wiz-icon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
Question

The decomposition of hydrogen peroxide in an aqueous solution is a reaction of first order. It can be followed by titrating 10 mL portions of reaction mixture at various times from the beginning of reaction against a standard solution of KMnO4. Volume of KMnO4 solution used in each case is proportional to the remaining concentration of H2O2. From the following data, calculate the rate constant of the reaction.
Time (in seconds)06001200KMnO4 sol. used (in mL)22.813.88.20

A
k=8.44×104sec1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
k=8.44×102sec1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
k=2.44×106sec1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
k=2.22×104sec1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A k=8.44×104sec1
Given that decomposition of H2O2 in an aqueous solution follows first order kinetics.
For first order reactions,
k=1tln[A]0/[A]t
where, k is the rate constant,
t= time
[A]0= Initial concentration of reactant (i.e. H2O2)
[A]t= concentration of reactant at t=t
Now, concentration of H2O2 present at any time is proportional to volume of KMnO4 used in titration.
Now, given that
at, t=0 volume of KMnO4 used
22.8 mL[A]0
At t=1200s Volume of KMnO4 used8.20 mL[A]t
So,
k=1/1200×ln(22.8/8.2)1/1200=8.33×104sec1
This approximation works given the options in the question, the nearest one is 8.44×104sec1

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Rate Constant
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon