(13.39) A reaction has a rate constant of 1.24×10−2 /s at 400. K and 0.689 /s...

(13.39) A reaction has a rate constant of 1.24×10⁻² /s at 400. K and 0.689 /s at 450. K.

1. Determine the activation barrier for the reaction.

2. What is the value of the rate constant at 425 K?

Expert Solution

1. Determine the activation barrier for the reaction.

activation barrier or Ea also can be calculated from the values of k at only two temperatures by the formula;

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

T1 = 400 K
k1 = 1.24 x 10^-2 s^-1

T2 = 450 K
k2 = 0.689 s^-1

R = 8.3145 J/K.mol (Gas constant)

ln(0.689 s^-1 / 1.24 x 10^-2 s^-1) = Ea / 8.3145 (1/400 - 1/450)

4.02 = Ea / 8.3145 (2.5*10^-3- 2.22*10^-3)

4.02 = Ea / 8.3145 (2.8 x 10^-4)
Ea = (4.02 x 8.3145) / 2.8 x 10^-4

= 1.2x 10^5 J/mol

2. What is the value of the rate constant at 425 K?

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

T1 = 400 K
k1 = 1.24 x 10^-2 s^-1

T2 = 425 K
k2 = ? s^-1

R = 8.3145 J/K.mol (Gas constant)
Ea= 1.2x 10^5 J/mol

ln k2- ln1.24 x 10^-2 s^-1) = 1.2x 10^5 J/mol / 8.3145 (1/400 - 1/425)

ln k2+ 4.39 s^-1 = 1.2x 10^5 J/mol / 8.3145 (2.5*10^-3 -2.35*10^-3)
ln k2+ 4.39 s^-1 = 1.2x 10^5 J/mol / 8.3145 (1.5*10^-4 )

ln k2+ 4.39 s^-1 = 2.16

ln k2 = -2.23

k2 = 0.108 s-1

queen_honey_blossom answered 2 years ago

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