system
AP Chemistry - DrBob222, Saturday, January 12, 2008 at 11:16pm
2SO2 + O2 ==> 2SO3
SO2 = 8.00 mols and (SO2) = 8.00/2 = 4.00M
O2 = 10.00 mols and (O2) = 10.00/2 = 5.00 M
(SO3) = 2x
(O2) = 5.00-x
(SO2) = 4.00-2x
Plug into the equilibrium constant expression and solve for x. Then you can calculate the other concentrations.
For part b, I'm sure you know this is based on the principle of Le Chatelier. Let me know what you don't understand about this and I can help you through it BUT I don't want to just give you the answers. For biii it may help if you write the equation as
2SO2 + O2 ==> 2SO3 + heat (since the problem states that the enthalpy is negative for the production of SO3). So if enthalpy is negative, that means delta H is minus and that means heat is given off so heat goes on the product side.
AP Chemistry - S, Sunday, January 13, 2008 at 11:10am
since there are 10.00 moles of sulfur trioxide, shouldn't the initial concentration of SO3 = 5.00 M? (you have it posted as O2 = 5.00 M)
And for part b, I'm having problems on letter iii. I don't understand how the equilibrium constant would be affected by temperature. Would the constant increase because the molecules move faster in a raised temperature, causing more production of SO3?
AP Chemistry - DrBob222, Sunday, January 13, 2008 at 1:34pm
I see I can't read. It was late last night but I didn't know it was that late. Please excuse me and thank you for pointing out the obvious error.
On part biii. T almost ALWAYS affects K. In fact, the K of a reaction is evaluated at a particular T and it isn't good for other temperatures. As I pointed out last night, the reaction is exothermic since the problem states that the enthalpy is negative with respect to the production of SO3. Therefore, we may write the equaion as
2SO2 + O2 ==> 2SO3 + heat.
Now treat heat the same way you would a change in concentration of one of the reagents; i.e.,an increase in heat makes the reaction shift in such as way as to use up the added heat. That means it will shift to the left. Shifting to the left means (SO2) and (O2) will become larger and (SO3) will become smaller. Check me out on that. What does that do to K?
AP Chemistry - S, Sunday, January 13, 2008 at 1:38pm
so since K= products/reactants, K will get smaller?
AP Chemistry - DrBob222, Sunday, January 13, 2008 at 3:34pm
AP Chemistry - S, Sunday, January 13, 2008 at 3:56pm
I definitely understand this concept better now. Thank you so much for your help!
AP Chemistry - DrBob222, Sunday, January 13, 2008 at 4:17pm
You're welcome. Come back anytime.
AP Chemistry - frankie, Wednesday, December 10, 2008 at 11:39am
which one of the following reactions is not balanced (1)2SO2+O2-2so2,(2)2co+02-2co2,(3)2kno3+10-5k2o+n2,(4)SF4+3H2O-H2so3+4HF