CBSE Class 12 Chemistry 2017 Outside Delhi Set 1 Solved Paper

Question : 15 of 26

Marks: +1, -0

A first order reaction takes 20 minutes for

25 \%

decomposition. Calculate the time when

75 \%

of the reaction will be completed.
(Given :

\log 2=0.3010, \log 3=0.4771, \log 4=

0.6021)

Solution:

t\;=\;{2.303}/{k} \log \;{[ {\A}]^0}/{[ {\A}]}

20 {\min}\;=\;{2.303}/{k} \log \;{100}/{75}

t\;=\;{2.303}/{k} \log \;{100}/{25}

Divide (i) equation by (ii)

\;{20}/{t}\;=\;{\;{2.303}/{k} \log {100}/{75}}/{{2.303}/{k} \log \;{100}/{25}}

\;=\;{\log 4 \/ 3}/{\log 4}

\;{20}/{t}\;=0.1250 \/ 0.6021

t\;=96.3 {\min}

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