Question

A reaction has a rate constant of 0.0117 s-1 at 400.0 k and 0.689 s-1 at 450.0 k. calculate the activation energy in kilojoules for this reaction.

Answer 1

By definition we have to:
 LOG (k2 / k1)=(-Ea/R)*(1/T1-1/T2)
 Where,
 k1 = 0.0117 s-1
 K2 = 0.689 s-1
 T1 = 400.0 k
 T2 = 450.0 k
 R is the ideal gas constant
 R = 8.314 KJ / (Kmol * K)
 Substituting
 ln (0.0117/0.689)=-Ea/(8.314)*((1/400)-(1/450))
 Clearing Ea:
 Ea = 122 kJ
 answer
 the activation energy in kilojoules for this reaction is
 Ea = 122 kJ

Answer 2

Answer: The activation energy of the reaction is 122.007 kJ.

Explanation:

Rate constant at $T_1,K_1=0.0117 s^{-1}$

$T_1=400 K$

Rate constant at $T_2,K_2=0.689 s^{-1}$

$T_2=450 K$

Activation of the energy is given by formula:

$\log\frac{K_2}{K_1}=\frac{E_a}{2.303\times R}\times \frac{T_2-T_1}{T_1T_2}$

$\log\frac{0.689 s^{-1}}{0.0117 s^{-1}}=\frac{E_a}{2.303\times 8.314 J/K mol}}\times \frac{450 K-400 K}{450 K\times 400 K}$

On solving for $E_a$

$E_a=122,007.88 Joules=122.007 kilo-Joules (1000 J = 1kJ)$

The activation energy of the reaction is 122.007 kJ.