1
$\begingroup$

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

  • $0.030$
  • $3.1x10^{-3}$
  • $0.013$
  • $3.0$
  • $0.14$
$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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,

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.