Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  •  For a zero order reaction, <img alt="\mathrm{K}=2 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}" src="/latex-image/?%5Cmathrm%7BK%7D%3D2%20%5Ctimes%2010%5E%7B-3%7D%20%5Cmathr

 For a zero order reaction, \mathrm{K}=2 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} . If initial concentration of the reactant is 1.0 mol \mathrm{L^{-1}}  the concentration after 5 minutes would be

Option: 1

1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \\


Option: 2

0.6 \mathrm{~mol} \mathrm{~L}^{-1} \\


Option: 3

0.4 \mathrm{~mol} \mathrm{~L}^{-1}


Option: 4

1.0 \mathrm{~mol} \mathrm{~L}^{-1}


Answers (1)

best_answer

For a zero order reaction, concentration of reactant at any given time \mathrm{t, A=A o-K t \quad} where Ao is initial concentration and \mathrm{t}  is time elapsed So after 5 minutes, \mathrm{A=1.0 \mathrm{~mol} \mathrm{~L}^{-1}-2.0 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \times 5 \times 60 \mathrm{~s}}
\mathrm{ \Rightarrow A=1-0.6 \mathrm{~mol} \mathrm{~L}^{-1} } 
\mathrm{ \Rightarrow A=0.40 \mathrm{~mol} \mathrm{~L}^{-1} }.

Posted by

chirag

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

75% criteria for JEE Main 2025 not removed; eligibility criteria for admission to IITs
anam-khan

BTech Admissions 2025 LIVE: AEEE phase 1 exam date out, JEE Mains registration ongoing; BITSAT, VITEEE updates
anam-khan

JEE Main 2025 Eligibility Criteria: What is state code of eligibility? Here’s list of qualifying exams

Latest Question