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 25.4 seconds Calculate k in the rate law.

A clock reaction is run at 18 ºC with the following initial concentrations

[I^-]               [BrO₃^-]            [H⁺]                         [S₂O₃^2-]
0.002                  0.008              0.02                          0.0001  

Then the experiment is repeated at 31 ºC, the rate constant k is found to be 2.3 times larger.

Calculate the activation energy

Answer 1

Q) Then the experiment is repeated at 31 ºC, the rate constant k is found to be 2.3 times larger.

Calculate the activation energy?

Ans:

Arrhenius equation can be written as

In (k2/k1) = Ea/R (1/T1 - 1/T2)

Ea = R ln (k2/k1) / (1/T1 - 1/T2) -- Eq (1)

Given that rate constant k is found to be 2.3 times larger.

Hence, Initial rate of rection = k1

Final rate of reaction k2 = 2.3 k1

Initial temperature T1 = 18 oC = 18 + 273 K = 291 K

Final temperature T2 = 31 oC = 31 + 273 K = 304 K

substitute all these velues in eq (1),

Ea = R ln (k2/k1) / (1/T1 - 1/T2) -- Eq (1)

= (8.314 J/K/mol) [ In (2.3k1/k1)] / [(1/291) - (1/304)]

= 47122 J/mol

= 47122 J/mol

Ea = 47122 J/mol

Therefore, activation energy of the reaction = 47122 J/mol