1) If the rate law for the clock reaction is: Rate = k [ I - ] [ BrO 3 - ] [H +
ID: 775762 • Letter: 1
Question
1) If the rate law for the clock reaction is:
Rate = k [ I-] [ BrO3-] [H+]
A clock reaction is run with the following initial concentrations:
[I-] [BrO3-] [H+] [S2O32-]
0.002 0.008 0.02 0.0001
The reaction time is 29 seconds
Calculate k in the rate law:
2)If the rate law for the clock reaction is:
Rate = k [ I-] [ BrO3-] [H+]
A clock reaction is run at 21 C with the following initial concentrations
[I-] [BrO3-] [H+] [S2O32-]
0.002 0.008 0.02 0.0001
Then the experiment is repeated at 33 C, the rate constant k is found to be 2.1 times larger.
Calculate the activation energy
Explanation / Answer
1). The rate law for the clock reaction is:
Rate = k [ I-] [ BrO3-] [H+]
time period = 29 sec
rate = [S2O32-] / t
= 0.0001 / 29
= 3.45x10-6 M/s
Now, Rate = k [ I-] [ BrO3-] [H+]
3.45x10-6 = k * 0.002*0.008*0.02
3.45x10-6 = k * 3.2x10-7
k = 10.78 M-2 s-1
2). T1 = 21 oC = 294 K
T2 = 33 oC = 306 K
k2 = 2.1*K1
ln (K2 / K1) = - Ea /R (1 / T1 - 1 / T2)
ln 2.1 = (- Ea / 8.314)* (1 / 294 - 1 / 306)
0.7419 = (- Ea / 8.314) * (0.00340 - 0.00326)
0.7419 = (- Ea / 8.314) * 0.00014
Ea = - 44.058 kJ/ mol
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