1st, 2nd,3rd Order

Themes > Science > Chemistry > General Chemistry > Solution and Solubility > Solubility and Redox Reactions > Chemical Kinetics and Rate Laws > 1st, 2nd,3rd Order

_{To determine order, a table of experimentally measured concentrations and
times must be presented. Below is an example of the data that must be
presented to you to solve the question of what order the reaction is.

Reaction:

BrO₃^- + 5Br^- + 6H⁺ -----> 3
Br₂ + 3H₂O

Experiment
[BrO₃^-]
[Br^-]
[H⁺]
Initial Rate

1
.1
.1
.1
8 * 10 ^-4

2
.2
.1
.1
1.6 * 10 ^-3

3
.2
.2
.1
3.2 * 10 ^-3

4
.1
.1
.2
3.2 * 10 ^-3

To solve this problem, the first step is to
find 2 concentrations in BrO₃^- that are the same and
2 concentrations in Br^- that are the same for the same
experiment number.

In experiment 1 and experiment 4, BrO₃^-
has a concentration of .1 and Br^- has a concentration of .1.
Now look at the concentrations of H⁺ and decide how it changes.
The concentration doubles. Then you have to compare it to the quotient of
the initial rates. So:

2ⁿ = 3.2 * 10^-3
/ 8 * 10^-4

n = 2, so the order for BrO₃^- is second
Next, we need to find 2 experiments where
the concentrations of BrO₃^- in each experiment are
the same and the concentrations of H⁺ in each experiment are
the same.

In experiment 2 and experiment 3, BrO₃^-^{has a concentration of .2 and H⁺ has a concentration of .1. Now
look at the concentration of Br^- and decide how it changes. The
concentration doubles. Then you have to compare it to the quotient of the
initial rates. So:}
2^p = 3.2 * 10 ^-3
/ 1.6 * 10 ^-3

p = 1, so the order of Br^- is 1
Finally, we must do this same procedure
again for BrO₃^-. We must find 2 experiments where
the concentrations of Br^- in each experiment are the same and
teh concentrations of H⁺ in each experiment are the same.

In experiment 1 and experiment 2, Br^-
has a a concentration of .1 and H⁺ has a concentration of .1.
Now look at the concentrations of H⁺ and decide how it changes.
The BrO₃^- concentration doubles. Then you have to
compare it to the quotient of the initial rates. So:

2^q = 1.6 * 10 ^-3
/ 8 * 10 ^-4

q = 1, so the order of BrO₃^- is 1.
Finally, the general formula of the rate
law for the reaction of BrO₃^- + 5Br^- + 6H⁺
-----> 3 Br₂ + 3H₂O is:

Rate = k [BrO₃^-]¹
[Br^-]¹ [H⁺]²}

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