3
$\begingroup$

The question is:

If $2.0 \cdot 10^{-4}$ moles of dye in $\pu{50 mL}$ of solution is consumed in 188 seconds, what is the average rate of consumption of dye in $\pu{mol L-1 s-1}$?

I am not sure if I have done this correctly. I use the rate of reaction formula which is $\text{Rate} = -\frac{\mathrm{d[A]}}{\mathrm dt}$ There is a negative sign because the dye is being consumed and it is disappearing.

Given:

  • [dye] = $2.0 \cdot 10^{-4}$ moles in $\pu{50 mL}$

  • volume of solution = $\pu{50 mL}$ = $\pu{0.05 L}$

  • $t = \pu{188 s}$

Since we have $2.0 \cdot 10^{-4}$ moles of dye per $\pu{50 mL}$ of solution, that means in $\pu{1 L}$ of solution, we get $2*(2.0 \cdot 10^{-4})$ moles of dye = $4.0 \cdot 10^{-4}$ moles

\begin{align} \text{Rate of consumption/disappearance} &= \frac{-4.0 \cdot 10^{-4} \, \pu{mol}}{\pu{188 s}} \\ &= -2.1 \cdot 10^{-6} \pu{mol L-1 s-1} \end{align}

Is my answer correct? Please point out my mistakes if I did it incorrectly.

Improve this question
$\endgroup$
4
$\begingroup$

Since we have $2.0 \times 10^{-4}\text{ mol}$ per $50\text{ mL}$ of solution, to get the amount of substance per liter, you multiply by:

$$ \frac{1000\text{ mL}}{50\text{ mL}}=20$$

So we get $4.0 \times 10^{-3}\text{ mol L}^{-1}$ as Klaus says.

You're being asked to find a rate of dye consumption. This should be a positive number, if the dye is being consumed - which it is. If dye was being consumed at a negative rate, more dye would appear over time!

Imagine we begin with $c = 1.0\text{ mol L}^{-1}$. We end with $c = 0.5\text{ mol L}^{-1}$. Thus we might say that $\Delta c = -0.5\text{ mol L}^{-1}$. We see that the change is negative, signifying that dye has been consumed. If this occured over the course of 180 seconds, we might write:

$$\text{average rate of consumption} = -\frac{\Delta c}{\Delta t} = -\frac{-0.5\text{ mol L}^{-1}}{180\text{ s}} = 2.77\times 10^{-3}\text{ mol L}^{-1}\text{ s}^{-1}$$

So even though the change in dye concentration is negative, the rate of dye consumption should be a positive number.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thank you for providing me a very clear explanation and pointing out the rate of consumption part! And I later realized that I messed up the conversion between mL to L. $\endgroup$ – Jesse Feb 26 '14 at 6:53
0
$\begingroup$

Since we have 2.0*10^-4 moles of dye per 50 mL of solution, that means in 1 L of solution, we get 2*(2.0*10^-4 moles) of dye = 4.0*10^-4 moles

Unless I'm too tired, the concentration is $ 4.0 \times 10^{-3}\ \mathrm{mol} \cdot \mathrm{l}^{-1}$

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Thank you for pointing this out! I guess I was too tired so I messed up the conversion. $\endgroup$ – Jesse Feb 26 '14 at 6:54
  • $\begingroup$ @Jesse No problem! Honestly, my first mental arithmetic guess was also off by one order of magnitude :D $\endgroup$ – Klaus-Dieter Warzecha Feb 26 '14 at 7:01

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.