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NCERT Class XII Chemistry
Chapter - Chemical Kinetics
Questions with Solutions

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Question : 18 of 39
Marks: +1, -0
For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Solution:  
99%\% completion means that x=99%x = 99 \% of [R0][R_{0}]
or, [R]=[R0]−0.99[R0][R] = [R_{0}] - 0.99 [R_{0}]
=0.01[R0]= 0.01 [R_{0}]
For first order reaction, t=2.303klog⁡[R0][R]t = \frac{2.303}{k} \log \frac{[R_{0}]}{[R]}
∴    t99%=2.303klog⁡[R0]0.01[R0]\therefore\;\; t_{99\%} = \frac{2.303}{k} \log \frac{[R_{0}]}{0.01 [R_{0}]}
=2.303klog⁡102=2×2.303k= \frac{2.303}{k} \log 10^{2} = 2 \times \frac{2.303}{k}
90%\% completion means that [R]=[R0]−0.90R0][R] = [R_{0}] - 0.90 R_{0} ]
=0.1[R0]= 0.1 [R_{0}]
∴    t90%=2.303klog⁡[R0]0.1[R0]\therefore\;\; t_{90\%} = \frac{2.303}{k} \log \frac{[R_{0}]}{0.1 [R_{0}]}
=2.303klog⁡10=2.303k= \frac{2.303}{k} \log 10 = \frac{2.303}{k}
∴    t99%t90%=(2×2.303k)/(2.303k)=2\therefore\;\; \frac{t_{99\%}}{t_{90\%}} = \left( \frac{2 \times 2.303}{k} \right) / \left( \frac{2.303}{k} \right) = 2
or, t99%=2×t90%t_{99\%} = 2 \times t_{90\%}
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