# Finding the average rate of consumption

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.

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.

• 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. Feb 26, 2014 at 6:53

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 $4.0 \times 10^{-3}\ \mathrm{mol} \cdot \mathrm{l}^{-1}$

• Thank you for pointing this out! I guess I was too tired so I messed up the conversion. Feb 26, 2014 at 6:54
• @Jesse No problem! Honestly, my first mental arithmetic guess was also off by one order of magnitude :D Feb 26, 2014 at 7:01