Finding [A] when looking for the velocity constant


> 
> Some reaction of first order on $[A]$ has the data:
> 
> 
> [![enter image description here](https://i.stack.imgur.com/T6wAW.png)](https://i.stack.imgur.com/T6wAW.png)
> 
> 
> What is the velocity constant for this reaction? The units are
>  $s^{-1}$.
> 
> 
> 




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Well, since the units are $s^{-1}$, then the reaction must be of global order one. Meaning that $A$ is the only reagent, yes?


Anyway, the velocity equation is


$$V = k \cdot [A]^1$$


Let's pick the first experiment...


$$\left (-\frac{0.40 - 1.60}{10} \right ) = k \cdot [A]$$


Huh. I don't actually know how to find $[A]$. Should I add $1.6 + 0.40$? Subtract them maybe? Or pick just one?


I tried all three of them and still couldn't get the answer. The options were


* $0.030$
* $3.1x10^{-3}$
* $0.013$
* $3.0$
* $0.14$




For a first-order reaction, the half-life is constant and is given by the equation 


$$t\_{1/2} = \frac{\ln 2}{k}$$


In $10~\mathrm{s}$, two half-lives have passed (the concentration of $\ce{A}$ drops to a quarter), so $t\_{1/2} = 5~\mathrm{s}$. Accordingly,


$$\begin{align}
k &= \frac{\ln 2}{5~\mathrm{s}} \\
&= 0.1386~\mathrm{s^{-1}}
\end{align}$$


 4

A way to get an approximate value of k is to divide by the average value of [A] over the time interval. In your example, it would be 1 (over the first time interval). So you would get an approximate value of 0.12 for k. This approximate value is closest to the 0.14 in your multiple choices.


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