Question

The rate constant k for a certain reaction is measured at two different temperatures:

temperature k

376.0 °c 4.8 x 108

280.0 °C 2.3 x 10 8

Assuming the rate constant obeys the Arrhenius equation, calculate the activation energy Ea for this reaction.

Round your answer to 2 significant digits.

Answer 1

Ea=5.5 Kcal/mole

Step-by-step explanation:

Let rate constant are
`K_1` and
`K_2` at temperature
`T_1` and
`T_2`

By using Arrhenius equation at two different two different temperature,

`Log K_1/K_2 =E_a/2.303R*(1/T_2 -1/T_2 );T_1=273+376=649K ;K_1=4.8*10^8;T_2=273+280=553K ;K_2=2.3*10^8;R=2 cal/(mole.K);Log (4.8*10^8)/(2.3*10^8 )=E_a/2.303R*(1/553K-1/649); Log 4.8/2.3=E_a/2.303R*96/358897 ;0.32=E_a/2.303R*96/358897;E_a=(0.32*2.303R*358897)/96;`

By putting value of R=2 cal/mole.K

`E_a=5510.265cal/mole;`

By rounding off upto 2 significant figure;

`E_a=5.5Kcal/mole;`