Exercise 8 | Chemical Kinetics: Rate Laws

Oxalic acid decomposes at high temperature according to the following chemical reaction:

H2C2O(g) → CO(g) + HCHO(g)


The total pressure after 2.00 x 104s is measured for 3 experiments with different initial pressure of oxalic acid.

Exp 1: P0H2C2O4 = 5.0 Torr (at t = 0) ⇒ Ptot = 7.2 Torr (at t = 2.00 x 104 s)

Exp 2: P0H2C2O4 = 7.0 Torr ⇒ Ptot = 10.0 Torr

Exp 3: P0H2C2O4 = 8.4 Torr ⇒ Ptot = 12.0 Torr

 

Determine the rate law and the value of the rate constant of this reaction.

Rate law = - [H2C2O4]t = k [H2C2O4]α

 

Ideal gas law:

PVA = nART ⇒ P = nART/VA ⇒ P = [A]RT

⇒ Rate law = - 1RT PH2C2O4t = k PH2C2O4RTα

 

Ptot = PH2C2O4 + PCO2 + PHCHO2

One mole of CO2(g) and one mole of HCHO2(g) are produced from every mole of H2C2O4(g) that react:

PCO2 = PHCHO2 = P0H2C2O4 - PH2C2O4
⇒ Ptot = PH2C2O4 + 2 PCO2
⇒ Ptot = PH2C2O4 + 2 (P0H2C2O4 - PH2C2O4)
⇒ Ptot = 2 P0H2C2O4 - PH2C2O4
⇒ PH2C2O4 = 2 P0H2C2O4 - Ptot

 

Let’s calculate PH2C2O4 for the 3 experiments:

Exp 1: P0H2C2O4 = 5.0 Torr (at t = 0) ⇒ PH2C2O4 = 2.8 Torr (at t = 2.00 x 104 s)

Exp 2: P0H2C2O4 = 7.0 Torr ⇒ PH2C2O4 = 4.0 Torr

Exp 3: P0H2C2O4 = 8.4 Torr ⇒ PH2C2O4 = 4.8 Torr

 

If the reaction is a first-order reaction:
α = 1 and - PH2C2O4t = k P H2C2O4
After integrating: PH2C2O4/P0H2C2O4 = e – kt

If the reaction is a first-order reaction, the ratio PH2C2O4/P0H2C2O4 at t = 2.00 x 104 s should be constant.

Exp 1: PH2C2O4/P0H2C2O4 = 0.56

Exp 2: PH2C2O4/P0H2C2O4 = 0.57

Exp 3: PH2C2O4/P0H2C2O4 = 0.57

The reaction is a first-order reaction and the rate of reaction = k P H2C2O4

Let’s calculate k:

ln PH2C2O4P0H2C2O4 =  – kt

⇒ k = - ln PH2C2O4P0H2C2O4t = 2.9 x 10-5 s-1