Exercises

Exercise 1. Consider the plot below. What is the average rate for the disappearance of CO₂ over the first 40 minutes of the reaction? What is the average rate between 60 and 80 minutes?

At t = 0, the concentration is 0.65 M. At t = 40, the concentration is 0.41 M.

\(\displaystyle -\frac{0.41\;M\;-\;0.65\;M}{40\;min\;-\;0\;min}\;=\;\mathbf{6.0\times 10^{-3}\;M/min}\)
 
At t = 60 min, the molarity is 0.41 M. At t = 80 min, the molarity is 0.35 M.

\(\displaystyle -\frac{0.35\;M\;-\;0.41\;M}{80\;min\;-\;60\;min}\;=\;\mathbf{3.0\times 10^{-3}\;M/min}\)
 

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Exercise 2. Consider the table below for the disappearance of C₄H₉Cl.

Calculate the average rate in the time interval 100 – 400 seconds.

\(\displaystyle -\frac{0.090\;M\;-\;0.164\;M}{400\;s\;-\;100\;s}\;=\;\mathbf{2.5\times 10^{-4}\;M/s}\)

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Exercise 3. Write the rate expression in terms of products and reactants for the following reaction:

3 I^– (aq) + H₃AsO₄ (aq) + 2 H⁺ (a) → I₃^– (aq) + H₃AsO₃ (aq) + H₂O (l)

\(\displaystyle -\frac{\Delta [\mathrm{I^-}]}{3\Delta t}\;=\;-\frac{\Delta [\mathrm{H_3AsO_4}]}{\Delta t}\;=\;-\frac{\Delta [\mathrm{H^+}]}{2\Delta t}\;=\;\frac{\Delta [\mathrm{I_3^-}]}{\Delta t}\;=\;\frac{\Delta [\mathrm{H_3AsO_3}]}{\Delta t}\;=\;\frac{\Delta [\mathrm{H_2O}]}{\Delta t}\)

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Exercise 4. Consider the following chemical reaction:

4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g)

What is the rate of disappearance of NH₃ if the rate of formation of NO is 2.6 x 10^-5 M/s? What is the rate of the disappearance of O₂ in the same time interval?

\(\displaystyle -\frac{\Delta [\mathrm{NH_3}]}{4\Delta t}\;=\;\frac{\Delta [\mathrm{NO}]}{4\Delta t}\)
 
Multiply both sides by 4

\(\displaystyle -\frac{\Delta [\mathrm{NH_3}]}{\Delta t}\;=\;\frac{\Delta [\mathrm{NO}]}{\Delta t}\)
 
\(\displaystyle -\frac{\Delta [\mathrm{NH_3}]}{\Delta t}\;=\;2.6\times 10^{-5}\;M/s\)
 
\(\displaystyle \frac{\Delta [\mathrm{NO}]}{4\Delta t}\;=\;-\frac{\mathrm[\Delta O_2]}{5\Delta t}\)
 

Multiply both sides by 5

\(\displaystyle \frac{5\Delta [\mathrm{NO}]}{4\Delta t}\;=\;-\frac{\mathrm[\Delta O_2]}{\Delta t}\)
 
\(\displaystyle -\frac{\mathrm[\Delta O_2]}{\Delta t}\;=\;\frac{5}{4}\times\;(2.6\times 10^{-5}\;M/s)\;=\;3.3\times 10^{-5}\;M/s\)
 
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