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⁺]                         [S₂O₃^2-]
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⁺]                         [S₂O₃^2-]
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

Answer 1

1). The rate law for the clock reaction is:

Rate = k [ I^-] [ BrO₃^-] [H⁺]

time period = 29 sec

rate = [S₂O₃^2-] / 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