4.38 The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 * 1010s –1, calculate k at 318K and Ea.
4.38 The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 * 1010s –1, calculate k at 318K and Ea.
4.38 We know, time t = (2.303/k) * log ( [R]0/ [R])
Where, k- rate constant
[R]0-Initial concentration
[R]-Concentration at time 't'
At 298K, If 10% is completed, then 90% is remaining. t = (2.303/k) * log ( [R]0/0.9 [R]0)
t = (2.303/k) * log (1/0.9) t = 0.1054 / k
At temperature 308K, 25% is completed
Similar Questions for you
Kindly go through the solution
Ea = 216.164kJ/mol 216
Reaction rate is used to measure how fast or slow reactions occur per unit time. The rate constant is a proportionality factor that remains constant for every reaction.
Yes, in elementary reactions, order and molecularity can be the same, but this is not always the case because order is an experimental quantity, and molecularity is a theoretical concept.
Reaction Kinetics, also known as chemical kinetics, is the study of the rate of chemical reaction and the factors affecting the reaction rate, such as temperature, concentration, and catalyst.
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else.
On Shiksha, get access to
Learn more about...

Chemistry Ncert Solutions Class 12th 2023
View Exam DetailsMost viewed information
SummaryDidn't find the answer you were looking for?
Search from Shiksha's 1 lakh+ Topics
Ask Current Students, Alumni & our Experts
Have a question related to your career & education?
See what others like you are asking & answering