Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

For a first order reaction mathrm{A} ightarrow mathrm{B}, the rate constant, mathrm{k}=5.5 imes 10^{-14} mathrm{~s}^{-1}. The time required for 67% completion of reaction is mathrm{x imes 10^{-1}} times the half life of reaction. The value of x is ________ (Nearest integer) 

mathrm{ ext { (Given : } log 3=0.4771 ext { ) }}

Option: 1

16


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

For the first order reaction,

\mathrm{kt=ln\left ( \frac{a_0}{a_t} \right ) ~and ~t_{1/2}=\frac{ln2}k}

Given, \mathrm{ t_{67%}=x \times 10^{-1} \times t_{1/2}}

So, we can write 

\mathrm{\Rightarrow \frac{\ln \left(\frac{a_0}{a_0-0.67 a_0}\right)}{k}=x \times 10^{-1} \times \frac{\ln 2}{k} \\}

\mathrm{\Rightarrow \frac{\ln \left(\frac{a_0}{0.33 a_0}\right)}{k}=x \times 10^{-1} \times \frac{\ln 2}{k} \\}

\mathrm{\Rightarrow \frac{\ln 3}{\ln 2}=x \times 10^{-1} }

\mathrm{\Rightarrow \frac{\log 3}{\log 2}=x \times 10^{-1}}

\mathrm{ \Rightarrow \frac{0.4771}{0.3010}=x \times 10^{-1} \\}

\mathrm{\Rightarrow \quad x \times 10^{-1}=1.585 \\}

\mathrm{\Rightarrow \quad x=15.85 \\}

\mathrm{\Rightarrow \quad x=16 \text { (Nearest Integer })}

Answer is 16

Posted by

Pankaj Sanodiya

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: 11 from Kota coaching centres among 24 students with 100 percentile
anam-khan

Low JEE Main rank? Explore alternative paths for engineering admission
anam-khan

Gender-gap in JEE Main 2025: Female participation remains around 31%, overall pool drops by 7.09%

Latest Question