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NCERT Class XII Chemistry
Chapter - Chemical Kinetics
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Question : 25 of 39
Marks: +1, -0
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t1/2_{1/2} = 3.00 hours. What fraction of sample of sucrose remains after 8 hours ?
Solution:  
Sucrose decomposes according to first order rate law, hence
k=2.303tlog⁡[R0][R],k = \frac{2.303}{t} \log \frac{[R_0]}{[R]},
t1/2=3 hrs,t_{1/2} = 3 \text{ hrs},
t=8 hrs,[R][R0]=?t = 8 \text{ hrs}, \frac{[R]}{[R_0]} = ?
t1/2=3.0t_{1/2}=3.0 hrs,
∴  k=0.693t1/2\therefore\; k = \frac{0.693}{t_{1/2}}
=0.6933=0.231 hr−1= \frac{0.693}{3} = 0.231 \text{ hr}^{-1}
Hence, 0.231=2.3038log⁡[R0][R]0.231 = \frac{2.303}{8} \log \frac{[R_0]}{[R]}
or, log⁡[R0][R]=0.8024\log \frac{[R_0]}{[R]} = 0.8024
or, [R0][R]=\frac{[R_0]}{[R]} = Antilog (0.8024)=6.345(0.8024)=6.345
or, [R][R0]=16.345=0.158\frac{[R]}{[R_0]} = \frac{1}{6.345} = 0.158
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