# Finding [A] when looking for the velocity constant

Some reaction of first order on $[A]$ has the data:

What is the velocity constant for this reaction? The units are $s^{-1}$.

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

• $0.030$
• $3.1x10^{-3}$
• $0.013$
• $3.0$
• $0.14$
• dah... $\ce{[A]}$ is in the table. You're trying to solve for $k$. – MaxW Dec 14 '15 at 5:37
• @MaxW then it's none of the answers, as I mentioned. – Voldemort Dec 14 '15 at 5:39
• You should read Wikipedia article en.wikipedia.org/wiki/Rate_equation Your use of "velocity constant" is weird terminology. – MaxW Dec 14 '15 at 5:48
• Actually you should really just read a textbook. The questions you have posted are very common examples and should be covered in pretty much any general chemistry textbook or books tailored for the IB / A level syllabus (or its equivalent in the US). Right now, I get the feeling that you do not understand the topic very well and I think you would probably help yourself more by reading the topic instead of trying to do questions. – orthocresol Dec 14 '15 at 6:09

$$t_{1/2} = \frac{\ln 2}{k}$$
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}
• If you use the second time interval, you would get $0.03$, which would be wrong. I imagine it was just a coincidence that the first interval worked. – Voldemort Dec 14 '15 at 16:11