Sulfur trioxide decomposes into sulfur dioxide and oxygen in an equilibrium. You have a 3.00 L vessel that is charged with 0.755 mol of SO3. At equilibrium, the amount of SO3 is 0.250 mol.
2 SO3 (g) ----> 2 SO2 (g) +O2 (g)
A) calculate the equilibrium concentrations of SO2 and O2.
B) calculate the value of Kc for the equilibrium.
Please help me. I don't know where to start.
Calculate concentrations.
Set up an ICE chart.
Solve. I like to do the concentrations first but it may be easier to do it in mols first, THEN make the conversion. The only problem is that we ofte forget to mak the conversion.
.........2SO3 ==> 2SO2 + O2
I......0.755........0.....0
C.......-2x........2x.....x
E....0.755-2x.......2x....x
The problem tells you that 0.755-2x = 0.250. Solve for x, then evaluate 2x. Then convert all of the mols into concentrations in mols/L. You have mols and L = 3.00 L. Finally, substitute concentrations into the Kc expression and solve for Kc. Post your work if you get stuck.
1.66
mjj kjl;
To solve this problem, you need to use the information provided to calculate the equilibrium concentrations of SO2 and O2 in the 3.00 L vessel.
A) To calculate the equilibrium concentrations, you need to understand the stoichiometry of the reaction. The balanced equation for the reaction shows that 2 moles of SO3 react to form 2 moles of SO2 and 1 mole of O2.
Initially, the vessel is charged with 0.755 mol of SO3, but at equilibrium, the amount of SO3 is 0.250 mol. Therefore, the change in the amount of SO3 is given by: Δn(SO3) = 0.250 mol - 0.755 mol = -0.505 mol.
According to the stoichiometry, this means that Δn(SO2) = 2 * Δn(SO3) = -2 * 0.505 mol = -1.01 mol.
Similarly, Δn(O2) = 1 * Δn(SO3) = 1 * 0.505 mol = 0.505 mol.
Now, you can use the ideal gas law to relate the concentrations of the gases to the number of moles:
PV = nRT
Since the volume (V) and temperature (T) remain constant, we can rewrite the equation as:
P = (n/V)RT = C(RT)
Where P is the pressure and C is the molar concentration.
Since we know that the total volume of the vessel is 3.00 L, we can calculate the concentration of each gas at equilibrium:
C(SO3-equilibrium) = n(SO3-equilibrium) / V = 0.250 mol / 3.00 L = 0.0833 M
C(SO2-equilibrium) = n(SO2-equilibrium) / V = (0.755 mol - 1.01 mol) / 3.00 L = -0.085 M (mol/L)
C(O2-equilibrium) = n(O2-equilibrium) / V = 0.505 mol / 3.00 L = 0.168 M
Note that the negative sign for the concentration of SO2 implies that it is lower at equilibrium compared to the initial concentration.
B) Kc is the equilibrium constant and is determined by the concentrations of the products and reactants at equilibrium. It is defined as follows:
Kc = [SO2]^2 * [O2] / [SO3]^2
Using the equilibrium concentrations we found earlier:
Kc = (0.0833 M)^2 * (0.168 M) / (0.250 M)^2
Calculating this expression will give you the value of Kc for the equilibrium.
It is also worth noting that since pressure was not explicitly given in the problem, we assumed that the reaction is taking place at a constant temperature and pressure.