Integrated rate laws - 2 | Chemical Kinetics: Rate Laws

Rate law = - $\frac{∆\left[{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\right]}{∆\mathrm{t}}$ = k [H₂C₂O₄]^α

Ideal gas law:

PV_A = n_ART ⇒ P = n_ART/V_A ⇒ P = [A]RT

⇒ Rate law = - $\frac{1}{\mathrm{RT}}$ $\frac{∆{\mathrm{P}}_{\mathrm{H}2\mathrm{C}2\mathrm{O}4}}{∆\mathrm{t}}$ = k ${\left(\frac{{\mathrm{P}}_{\mathrm{H}2\mathrm{C}2\mathrm{O}4}}{\mathrm{RT}}\right)}^{\mathrm{\alpha }}$

P_tot = P_H2C2O4 + P_CO2 + P_HCHO2

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

P_CO2 = P_HCHO2 = P⁰_H2C2O4 - P_H2C2O4
⇒ P_tot = P_H2C2O4 + 2 P_CO2
⇒ P_tot = P_H2C2O4 + 2 (P⁰_H2C2O4 - P_H2C2O4)
⇒ P_tot = 2 P⁰_H2C2O4 - P_H2C2O4
⇒ P_H2C2O4 = 2 P⁰_H2C2O4 - P_tot

Let’s calculate P_H2C2O4 for the 3 experiments:

Exp 1: P⁰_H2C2O4 = 5.0 Torr (at t = 0) ⇒ P_H2C2O4 = 2.8 Torr (at t = 2.00 x 10⁴ s)

Exp 2: P⁰_H2C2O4 = 7.0 Torr ⇒ P_H2C2O4 = 4.0 Torr

Exp 3: P⁰_H2C2O4 = 8.4 Torr ⇒ P_H2C2O4 = 4.8 Torr

If the reaction is a first-order reaction:
α = 1 and - $\frac{∆{\mathrm{P}}_{\mathrm{H}2\mathrm{C}2\mathrm{O}4}}{∆\mathrm{t}}$ = k P _H2C2O4
After integrating: P_H2C2O4/P⁰_H2C2O4 = e ^{– kt}

If the reaction is a first-order reaction, the ratio P_H2C2O4/P⁰_H2C2O4 at t = 2.00 x 10⁴ s should be constant.

Exp 1: P_H2C2O4/P⁰_H2C2O4 = 0.56

Exp 2: P_H2C2O4/P⁰_H2C2O4 = 0.57

Exp 3: P_H2C2O4/P⁰_H2C2O4 = 0.57

The reaction is a first-order reaction and the rate of reaction = k P _H2C2O4

Let’s calculate k:

ln $\left(\frac{{\mathrm{P}}_{\mathrm{H}2\mathrm{C}2\mathrm{O}4}}{{{\mathrm{P}}^0}_{\mathrm{H}2\mathrm{C}2\mathrm{O}4}}\right)$ =  – kt

⇒ k = - $\frac{\ln \left({\displaystyle \frac{{\mathrm{P}}_{\mathrm{H}2\mathrm{C}2\mathrm{O}4}}{{{\mathrm{P}}^0}_{\mathrm{H}2\mathrm{C}2\mathrm{O}4}}}\right)}{\mathrm{t}}$ = 2.9 x 10^-5 s^-1