Consider the gas phase reaction 302(g) = 203(g). (a) At 298K and a total pressure of...

Question

Question

Consider the gas phase reaction 302(g) = 203(g). (a) At 298K and a total pressure of 1 atm, what are the partial pressures at

Answer 1

Answer #1

The initial moles of O2 = 3

Let the equilibrium moles of O3 = 2x

Then the equilibrium moles of O2 = 3-3x

According to ideal gas equation = PV = nRT

Let volume of the system = 1 L

Part a.

For O2

PO2 * 1 L = (3-3x) * 0.0821 L.atm/mol.K * 298 K

i.e. PO2 = 73.3974 - 73.3974x

For O3

PO3 * 1 L = 2x * 0.0821 L.atm/mol.K * 298 K

i.e. PO3 = 48.9316 x

But Ptot = PO2 + PO3 = 73.3974 - 73.3974x + 48.9316 x = 73.3974 - 24.4658x = 1 atm

i.e. x = 2.96 atm

i.e. PO2 = 3-3x = 3-(3*2.96) = 5.88 atm

And PO3 = 2x = 2*2.96 = 5.92 atm

Answer 2

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