MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 12
        For a first order reaction t0.75 is 138.6sec. it`s specific rate constant is (in s^-1)1. 10^-22. 10^-43. 10^-64. 10^-5
2 years ago

Answers : (2)

Gopal
54 Points
							
For first order reaction  half life and time required for 75% completion are related as  t0.75=2*t0.5     and specific rate constant is related with half life as  k=\tfrac{0.693}{t0.5 }
\therefore t0.5=138.6 /2 =69.25 sec
k=\tfrac{0.693}{69.25 }  ,
k=10 ^{^{-2}}
2 years ago
Ujwal dayma
15 Points
							
T0.75=138.6sec
T3/4=138.6sec
2t1/2=138.6sec
T1/2=138.6/2sec
T1/2=69.3
As it is a first order reaction so,
t1/2=0.693/k
Therefore,k=0.693/t1/2
0.693/69.3=
10-2
 
11 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 141 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details