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JEE Advanced 2012Paper 1
Exam
JEE Advanced
Year
2012
Paper
Paper 1
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Advanced 2012 · Paper 1 · Chemistry · Chemical Kinetics

An organic compound undergoes first-order decomposition. The time...

An organic compound undergoes first-order decomposition. The time taken for its decomposition to 1/1 / 8 and 1/101 / 10 of its initial concentration are t1/8andt1/10t_{1/8} and t_{1/10} respectively. What is the value of \left[\frac{t_{1/8}}{t_{1/10}}\right] \times 10?(log⁡102=0.3)? \left(\log _{10} 2=0.3\right)

Correct answer

0

Step-by-step solution

t1/8=2.303log⁡8k=2.303×3log⁡2kt1/10=2.303klog⁡10=2.303k[t1/8t1/10]×10=(2.303×3log⁡2k)(2.303k)×10=9\begin{array}{l} t_{1/8}=\frac{2.303 \log 8}{k}=\frac{2.303 \times 3 \log 2}{k} \\ t_{1/10}=\frac{2.303}{k} \log 10=\frac{2.303}{k} \\ {\left[\frac{t_{1/8}}{t_{1/10}}\right] \times 10=\frac{\left(\frac{2.303 \times 3 \log 2}{k}\right)}{\left(\frac{2.303}{k}\right)} \times 10=9} \end{array}
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