Rate Law
A detailed exploration of the concept of rate law in chemistry, focusing on its definition, importance, and application in reaction kinetics.
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Rate Law
For the following reaction, use the experimental data to determine the rate law.
2 BrO3–(aq) + 5 HSO3–(aq) → Br2(g) + 5 SO42–(aq) + H2O(l) + 3 H+(aq)
Solution:
Experiment
Initial
[BrO3]
(mmol•L−1)
Initial
[HSO3–]
(mmol•L−1)
Initial Rate of
Formation of Bromine
(mmol•L−1•s−1)
1 4.0 6.0 1.60
2 2.0 6.0 0.80
3 4.0 3.0 0.40
The general equation for the rate law can be expressed as:
Rate = k [BrO3]x. [HSO3⁻]y
From the first experiment, we have:
1.6 = k [4]x. [6]y
So k = 1.6 / [4]x. [6]y…………………………………………………………………………..1
Similarly from experiment 2 we have:
0.8 = k [2]x. [6]y
So k = 0.8 / [2]x. [6]y…………………………………………………………………………..2
Comparing eq 1 and 2:
1.6/ 0.8 = [4]x/ [2]x
=> 2 = [2]x
=> x = 1………………………………………………………………………………………3
Also from experiment 3 we have:
For the following reaction, use the experimental data to determine the rate law.
2 BrO3–(aq) + 5 HSO3–(aq) → Br2(g) + 5 SO42–(aq) + H2O(l) + 3 H+(aq)
Solution:
Experiment
Initial
[BrO3]
(mmol•L−1)
Initial
[HSO3–]
(mmol•L−1)
Initial Rate of
Formation of Bromine
(mmol•L−1•s−1)
1 4.0 6.0 1.60
2 2.0 6.0 0.80
3 4.0 3.0 0.40
The general equation for the rate law can be expressed as:
Rate = k [BrO3]x. [HSO3⁻]y
From the first experiment, we have:
1.6 = k [4]x. [6]y
So k = 1.6 / [4]x. [6]y…………………………………………………………………………..1
Similarly from experiment 2 we have:
0.8 = k [2]x. [6]y
So k = 0.8 / [2]x. [6]y…………………………………………………………………………..2
Comparing eq 1 and 2:
1.6/ 0.8 = [4]x/ [2]x
=> 2 = [2]x
=> x = 1………………………………………………………………………………………3
Also from experiment 3 we have:
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