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There are two compounds $\ce{A}$ and $\ce{B}$ that react with one common reagent $\ce{C}$ following first order kinetics. Suppose 99.9% of $\ce{A}$ must react before 0.1% of $\ce{B}$ has reacted then what is the minimum ratio of their respective rate constants?

My Attempt at the question:

We know that the reactions follow first order kinetics, hence I tried to find the time at which the given percentage of $\ce{A}$ and $\ce{B}$ is left using the formula $$ t = \frac{2.303}{k} \cdot \log\left(\frac{C_0}{C_1}\right)$$ where $t$ is the time, $k$ is the rate constant for each reaction and $C_0$ and $C_1$ are initial and final concentration respectively.

After this I tried to find the ratio of rate constants directly but couldn't arrive at the answer. I derived the above formula using the integrated rate law.

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  • $\begingroup$ Your approach seems right. You know the values of $A_{0}$ and $A_{t}$; and $B_{0}$ and $B_{t}$. You can equate the two $t$'s to compute the ratio between $k_{A}$ and $k_{B}$. $\endgroup$ – Zhe Oct 16 '16 at 12:31
  • $\begingroup$ I tried that. But not arriving on any answer. Try out the calculations and steps. It's tricky and confusing after a few steps. Try it out for yourself and tell me your approach $\endgroup$ – user36160 Oct 16 '16 at 14:06
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$$k_1t = 2.303 \log\left(\frac{1}{(1-0.999)}\right)$$ $$k_2t = 2.303 \log\left(\frac{1}{(1-0.001)}\right)$$

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    $\begingroup$ While this does somewhat answer the question, I find it not very enlightening or helpful. You should at least add a few sentences explaining how you arrived at these formulae. Since I encountered this in the VLQ review queue I am going to say it "Looks OK", but I do consider it much more appropriate to be a comment. $\endgroup$ – Martin - マーチン Oct 17 '16 at 4:56

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