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
2SO2 + O2 ==> 2SO3
Initial values:
SO2 = 8.00 mols and (SO2) = 8.00/2 = 4.00M
O2 = 10.00 mols and (O2) = 10.00/2 = 5.00 M
At equilibrium:
(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.
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?
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?
so since K= products/reactants, K will get smaller?
That's right!
I definitely understand this concept better now. Thank you so much for your help!
You're welcome. Come back anytime.
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
a) To find the concentrations of all species at equilibrium, we first need to determine the initial moles of each species present in the system. Given that we have 8.00 moles of sulfur dioxide (SO2) and 10.00 moles of sulfur trioxide (SO3) in a 2.00 L container, we can calculate the initial concentration (in moles per liter) of each species as follows:
Initial concentration of SO2 = moles of SO2 / volume of container
= 8.00 moles / 2.00 L
= 4.00 M (Molar)
Initial concentration of SO3 = moles of SO3 / volume of container
= 10.00 moles / 2.00 L
= 5.00 M
Since we know the equilibrium constant (K) for this reaction is 3.40, we can use the stoichiometry of the reaction to set up an ICE (Initial, Change, Equilibrium) table:
2SO2 (g) + O2 (g) <--> 2SO3 (g)
Initial: 4.00 M 0 M 5.00 M
Change: -2x -x +2x
Equilibrium: 4.00 - 2x - x 5.00 + 2x
Here, x represents the change in concentration for each species at equilibrium, as the reaction proceeds.
At equilibrium, the equilibrium concentrations are as follows:
Concentration of SO2 at equilibrium = 4.00 - 2x
Concentration of O2 at equilibrium = -x
Concentration of SO3 at equilibrium = 5.00 + 2x
b) Now, let's predict the effect of the following changes on the value of the equilibrium constant and on the number of moles of sulfur trioxide (SO3) present in the mixture at equilibrium:
i) Decreasing the volume of the system:
When the volume is decreased, according to Le Chatelier's principle, the equilibrium will shift in the direction that produces fewer moles of gas to relieve the pressure. As the reaction consumes 2 moles of SO2 and 1 mole of O2 to produce 2 moles of SO3, reducing the volume will increase the pressure, favoring the reverse reaction (decreasing the moles of gas). Consequently, the equilibrium concentration of SO3 will decrease, and the value of the equilibrium constant (K) will remain the same.
ii) Adding oxygen to the equilibrium mixture:
By adding additional O2, the concentration of O2 will increase, which will shift the equilibrium to the right to use up the excess O2. As a result, more SO3 will be produced while SO2 will be consumed to achieve a new equilibrium. The concentration of SO3 at equilibrium will increase, and the value of the equilibrium constant (K) will remain the same.
iii) Raising the temperature of the system:
If the temperature is increased, the reaction will shift in the endothermic direction to consume heat. Since the production of SO3 is an exothermic reaction (with negative change in enthalpy), increasing the temperature will favor the reverse reaction and decrease the concentration of SO3 at equilibrium. Consequently, the value of the equilibrium constant (K) will decrease.
Remember, these predictions are based on the principles outlined by Le Chatelier's principle and the understanding of how factors such as concentration, pressure, and temperature affect the position of the equilibrium.