Part 1: Sulfur dioxide reacts with chlorine at 227 oC: SO2(g) +Cl2(g) SO2Cl2(g)

Question

Part 1:

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 g/mL. You will determine if the system has reached equilibrium. First, what is Kc (in L/mol)? (A g is 10-6 g.) Part 2: Next determine all initial concentrations. What is the initial sulfur dioxide concentration (in mol/L or M)? Part 3: Determine all concentrations after 15 minutes. What is the chlorine concentration? Part 4: What is Q after 15 minutes? Part 5:Has the system reached equilibrium? Part 6: Calculate the mass (in g) of SO2Cl2 expected at equilibrium.

Explanation / Answer

SO2(g) +Cl2(g) ? SO2Cl2(g)

(1) 0.051 atm^-1 = Kc (.08206L*atm/mol*K x 500.15K)^-1
0.051 atm^-1 = Kc (0.0244 mol/Latm)
Kc = 0.051 atm^-1 / .0244 mol/Latm
Kc = 2.09 L/mol

(2)[SO2] = 1/64 = 0.0156 M
[Cl2] = 1/71 = 0.0141 M

(3) [SO2Cl2] = (45.5 * 10^-3)/135 = 0.337 * 10^-3 M
therefore [SO2] = 0.0156 - 0.337 * 10^-3 = 0.0152 M
[Cl2] = 0.0141 - 0.337 * 10^-3 = 0.0137 M

(4) Q = (0.337 * 10^-3)/(0.0152)(0.0137) = 1.618 L/mol

(5) Kc = x/(0.0156-x)(0.0141-x)
therefore x = 4.332 * 10^-4 M
mass of SO2Cl2 = 4.332 * 10^-4 * 135 = 0.0584g

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