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CBSE Class 12 Chemistry 2017 Outside Delhi Set 1 Solved Paper

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Question : 15 of 26
Marks: +1, -0
A first order reaction takes 20 minutes for 25%25 \% decomposition. Calculate the time when 75%75 \% of the reaction will be completed.
(Given : log⁡2=0.3010,log⁡3=0.4771,log⁡4=\log 2=0.3010, \log 3=0.4771, \log 4= 0.6021)0.6021)
Solution:  
t  =  2.303klog⁡[A]0[A]t\;=\;\frac{2.303}{k} \log \frac{[A]^0}{[A]}
20 min  =  2.303klog⁡1007520 \text{ min}\;=\;\frac{2.303}{k} \log \frac{100}{75}
t  =  2.303klog⁡10025t\;=\;\frac{2.303}{k} \log \frac{100}{25}
Divide (i) equation by (ii)
20t  =  2.303klog⁡100752.303klog⁡10025\frac{20}{t}\;=\;\frac{ \frac{2.303}{k} \log \frac{100}{75} }{ \frac{2.303}{k} \log \frac{100}{25} }
  =  log⁡43log⁡4\;=\;\frac{\log \frac{4}{3}}{\log 4}
20t=0.12500.6021\frac{20}{t} = \frac{0.1250}{0.6021}
t=96.3 mint = 96.3 \text{ min}
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