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

enter image description here

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

  • $0.030$
  • $3.1x10^{-3}$
  • $0.013$
  • $3.0$
  • $0.14$
  • $\begingroup$ dah... $\ce{[A]}$ is in the table. You're trying to solve for $k$. $\endgroup$ – MaxW Dec 14 '15 at 5:37
  • $\begingroup$ @MaxW then it's none of the answers, as I mentioned. $\endgroup$ – Voldemort Dec 14 '15 at 5:39
  • $\begingroup$ You should read Wikipedia article en.wikipedia.org/wiki/Rate_equation Your use of "velocity constant" is weird terminology. $\endgroup$ – MaxW Dec 14 '15 at 5:48
  • $\begingroup$ 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. $\endgroup$ – orthocresol Dec 14 '15 at 6:09

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,

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

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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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  • $\begingroup$ 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. $\endgroup$ – Voldemort Dec 14 '15 at 16:11
  • $\begingroup$ No way. if you use the 2nd time interval, you get 0.03/0.25 = 0.12, the exact same result. The 0.25 is obtained as (0.4+0.1)/2 $\endgroup$ – Chet Miller Dec 14 '15 at 16:29

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