Chem equilibrium module 2

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Chem equilibrium module 2

Postby JulieAljamal1E » Fri Jan 11, 2019 9:33 am

For the last question on module 2, “The reaction 2 SO2 (g) + O2 (g) ⇌ 2 SO3 (g) occurs in a 1.00 L flask at 312 K and at equilibrium the concentrations are 0.075 mol.L-1 SO2 (g), 0.537 mol.L-1 O2 (g), and 0.925 mol.L-1 SO3 (g). Calculate their respective partial pressures at 312 K using R = 8.206 × 10-2 L.atm.K-1.mol-1”, do you use an ICE table to solve it. How do you solve?

Chase Yonamine 1J
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Joined: Fri Sep 28, 2018 12:17 am

Re: Chem equilibrium module 2

Postby Chase Yonamine 1J » Fri Jan 11, 2019 10:12 am

No, you do not need to use the ice table.
You should use the ideal gas law to convert from concentration to partial pressure of reactants and products at equilibrium.
divide by v to get P=(n/v)(RT) this is the same as P=(concentration)(RT) where R is 8.206 × 10-2 L.atm.K-1.mol-1 and temperature in this problem is 312K.
Just plug these into the equation to get:
PSO2= (0.075 mol.L-1)(8.206 × 10-2 L.atm.K-1.mol-1)(312K)= 1.92atm
PO2=(0.537 mol.L-1)(8.206 × 10-2 L.atm.K-1.mol-1)(312K)=13.7atm
PSO3= (0.925 mol.L-1)(8.206 × 10-2 L.atm.K-1.mol-1)(312K)=23.7atm

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