Question

The rate constant for a certain chemical reaction is 77.0 s-1 at a temperature of 20.0°C. If the activation energy for this reaction is 52. kJ/mole, what is the rate constant at a temperature of 30.0°C?

Answer 1

`155.65\ s^(-1)` is the rate constant at a temperature of 30.0 °C.

Step-by-step explanation:

Using the expression,

`\ln (k_(1))/(k_(2)) =-(E_(a))/(R) \left ((1)/(T_1)-(1)/(T_2) \right )`

Wherem

`k_1\ is\ the\ rate\ constant\ at\ T_1`

`k_2\ is\ the\ rate\ constant\ at\ T_2`

`E_a` is the activation energy

R is Gas constant having value = 8.314 J / K mol

Thus, given that,
`E_a` = 52 kJ/mol = 52000 J/mol (As 1 kJ = 1000 J)

`k_2=?`

`k_1=77.0s^(-1)`

`T_1=20\ ^0C`

`T_2=30\ ^0C`

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (20 + 273.15) K = 293.15 K

T = (30 + 273.15) K = 303.15 K

`T_2=303.15\ K`

So,

`\ln (77.0)/(k_2)\:=-(52000)/(8.314)* \left((1)/(293.15)-(1)/(303.15)\right)`

`(77)/(k_2)=e^{-(52000)/(8.314)\left((1)/(293.15)-(1)/(303.15)\right)}`

`k_2=77e^{(520000)/(738852.06466)}`

`k_2=155.65\ s^(-1)`

`155.65\ s^(-1)` is the rate constant at a temperature of 30.0 °C.