Question

The following shows the gas phase decomposition of hydrogen peroxide (H2O2) at 400 °C: 2H2O2(g)2H2O(g) + O2(g)

This reaction is second order in H2O2 with a rate constant of 0.650 M-1s-1 and the initial concentration of H2O2 is 0.600 M.
a) What is the concentration of H2O2 after 40 second? (extra credit) What is the concentration of O2

at that time?
b) Calculate the first and second half-life of this reaction. Are the values same or different?

Explain your answer.

Answer 1

The reaction:

2H₂O₂ (g) 2H₂O (g) + O₂ (g)

Initial concentration of H₂O₂ = 0.6 M

Rate constant = 0.650 M^-1sec^-1

Rate of the reaction is given by,

-dC/dt = kC² ,

where C represents the concentration of reactant , here it is H₂O₂

-dc/C² = kdt

On integrtaing,

1/C = kt + A

at t =0 , C = C₀

1/C₀ = A

(1/C) - (1/C₀) = kt

Given that us k = 0.650 M^-1sec^-1 , and C₀ = 0.6 M and t=40 sec. Substituting all these values give,

(1/C) - (1/0.6) = 0.650*40

(1/C) = 25 + 1.667

C = 0.0375 M

Concentration of H₂O₂ after 40 sec = 0.0375 M

From reaction stoichiometry, each mole of H₂O₂ gives one mole of O₂ . So, the concentration of O₂ is equal to the amount of consumption of H₂O₂ .

Amount of H₂O₂ consumed = C₀ - C = 0.6 - 0.0375 = 0.5625 M

Concentraton of O₂ = 0.5625 M

b) The integrated equation for second order reaction is :

(1/C) - (1/C₀) = kt

First half life calculation,

The initial concentration = C₀

The final concentration = C = Half of initial concentration = 0.5*C₀

t = t_1/2, Substituting all this in the equation,

(1/0.5*C₀) - (1/C₀) = kt_1/2

kt_1/2 = 1/C₀.

t_1/2 = 1/(kC₀) = 1/(0.650*0.6) = 2.564 sec

Second Half Life calculation.

Half-life time is given as,

kt_1/2 = 1/C₀.

Here the initial concentration C₀ is half of the what it was at t=0. The second half life time is calculated after one-half of the reactants have reacted. Out initial concentration was 0.600 M. So the concentration has reduced to 0.5*0.60 = 0.3 M. This is the initial concentration for second half life.

kt_1/2 = 1/C₀.

t_1/2 = 1/(k*C₀) = 1/(0.65*0.3) = 5.128 sec

Second half-life is more than the first half-life for the above reaction.It can be seen that the half life depends inversely on the initial concentration of the reactant. More initial concentration smaller half-life and less initial concentration gives longer half-life time.