Question

Consider the following for the reaction at 300 K:3 ClO– (aq)  ClO3– (aq) + 2 Cl– (aq)ExperimentInitial [ClO–] (M)Initial Rate of Formation of ClO3– (aq)(M/min)10.4521.048 × 10–420.9034.183 × 10–4(7) (4 pts) What is the order of the reaction with respect to ClO– (aq)?A) 0B) 1C) 2(8) (4 pts) For experiment #2, what is the initial rate of consumption of ClO–

Answer 1

Answer:

Second order

Δ[ClO⁻]/Δt = -  4.183 x 10⁻⁴ M/min

Explanation:

Given the data:

Experiment #         [ClO–] (M)   Initial Rate of Formation of ClO3– (M/min)

         1                      10.452                     1.048 x 10⁻⁴

         2                     20.903                    4.183 x 10⁻⁴

we need to determine the order of the reaction with respect to ClO⁻.

We know the rate law for this reaction will have the form:

Rate = k [ClO⁻]^n

where n is the order of the reaction. Thus, what we need to do is to study the dependence of the initial rate on n for the experiment.

If the reaction were zeroth order the rate would not change, so we can eliminate n= 0

If the reaction were first order, doubling the concentration of   [ClO–] , as it was done exactly in experiment # 2, the initial rate should have doubled, which is not the case.

If the reaction were second order n: 2, doubling the concentration of  [ClO–] , should quadruple the initial rate of formation of ClO3–, which is what it is observed experimentally. Therefore the reaction is second order respect to ClO–.

The initial rate of consumption of ClO⁻ is the same as the rate of formation of ClO₃⁻ since:

Δ = - Δ[ClO⁻]/Δt =  + Δ[ClO₃⁻]/Δt = + 1/2 [Cl⁻] /Δt

where t is the time.

from the coefficients of the balanced chemical equation.

- Δ[ClO⁻]/Δt =  + Δ[ClO₃⁻]/Δt  = + 1/2 [Cl⁻ ] = rate

Δ[ClO⁻]/Δt = -  4.183 x 10⁻⁴ M/min