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kondaur [170]
3 years ago
11

A second-order reaction has a rate law: rate = k[a]2, where k = 0.150 m−1s−1. If the initial concentration of a is 0.250 m, what

is the concentration of a after 5.00 minutes? Be sure to pay attention to the units in this problem so that they cancel out. G
Chemistry
1 answer:
Cerrena [4.2K]3 years ago
7 0

Rate law for the given 2nd order reaction is:

Rate = k[a]2

Given data:

rate constant k = 0.150 m-1s-1

initial concentration, [a] = 0.250 M

reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s

To determine:

Concentration at time t = 300 s i.e. [a]_{t}

Calculations:

The second order rate equation is:

1/[a]_{t} = kt +1/[a]

substituting for k,t and [a] we get:

1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M

1/[a]t = 49 M-1

[a]t = 1/49 M-1 = 0.0204 M

Hence the concentration of 'a' after t = 5min is 0.020 M



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Consider the following mechanism for the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution. H+ + H2O2 ? H3
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<u>Explanation:</u>

Rate law is the expression which is used to express the rate of the reaction in terms of the molar concentration of reactants where each term is raised to the power their stoichiometric coefficient respectively from a balanced chemical equation.

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

The chemical equation for the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution follows:

2H^++2Br^-+H_2O_2\rightarrow Br_2+2H_2O

The intermediate reaction of the mechanism follows:

<u>Step 1:</u>  H^++H_2O_2\rightleftharpoons H_3O_2^+;\text{ (fast)}

<u>Step 2:</u>  H_3O_2^++Br^-\rightarrow HOBr+H_2O;\text{(slow)}

<u>Step 3:</u>  HOBr+H^++Br^-\rightarrow Br_2+H_2O;\text{(fast)}

As, step 2 is the slow step. It is the rate determining step

Rate law for the reaction follows:

\text{Rate}=k[H_3O_2^+][Br^-]          ......(1)

As, [H_3O_2^+] is not appearing as a reactant in the overall reaction. So, we apply steady state approximation in it.

Applying steady state approximation for [H_3O_2^+] from step 1, we get:

K=\frac{[H_3O_2^+]}{[H^+][H_2O_2]}  

[H_3O_2^+]=K[H^+][H_2O_2]

Putting the value of [H_3O_2^+] in equation 1, we get:

\text{Rate}=k.K[H^+][H_2O_2][Br^-]\\\\\text{Rate}=k'[H+][H_2O_2][Br^-]

Hence, the rate law for the reaction is \text{Rate}=k'[H+][H_2O_2][Br^-]

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