15: Sulfur dioxide reacts with chlorine at 227 oC: SO2(g) +Cl2(g) SO2Cl2(g) Kp f

Question

15: 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.)

P2: Next determine all initial concentrations. What is the initial sulfur dioxide concentration (in mol/L or M)?

P3: Determine all concentrations after 15 minutes. What is the chlorine concentration?

P4: What is Q after 15 minutes?

P5: Calculate the mass (in g) of SO2Cl2 expected at equilibrium.

In depth explanations would be VERY appreciated, I feel like iI'm lost with this topic, will rate highly!

Explanation / Answer

To convert Kp to Kc you will have to use the universal gas equation:

pV = nRT

And if you work it out you will find.

Kc = RT*Kp (T = 500 K)

P2) The molar mass of SO2 is 64 g/mol.

So 1 g = (1g/64 g/mol) = 1.56x10-2 mol.

volume is 1.00 L. [SO2] = 1.56x10-2 M
1 g of Cl2: [Cl2] = 1.41x10-2 M
[SO2Cl2] = 0 M

P3) 45.5 g/mL = 0.0455 g/L => 3.37x10-4 M.

Subtract from initial concentrations.

1.41x10-2 M -  3.37x10-4 M = 1.3753x10-2 M

P4) find mass of SO2Cl2:

(1.00 L ) (1000 ml / L) (45.5 ug / ml) = 45,500 ug of SO2Cl2
find grams of SO2Cl2:
(45,500 ug of SO2Cl2 ) (1 gram /1,000,000 ug) = 0.0455 grams of SO2Cl2

use molar masses to find moles:
(1.00 g of SO2) / (64.06 g/mol SO2) = 0.01561 moles of SO2
(1.00 g of Cl2) / (70.91 g/mol Cl2) = 0.01410 moles of Cl2
(0.0455 grams of SO2Cl2) / (134.97 g/mol SO2Cl2) = 0.0003371 mol of SO2Cl2
SO2(g) + Cl2(g) SO2Cl2(g)
0.01561 & 0.01410 <=> none initially
after 15 minutes, it produced 0.0003371 mol of SO2Cl2
which it could only do at a loss of :
SO2(g) .... + .... Cl2(g) .... .... ... <=> ... SO2Cl2
0.01561 - 0.0003371 & 0.01410 - 0.0003371 --> 0.0003371
leaving us with:
SO2(g) + Cl2(g) <=> SO2Cl2
0.01527mol & 0.01376 mol <=> 0.0003371 mol
mol are converted into pressures using PV=nRT
P = nRT /V
since RT/V = (0.08206 L-atm/mol-K)(500K) / (1.00L) = 41.03
P = (41.03) (n)
Qp = [SO2Cl2] / [SO2] [Cl2]

Qp = [0.0003371 (41.03) ] / [0.01527 (41.03)] [0.01376 (41.03)]
canceling out a numerato'sr with a denominator's "(41.03)", simplifies it to:
Qp = [0.0003371] / [0.01527] [0.01376 (41.03)]
Qp = 0.0391

P5)
Calculate the mass (in g) of SO2Cl2 expected at equilibrium.

SO2(g) + Cl2(g) SO2Cl2(g)
0.01561mol & 0.01410mol <=> no mol initially

at equilib:
SO2(g) .... ... + .... Cl2(g) <=> . SO2Cl2
(0.01561 - X) mol & (0.01410 - X) mol <=> X mol
mol are converted into pressures using PV=nRT
P = nRT /V
since RT/V = (0.08206 L-atm/mol-K)(500K) / (1.00L) = 41.03
P = (41.03) (n)
Kp = [SO2Cl2] / [SO2] [Cl2]
0.051 = [(X) (41.03) ] / [(0.01561 - X) (41.03)] [(0.01410 - X) (41.03)]
canceling out a numerato'sr with a denominator's "(41.03)" , implifies it to:
0.051 = [(X)] / [(0.01561 - X)] [(0.01410 - X) (41.03)]
0.051 = [(X)] / [(0.01561 - X)] (0.5785 - 41.03 X)
0.051 = [(X)] / (0.00903 - 1.219X + 41.03 X^2 )
0.051 (0.00903 - 1.219X + 41.03 X^2 ) = (X)
0.0004605 - 0.06217X + 2.0925 X^2 = X
2.0925 X^2 - 1.06217X + 0.0004605 = 0
X = 0.00434 moles of SO2Cl2 made
using molar mass
( 0.000434 moles of SO2Cl2) (134.97 g/mol SO2Cl2) = 0.0586 grams SO2Cl2 at equilib

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