Chem equilibrium module 2

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JulieAljamal1E
Posts: 71
Joined: Fri Sep 28, 2018 12:24 am

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
Posts: 62
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.
Pv=nrt
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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