| ⇦ |
| ⇨ |
The rate constant of the reaction A → B is 0.6 ˣ 10⁻³ mol per second. If the concentration of A is 5 M, then concentration of B after 20 minutes is
Options
(a) 3.60 M
(b) 0.36 M
(c) 0.72 M
(d) 1.08 M
Correct Answer:
0.72 M
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - Calcination is used in metallurgy to remove
- The crystal structure of solid Mn(II) oxide is
- An atom has electronic configuration 1s²2s²2p⁶3s²3p⁶3d³4s², you will place it in
- As the alkaline earth metals (except Be) tend to lose their valence electrons
- Correct order of first IP among following elements Be, B, C, N, O is
Topics: Electrochemistry and Chemical Kinetics
(87)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Calcination is used in metallurgy to remove
- The crystal structure of solid Mn(II) oxide is
- An atom has electronic configuration 1s²2s²2p⁶3s²3p⁶3d³4s², you will place it in
- As the alkaline earth metals (except Be) tend to lose their valence electrons
- Correct order of first IP among following elements Be, B, C, N, O is
Topics: Electrochemistry and Chemical Kinetics (87)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends

Reaction is of zero order as the unit of rate constant is mol L⁻¹ S⁻¹
.·. Concentration of B =k × t =0.6×10⁻³ × 20 × 60 = 0.72 M
Let at time t=0 (A)= a and (B)=0
At time t=(20*60)sec (A)= (a-x) and (B)=x then
For 1st order reaction
(A’) =(A) – kt
: a-x = a – 0.6*10-3*20*60
Solving it will give
X= 0.72 M