Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The reaction between  and <img alt="\mathrm{Y}" src="https://entrancecorner.oncodecogs.com/gif.la

The reaction between \mathrm{X} and \mathrm{Y} is first order with respect to \mathrm{ X} and zero order with respect to\mathrm{ Y.}

 \mathrm{Experiment }

 \mathrm{ \frac{[\mathrm{X}]}{\mathrm{mol} \mathrm{L}^{-1}}} \frac{[\mathrm{Y}]}{\mathrm{mol} \mathrm{L}^{-1}} \mathrm{\frac{\text { Initial rate }}{\operatorname{mol~L}^{-1} \min ^{-1}}}

 I 

 \mathrm{0.1 } \mathrm{0.1} \mathrm{2 \times 10^{-3}}
II \mathrm{L} \mathrm{0.2} \mathrm{4 \times 10^{-3}}
III \mathrm{0.4 } \mathrm{0.4} \mathrm{M \times 10^{-3}}
IV \mathrm{0.1} \mathrm{0.2} \mathrm{2 \times 10^{-3}}

 Examine the data of table and calculate ratio of numerical values of  \mathrm{M} and \mathrm{L}. (Nearest Integer) 

Option: 1

40


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Reaction is first order with \mathrm{X} and Zero order with \mathrm{Y}, So,
\mathrm{rate= K[X]^{1}[Y]^{0}}
\mathrm{rate\, \alpha \, [X]^{1}}

Now,
From Experiment I and II.
\mathrm{\frac{r_{II}}{r_{I}}= \frac{4\times 10^{-3}}{2\times 10^{-3}}= \frac{k[X]_{II}}{k[X]_{I}}= \frac{[X]_{II}}{[X]_{I}}= \frac{L}{0.1}}
\mathrm{L= [X]_{II}= 2 \times0.1= 0.2}

Now, From experiment III and IV
\mathrm{\frac{r_{III}}{r_{IV}}= = \frac{k[X]_{III}}{k[X]_{IV}}= \frac{[X]_{III}}{[X]_{IV}}}
\mathrm{\frac{M\times10^{-3}}{2\times10^{-3}}= \frac{0.4}{0.1}}
\mathrm{M= 8}

Now, ratio of numerical values of M and L
\mathrm{\frac{M}{L}= \frac{8}{0.2}= 40}

Ans = 40

Posted by

himanshu.meshram

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

JEE Main 2025: BTech ECE cut-offs for female candidates go up at top NITs
anam-khan

JEE Main 2025 Session 2 City Slip (OUT) Live: Admit card likely on March 29 at jeemain.nta.nic.in; cut-offs
anam-khan

AESL expands Aakash Digital 2.0 to offer AI-powered coaching for NEET, JEE, Olympiads

Latest Question