At 40 degrees Celsius CH2O2(aq) will decompose according to the following reaction: 2H2O2(aq) 2H2O(l) + O2(g)
The following data were collected for the concentration of H2O2 at various times.
Times(s) [H2O2](M)
0 1.000
2.16x10^4 0.500
4.32x10^4 0.250
a. Calculate the average rate of decomposition of H2O2 between 0 and 2.16 x 10^4 s. Use this rate to calculate the average rate of production of O2(g) over the same time period.
b. What are these rates for the time period 2.16 3 x10^4 s to 4.32 x 10^4 s?

A. Rate of decomposition of H₂O₂ = 2.31 × 10-⁵ M/s

Average rate of production of O₂ = 1.16 × 10-⁵ M/s

B. Rate of decomposition of H₂O₂ = 1.16 × 10-⁵ M/s

Average rate of production of O₂ = 5.8 × 10-⁶ M/s

Explanation:

The rate of a reaction is defined as the number of moles of reactant converted or the product formed per unit time.

Reaction rate = change in the concentration of product or reactant / time taken for the change.

A. i) For the decomposition of H₂O₂, the reaction equation is given below:

2 H₂O₂ (aq) > 2 H₂O (l) + O₂ (g)

a. the average rate of decomposition of H2O2 between 0 and 2.16 x 10^4 s is calculated as follows :

Rate of decomposition of H₂O₂ = change in concentration / change in time

Rate of decomposition of H₂O₂ = (0.500 - 1.000) M / (2.16 × 10⁴ - 0) s = 2.31 × 10-⁵ M/s

ii) The mole ratio of O₂ to H₂O₂ = 1 : 2 = 0.5

Average rate of production of O₂ = 0.5 × rate of decomposition of H₂O₂

Average rate of production of O₂ = 0.5 × 2.31 × 10-⁵ M/s = 1.16 × 10-⁵ M/s

B. The rate of decomposition of H₂O₂ for the time period 2.16 3 x10^4 s to 4.32 x 10^4 s is calculated as follows:

Rate of decomposition of H₂O₂ = change in concentration / change in time

Rate of decomposition of H₂O₂ = (0.250 - 0.500) M / ( 4.32 × 10⁴ - 2.16 × 10⁴) s = 1.16 × 10-⁵ M/s

ii) The mole ratio of O₂ to H₂O₂ = 1 : 2 = 0.5

Average rate of production of O₂ = 0.5 × rate of decomposition of H₂O₂

Average rate of production of O₂ = 0.5 × 1.16 × 10-⁵ M/s = 5.8 × 10-⁶ M/s

the ratio of the conjugate base to the acid required to attain the desired ph.

the answer is d.) potassium