Question

For a certain reaction, the activation energy is 52.1 kJ/mole. By what ratio will the rate constant change if the temperature is decreased from 175C to 75C?

Answer 1

The value of the ratio of the rate constant is
`4.02*10^(-3)`

Step-by-step explanation:

Given that,

Energy = 52.1 kJ/mole

Initial temperature = 175°C

Final temperature = 75°C

We need to calculate the ratio of the rate constant change

Using formula of rate constant

`(k_(2))/(k_(1))=(E)/(R)((1)/(T_(2))-(1)/(T_(1)))`

Where. T₁= initial temperature

T₂ = final temperature

E = energy

R = gas constant

Put the value into the formula

`(k_(2))/(k_(1))=(52.1)/(8.31)((1)/(273+75)-(1)/(273+175))`

`(k_(2))/(k_(1))=0.00402`

`(k_(2))/(k_(1))=4.02*10^(-3)`

Hence, The value of the ratio of the rate constant is
`4.02*10^(-3)`