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

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

Given:

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

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

  • $t = 188$ s

Since we have $2.0 \cdot 10^{-4}$ moles of dye per $50$ mL of solution, that means in $1$ 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} \, \text{moles}}{188 \, \text{s}} \\ &= -2.1 \cdot 10^{-6} \, \text{mol}/\text{L}\cdot \text{s} \end{align}

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

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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.

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  • $\begingroup$ 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
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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}$

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    $\begingroup$ 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

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