Sulfur dioxide reacts with chlorine at 227 oC:

SO2(g) +Cl2(g) ↔ SO2Cl2(g)

Kp for this reaction is 5.1 x 10-2 atm-1. Initially, 1.00 g each of SO2 and Cl2 are placed in a 1.00 L reaction vessel. After 15 minutes, the concentration of SO2Cl2 is 45.5 μg/mL. You will determine if the system has reached equilibrium. First, what is Kc (in L/mol)? (A μg is 10-6 g.)

Next determine all initial concentrations. What is the initial sulfur dioxide concentration (in mol/L or M)?

Determine all concentrations after 15 minutes. What is the chlorine concentration?

What is Q after 15 minutes?

Kp = Kc(RT)^delta n.

Solve for Kc.

Convert 1 g SO2 and Cl2 to mols/L
Calculate Qc and compare that with Kc.
Post your work if you get stuck.

To find Kc, we need to convert Kp to Kc using the ideal gas law equation:

Kp = Kc(RT)^(Δn)

Where Δn is the change in the number of moles between products and reactants. In this case, Δn = 1 - (2 + 1) = -2.

R is the ideal gas constant, which is 0.0821 L·atm/(K·mol), and T is the temperature in Kelvin. Given that the temperature is 227 oC, we convert it to Kelvin:

227 oC + 273.15 = 500.15 K

Now we can calculate Kc:

Kc = Kp / (RT)^(Δn)
= (5.1 x 10^(-2)) / (0.0821 * 500.15)^(-2)
= 5.1 x 10^(-2) / (20.56)^(-2)
= 5.1 x 10^(-2) / 0.037938
≈ 1.34 M

So, Kc is approximately 1.34 M.

Next, let's calculate the initial concentration of sulfur dioxide (SO2) in mol/L or M. We are given that 1.00 g of SO2 is placed in a 1.00 L reaction vessel. The molar mass of SO2 is approximately 64.07 g/mol.

Number of moles of SO2 = mass / molar mass
= 1.00 g / 64.07 g/mol
≈ 0.0156 mol

Since we have 1.00 L of solution, the initial concentration of SO2 is:

Initial concentration of SO2 = number of moles / volume
= 0.0156 mol / 1.00 L
≈ 0.0156 M

Therefore, the initial concentration of SO2 is approximately 0.0156 M.

To find the concentration of chlorine (Cl2) after 15 minutes, we need to use the given information that after 15 minutes, the concentration of SO2Cl2 is 45.5 μg/mL.

First, we convert the concentration of SO2Cl2 from μg/mL to g/L:

45.5 μg/mL = 45.5 x 10^(-6) g/mL = 45.5 x 10^(-6) g / 1000 mL
= 4.55 x 10^(-8) g/L

Since the volume of the solution is 1.00 L, the concentration of SO2Cl2 after 15 minutes is:

Concentration of SO2Cl2 = mass / volume
= 4.55 x 10^(-8) g / 1.00 L
≈ 4.55 x 10^(-8) M

So, the concentration of SO2Cl2 after 15 minutes is approximately 4.55 x 10^(-8) M.

Lastly, we need to calculate Q after 15 minutes. Q is the reaction quotient and can be calculated by substituting the concentrations of the species at any point in the reaction. In this case, since we are given the concentrations after 15 minutes, we can use those values.

Q = [SO2Cl2] / ([SO2] * [Cl2])
= (4.55 x 10^(-8)) / (0.0156 * [Cl2])

Since we are asked to find the chlorine concentration (Cl2), we can rearrange the equation as:

[Cl2] = (4.55 x 10^(-8)) / (0.0156 * Q)

Now we need to determine the value of Q. Given that Q is calculated after 15 minutes, Q = [SO2Cl2] / ([SO2] * [Cl2]) = (4.55 x 10^(-8)) / (0.0156 * [Cl2]). The value of Q can be calculated by substituting the known values.

Q = (4.55 x 10^(-8)) / (0.0156 * [Cl2])

Please provide the value of [Cl2] to calculate the value of Q after 15 minutes.