Question

A reaction has a rate constant of 0.000122s-1 at27oC and 0.228s-1 at 77oC. a)Determine the activation energy of the reaction. b)What is the value of the rate constant at17oC

Answer 1

a)
`Ea=-131.522x10^(3)(J)/(mol)`

b)
`k_3=1.98x10^(-5)s^(-1)`

Step-by-step explanation:

Hello,

a) In this case, for this kinetics problem, we can consider the temperature-dependent Arrhenius equation:

`ln((k_2)/(k_1) )=(Ea)/(R)((1)/(T_2)-(1)/(T_1) )`

In such a way, for the given temperatures and rate constant, we compute the activation energy as follows:

`ln((0.228s^(-1))/(0.000122s^(-1)) )=(Ea)/(R)((1)/((77+273)K)-(1)/((27+273)K) )\\\\7.533=(Ea)/(R)*-4.762x10^(-4)K^(-1)\\ \\Ea=R(7.533)/(-4.762x10^(-4)K^(-1)) \\`

`Ea=8.314(J)/(mol*K)*-15819.3K\\\\Ea=-131.522x10^(3)(J)/(mol)`

b) In this case, we use the previously computed activation energy in order to compute the rate constant at the asked 17°C:

`k_3=k_1exp((Ea)/(R)((1)/(T_2)-(1)/(T_1) ))\\\\k_3=0.000122s^(-1)exp[(-131.522x10^3(J)/(mol) )/(8.314(J)/(mol*K))*((1)/((17+273)K) -(1)/((27+273)K) )]\\\\k_3=1.98x10^(-5)s^(-1)`

Best regards.