Kc = 0.040 for the system below at 450 oC. If a reaction is initiated with 0.40 mole of Cl2 and 0.40 mole of PCl3 in a 2.0 liter container, what is the equilibrium concentration of Cl2 in the same system? PCl5(g) ⇌ PCl3(g) + Cl2(g)

Fundamentals Of Analytical Chemistry
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Chapter6: Random Errors In Chemical Analysis
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Kc = 0.040 for the system below at 450 oC. If a reaction is initiated with 0.40 mole of Cl2 and 0.40 mole of PCl3 in a 2.0 liter container, what is the equilibrium concentration of Cl2 in the same system?

PCl5(g) ⇌ PCl3(g) + Cl2(g)

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Expert Solution
Step 1
Given :
Kc = 0.040
Step 2
Calculations:
PCI3 (g) + Cl2(g)=PC15(g) Initial :
0.2
0.2
OAt equi: 0.20-x
0.20-х х Кс %3DPCІЗ]
[Cl2][PCI5]
0.040x = 0.04 + x2 - 0.40x 0= x2 + 0.04 - 0.44 x x = 0.311652, 0.128348[Cl2] = 0.20-
Since Kc is given for reverse reaction. 0.040 = [0.20-
x]2x
x = 0.20- 0.128348 =0.0717M
Transcribed Image Text:Expert Solution Step 1 Given : Kc = 0.040 Step 2 Calculations: PCI3 (g) + Cl2(g)=PC15(g) Initial : 0.2 0.2 OAt equi: 0.20-x 0.20-х х Кс %3DPCІЗ] [Cl2][PCI5] 0.040x = 0.04 + x2 - 0.40x 0= x2 + 0.04 - 0.44 x x = 0.311652, 0.128348[Cl2] = 0.20- Since Kc is given for reverse reaction. 0.040 = [0.20- x]2x x = 0.20- 0.128348 =0.0717M
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