Transcript Rate Laws - Schoolwires

How to solve
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This method requires that a reaction be run
several times.
The initial concentrations of the reactants are
varied.
The reaction rate is measured just after the
reactants are mixed.
Eliminates the effect of the reverse reaction.
A + 2B
[A]
.240
.240
.360
.120
.240
.0140
3C + D
[B]
.480
.120
.240
.120
.0600
1.35
Initial Rate
8.00
2.00
9.00
0.500
1.00
?
Rate = k[A]x[B]y
Take any two experiments where the [ ] of one
species is held constant.
8.00 = k[.240]x[.480]y
2.00 = k[.240]x[.120]y
4.00 = 4y
therefore, y = 1
Now repeat the process, holding the other [ ]
constant.
2.00 = k[.240]x[.120]y
.500 = k[.120]x[.120]y
4.00 = 2x
therefore, x = 2
Rate = k[A]2[B]
Determine the value of k with units:
8.00 = k[.240]2[.480]
290 L2/mol2 sec = k
Determine the initial rate for the last
experiment
rate = 290[.0140]2[1.35]
rate = .0767 mol/l sec
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For the reaction
BrO3- + 5 Br- + 6H+
3Br2 + 3 H2O
The general form of the Rate Law is Rate =
k[BrO3-]n[Br-]m[H+]p
We use experimental data to determine the
values of n, m, and p
Initial concentrations (M)
H+
0.10
Rate (M/s)
0.10
Br0.10
0.20
0.20
0.10
0.10
0.20
0.10
0.10
0.10
0.20
1.6 x 10-3
3.2 x 10-3
3.2 x 10-3
BrO3-
8.0 x 10-4
Now we have to see how the rate changes
with concentration
x=1
y=1
z=2
Rate Law = k[BrO3-][Br-][H+]2