Question

Exercise 13.97

The kinetics of the following second-order reaction were studied as a function of temperature:
C2H5Br(aq)+OH?(aq)?C2H5OH(l)+Br?(aq)

Temperature (?C)	k (L/mol?s)
25	8.81

Answer 1

We will use the relation,

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

a) First calculate the Ea for the reaction.

Lets consider the reading at 25 oC and 35 oC

T1 = 25 oC = 25 + 273 = 298 K

T2 = 35 oC = 35 + 273 = 308 K

k1 = 8.81 x 10^-5 L/mol.s

k2 = 0.000285 L/mol.s

R = gas constant = 8.314 J/mol.K

Ea = unknown

Feeding the values into the above equation we get,

ln(8.81 x 10^-5 / 0.000285 ) = -Ea/8.314 (1/298 - 1/308)

Solving for Ea,

Ea = 89.59 kJ/mol

b) Now let us calculate rate constant k at 15 oC = 288 K

Again taking reading at 25 oC and feeding the values we have,

ln(k/8.81 x 10^-5) = -89.59/8.314 (1/288 - 1/298)

Solving for k = 8.80 x 10^-5 L/mol.s

Therefore, rate constant at 15 oC = 8.80 x 10^-5 L/mol.s

b) First calculate rate constant k at 90 oC = 363 K

here, k1 is at 25 oC and k2 is at 90 oC

Again taking reading at 25 oC and feeding the values we have,

ln(8.81 x 10^-5/k) = -89.59/8.314 (1/288 - 1/363)

Solving for k = 8.87 x 10^-5 L/mol.s

Now the initial rate with given values,

[C2H5Br] = 0.155 M

[OH-] = 0.260 M

k = 8.87 x 10^-5 L/mol.s (calculated)

Feeding all the values we get,

rate = k[A][B]

rate = 8.87 x 10^-5 x 0.155 x 0.260

       = 3.57 x 10^-6 L/mol.s

thus, the intial rate will be 3.57 x 10^-6 L/mol.s