If the half-lives of the first order reaction at 350 K and 300 K are 2 and 20 seconds respectively, the activation energy of the reaction in kJ `"mol"

Question

If the half-lives of the first order reaction at 350 K and 300 K are 2 and 20 seconds respectively, the activation energy of the reaction in kJ `"mol"

If the half-lives of the first order reaction at 350 K and 300 K are 2 and 20 seconds respectively, the activation energy of the reaction in kJ `"mol"^(-1)` is :
A. `40.2`
B. `20.1`
C. `60.3`
D. `30.2`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

Related Questions

The rate constant for a first order reaction becomes six times when the temperature is raised from 350 to 400 K. Calculate the energy of activation fo

2 Answers 4 Views

For a reaction, the activation energy is zero. What is the value of rate constant at `300 K` if `k=1.6 xx 10^(6)s^(-1)` at 280 K `(R= 8.31 JK^(-1)mol^

2 Answers 4 Views

In the first order reaction, half of the reaction is completed in 100 seconds. The time for `99%` of the reaction to occur will be

2 Answers 4 Views

The rate of a reaction doubles when the tempeature changes from 300 K to 310 K. Activation energy for the reaction is: `(R = 8.314 JK^(-1) mol^(-1), l

2 Answers 5 Views

Calculate `DeltaG` for (i) `H_(2)O(l,2 atm , 300 K) to H_(2)O(g, 2 atm , 300 K ) ` (ii) `H_(2)O(l,P" "atm , 300 K) to H_(2)O(g, P" "atm , 300 K )` Cac

2 Answers 5 Views

In the decomposition of `H_2O_2` at 300 K, the energy of activation was found to be `18 kcal//"mol"` white it decreases to `6 kcal//"mol"` when the de

2 Answers 5 Views

The energy of activation for a reaction is `100 KJ mol^(-1)`. The peresence of a catalyst lowers the energy of activation by `75%`. What will be the e

2 Answers 8 Views

The activation energy of exothermic reaction `ArarrB` is 80 kJ `"mol"^(-1)` . The heat of reaction is `200 kJ "mol"^(-1)`. The activation energy for t

2 Answers 5 Views

A reaction takes place in three steps with individual rate constant and activation energy, `{:(,"Rate constant","Activation energy"),("Step 1",k_(1),E

2 Answers 5 Views

Lacto bacillus has generation time 60 min. at 300 K and 40 min. at 400 K. Determine activation energy in `(kJ)/(mol)` . (`R = 8.3 J K^ (–1) mol^(–1)`

2 Answers 9 Views

Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

`log((k_(2))/(k_(1)))=(E_(a))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]`
`log.((t_(1//2))_(t))/((t_(1//2))_(2))=(E_(a))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]`
`log. ((20)/(20))=(E_(a))/(2.303xx8.314)[(1)/(300)-(1)/(350)]`
`E_(a)=(1xx2.303xx8.314xx300xx350)/(50)J "mol"^(-1)`
`=40208J "mol"^(-1)`
`=40.2kJ"mol"^(-1)`

5 views Answered 2023-02-05 15:29