# How to calculate the ratio of rate constants of competing reactions?

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.

• 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 $k_{B}$.
– Zhe
Oct 16, 2016 at 12:31
• 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 Oct 16, 2016 at 14:06

For the minimum ratio of rate constants, we take the first instant of time that can be measured. Wherein, 99.9% of A has reacted and 0.1% of B has reacted.

Therefore for A

$$\dfrac{C_0}{C_1} = \dfrac{1}{1 - 0.999}$$

Similarly for B, we have

$$\dfrac{C_0}{C_1} = \dfrac{1}{1 - 0.001}$$

Therefore,

$$k_At = 2.303 \log\left(\frac{1}{(1-0.999)}\right)$$ $$k_Bt = 2.303 \log\left(\frac{1}{(1-0.001)}\right)$$