Finding rate constants for parallel reactions

A reactant $\ce{X}$ is converted into products $\ce{Y}$ and $\ce{Z}$
according to the following first order parallel reactions:


$\ce{X->[k\_1]Y}$


$\ce{X->[k\_2]Z}$


$\ce{[X]}$ has decreased from $\pu{1.00 M}$ at $t = 0$ to $\pu{0.549 M}$ after $\pu{10 min}$.
Calculate the rate constants $k\_1$ and $k\_2$, knowing that the yields of $\ce{Y}$ and $\ce{Z}$ obtained are $25.0$ and $\pu{75.0 mol\%}$, respectively.


So far, I have the ratio between $k\_1$ and $k\_2$ $=0.33$, which I found from the yield ratio but I'm unsure what to do with this ratio.


I also have derived the equation $[A] = [A]\_0 \cdot e^{-(k\_1+ k\_2)t}$




By some simple mathematics, you can reach the following equations (the first of which you have reached):


$[X]=[X]\_0\cdot e^{-(k\_1+k\_2)t}$


$[Y]=\frac{k\_1}{k\_1+k\_2}[X]\_0\cdot e^{-k\_1 t}$


$[Z]=\frac{k\_2}{k\_1+k\_2}[X]\_0\cdot e^{-k\_2 t}$


The $1:3$ ratio that you have arrived at is correct, i.e. $\dfrac{k\_1}{k\_2}=\dfrac{1}{3}$.


Substituting this in the equation for $\ce{X}$, we get


$[X]=[X]\_0\cdot e^{-4k\_1 t}$


All that is left to do now is to substitute and solve, which leads to the answer:


$(k\_1,k\_2)\equiv(0.015\ (min)^{-1},0.045\ (min)^{-1})$


There you go!


PS: I've used $[X]\_0$ for representing the initial concentration of $\ce{X}$.


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