Chemical Kinetics and Nuclear ChemistryHard
Question
The first order rate constant for a certain reaction increase from 1.667 × 10-6 s-1 at 727oC to 1.66 × 10-4s-1 at 1571oC. The rate constant at 1150oC assuming constancy of activation energy over the given temprature range is
[Given : log 19.9 = 1.299][2009]
[Given : log 19.9 = 1.299][2009]
Options
A.3.911 × 10-5s-1
B.1.139 × 10-5s-1
C.3.318 × 10-5s-1
D.1.193 × 10-5s-1
Solution
log 
or 2.030 log
2.030 log
2.030 × 2 =
.......(1)
∴
2.030log
=
......(2)
Dividing equation (2)by equation (1)
∴ log
2 ×
= 1.299
On taking Antilog., k3 = 19.9
∴ k3 = 19.9 × 1.667 × 10-6 = 3.318 × 10-5s-1
or 2.030 log
2.030 log
2.030 × 2 =
.......(1)∴
2.030log
=
......(2)Dividing equation (2)by equation (1)
∴ log
2 ×
= 1.299On taking Antilog., k3 = 19.9
∴ k3 = 19.9 × 1.667 × 10-6 = 3.318 × 10-5s-1
Create a free account to view solution
View Solution FreeTopic: Chemical Kinetics and Nuclear Chemistry·Practice all Chemical Kinetics and Nuclear Chemistry questions
More Chemical Kinetics and Nuclear Chemistry Questions
Which reaction will occur at the fastest rate in water...The rate constant of a reaction is given by K = 2.1 ×1010 exp (-2700 / RT) it means that is :...For the elementary reaction M → N, the rate of disappearance of M increases by a factor of 8 upon doubling the con...Select correct statement (s) :...An organic compound undergoes first order decomposition. The time taken for decomposition to $\left( \frac{1}{8} \right)...