JEE Advance - Chemistry (2012 - Paper 1 Offline - No. 2)
An organic compound undergoes first-order decomposition. The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are t1/8 and t1/10 respectively. What is the value of $$\left[ {{{{t_{1/8}}} \over {{t_{1/10}}}}} \right] \times 10$$? ($${\log _{10}}2 = 0.3$$)
Answer
9
Explanation
The expression for first-order reaction is
$$ t=\frac{2.303}{k} \log \frac{\left[\mathrm{A}_0\right]}{[\mathrm{A}]} $$
When the compound is decomposed to $1 / 8$ th of its initial value then the time taken is
$$ t_{1 / 8}=\left(\frac{2.303}{k}\right) \log \frac{1}{(1 / 8)}=\left(\frac{2.303}{k}\right) \log 8 $$ .........(1)
When the compound is decomposed to $1 / 10$ th of its initial value then the time taken is
$$ t_{1 / 10}=\left(\frac{2.303}{k}\right) \log \frac{1}{(1 / 10)}=\left(\frac{2.303}{k}\right) \log 10 $$ ........(2)
Dividing Eq. (1) by Eq. (2), we get
$$ \frac{t_{1 / 8}}{t_{1 / 10}}=\frac{\log 8}{\log 10}=\log \left(2^3\right)=3 \times 0.3=0.9 $$
So, the value of
$$ \frac{\left[t_{1 / 8}\right]}{\left[t_{1 / 10}\right]} \times 10=9 $$
$$ t=\frac{2.303}{k} \log \frac{\left[\mathrm{A}_0\right]}{[\mathrm{A}]} $$
When the compound is decomposed to $1 / 8$ th of its initial value then the time taken is
$$ t_{1 / 8}=\left(\frac{2.303}{k}\right) \log \frac{1}{(1 / 8)}=\left(\frac{2.303}{k}\right) \log 8 $$ .........(1)
When the compound is decomposed to $1 / 10$ th of its initial value then the time taken is
$$ t_{1 / 10}=\left(\frac{2.303}{k}\right) \log \frac{1}{(1 / 10)}=\left(\frac{2.303}{k}\right) \log 10 $$ ........(2)
Dividing Eq. (1) by Eq. (2), we get
$$ \frac{t_{1 / 8}}{t_{1 / 10}}=\frac{\log 8}{\log 10}=\log \left(2^3\right)=3 \times 0.3=0.9 $$
So, the value of
$$ \frac{\left[t_{1 / 8}\right]}{\left[t_{1 / 10}\right]} \times 10=9 $$
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