Chemistry - Chemical Equilibrium Question with Solution | TestHub
ChemistryChemical EquilibriumProblems involving KpMedium2 minQB
ChemistryMediumnumerical
The dissociation constant of SO2Cl2 for SO2Cl2(g) ⇌ SO2(g) + Cl2(g) is 1.0 atm at 127°C. What pressure (in atm) will be developed when 13.5 g of SO2Cl2 is vaporized at 127°C in a 3.28 L vessel originally full of SO2 gas at 127°C and 0.5 atm pressure? (S = 32, Cl = 35.5, O = 16; R = 0.082 L atm/mol/K)
Answer:
2.00
Solution:
Moles of SO2 initially = (0.5)(3.28)/[(0.082)(400)] = 0.05 mol Moles of SO2Cl2 = 13.5/135 = 0.10 mol
Let x mol of SO2Cl2 dissociate: SO2Cl2 = 0.10 - x, SO2 = 0.05 + x, Cl2 = x
RT/V = (0.082)(400)/3.28 = 10 atm per mol
KP = [x(0.05+x)/(0.10-x)](10) = 1.0
10x2 + 1.5x - 0.1 = 0, so x = 0.05 mol
Total moles = 0.05 + 0.10 + 0.05 = 0.20 mol
P = (0.20)(10) = 2 atm
Stream:JEESubject:ChemistryTopic:Chemical EquilibriumSubtopic:Problems involving Kp
⏱ 2mℹ️ Source: QB
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