The reaction rate is measured by determining how quickly a reactant is consumed or a product is formed during a chemical reaction. In other words, the change in the amount of a given reactant or product is measured experimentally over time. Different methods can be used to measure the amount of a substance. The optimal method is often chosen based on the state of the substance.
The following table presents different ways of calculating the reaction rate based on the state of the reactant or product and the variables measured to assess the change in its amount:
|
State of the reactant or product |
Measured variable |
Examples of measuring equipment |
Reaction rate formula |(r)| |
Examples of reaction rate units |
|---|---|---|---|---|
|
Solid |
Mass |(m)| |
Balance |
|r = \dfrac{\vert\Delta m\vert}{\Delta t}| |
Grams per second |(\text{g/s})| |
|
Liquid |
Mass |(m)| |
Balance |
|r = \dfrac{\vert\Delta m\vert}{\Delta t}| |
Grams per second |(\text{g/s})| |
|
Volume |(V)| |
Graduated cylinder |
|r = \dfrac{\vert\Delta V\vert}{\Delta t}| |
Millilitres per second |(\text{mL/s})| |
|
|
Gas |
Mass |(m)| |
Balance |
|r = \dfrac{\vert\Delta m\vert}{\Delta t}| |
Grams per second |(\text{g/s})| |
|
Volume |(V)| |
Gas burette |
|r = \dfrac{\vert\Delta V\vert}{\Delta t}| |
Millilitres per second |(\text{mL/s})| |
|
|
Pressure |(P)| |
Manometer |
|r = \dfrac{\vert\Delta P\vert}{\Delta t}| |
Kilopascals per second |(\text{kPa/s})| |
|
|
Molar concentration |(\text{[Substance]})| |
Spectrophotometer |
|r = \dfrac{\vert\Delta [\text{Substance}]\vert}{\Delta t}| |
Moles per litre per second |(\text{mol/L}\cdot\text{s})| |
|
|
Solution |
Molar concentration |
Spectrophotometer or pH-meter |
|r = \dfrac{\vert\Delta [\text{Substance}]\vert}{\Delta t}| |
Moles per litre per second |(\text{mol/L}\cdot\text{s})| |
Note: While all units presented in this table use seconds |(\text{s})| as the unit of time change |(\Delta t)|, reaction rate can be measured over minutes |(\text{min}),| hours |(\text{h}),| days |(\text{d})| or even years |(\text{yr}).|
The choice of the measured variable also depends on:
- the accessibility of the substances at different times of the experiment;
- the reaction speed that needs to be calculated: average reaction rate or instantaneous reaction rate.
Consider the following chemical reaction, where solid magnesium |(\text{Mg})| and aqueous sulphuric acid |(\text{H}_2\text{SO}_4)| react to form aqueous magnesium sulphate |(\text{MgSO}_4)| and hydrogen gas |(\text{H}_2)\!:|
|\text{Mg}_{\text{(s)}} + \text{H}_2\text{SO}_{4(aq)} → \text{Mg}\text{SO}_{4(aq)} + \text{H}_{2(g)}.|
The following table analyzes some of the methods of measuring the reaction rate in this experiment:
|
Substance |
Reaction rate formula |(r)| |
Method of measuring the reaction rate |
Possible use |
|---|---|---|---|
|
|\text{Mg}_{\text{(s)}}| |
|r = \dfrac{\vert\Delta m_{\text{Mg}}\vert}{\Delta t}| |
The initial and final mass of solid |\text{Mg}| is measured along with the time required to reach the final mass. Assuming that this reactant is consumed completely during the reaction, its final mass is |0\ \text{g}.| |
This is an easy method for determining the average reaction rate, but this method is not optimal for measuring the instantaneous reaction rate because obtaining data at regular intervals throughout the experiment is challenging. |
|
|\text{H}_2\text{SO}_{4(aq)}| |
|r = \dfrac{\vert\Delta [{\text{H}_2\text{SO}_4}]\vert}{\Delta t}| |
Sulfuric acid |(\text{H}_2\text{SO}_4)| is a strong acid. The pH of the solution can thus be measured at regular intervals in order to determine the molar concentration of protons |[\text{H}^+]|, followed by the concentration of sulfuric acid |[\text{H}_2\text{SO}_4]|. |
This method can be used for determining the average reaction rate or the instantaneous reaction rate. However, since the acidic solution of |(\text{H}_2\text{SO}_4)| is usually present in excess, it can be difficult to note any significant pH changes before the magnesium |(\text{Mg})| sample is fully consumed. |
|
|\text{Mg}\text{SO}_{4(aq)}| |
|r = \dfrac{\vert\Delta [{\text{Mg}\text{SO}_4}]\vert}{\Delta t}| |
A spectrophotometer is used to measure the absorbance of a coloured solution at a specific wavelength. The measure can be used to determine the molar concentration of the solution. |
This method cannot be used in this particular reaction because the solution remains colourless. |
|
|\text{H}_{2(g)}| |
|r = \dfrac{\vert\Delta V_{\text{H}_2}\vert}{\Delta t}| |
The |\text{H}_2| gas is collected using a gas burette and its volume is measured at regular intervals. Note that the mass of the gas or the pressure that it exerts can be measured instead. |
This method is optimal for determining the average reaction rate or the instantaneous reaction rate. |
It is often useful to determine the reaction rate in |\text{mol/s}| or |\text{mol/L}\cdot\text{s}.| Since concentration can only be measured for gaseous and aqueous solutions, other measurements can be converted to the desired values using the molar mass formula, Avogadro’s law and molar volume or the ideal gas law.
Consider the following chemical reaction, where |3.0\ \text{g}| of magnesium |\text{Mg}| reacts with |6.0\ \text{mol/L}| sulphuric acid |(\text{H}_2\text{SO}_4)| in excess: |\text{Mg}_{\text{(s)}} + \text{H}_2\text{SO}_{4(aq)} → \text{Mg}\text{SO}_{4(aq)} + \text{H}_{2(g)}.|
After |2.5\ \text{min},| the reaction is complete and all of the |\text{Mg}| is consumed. Determine the average reaction rate with respect to |\text{Mg}| in |\text{g/s}| and |\text{mol/s}.|
Reaction rate in |\text{g/s}\!:|
-
Identify the known values:
||\begin{align} m_{i(\text{Mg})}&= 3.0\ \text{g}\\ m_{f(\text{Mg})}&= 0.0\ \text{g}\\\\ t = 2.5\ \cancel{\text{min}}\times \dfrac{60\ \text{s}}{1\ \cancel{\text{min}}}&=150\ \text{s}\\\\ r &=\ ?\ \text{g/s} \end{align}|| -
Write the reaction rate formula that can be used in this case: ||r = \dfrac{\vert\Delta m_{\text{Mg}}\vert}{\Delta t}||
Note that: ||\Delta m_{\text{Mg}}=m_{f(\text{Mg})}- m_{i(\text{Mg})}|| -
Plug in the known values and calculate the answer: ||\begin{align}r &= \dfrac{\vert0.0\ \text{g} - 3.0\ \text{g}\vert}{150\ \text{s}}\\\\ r &= 0.020\ \text{g/s} =2.0\times10^{-2}\ \text{g/s} \end{align}||
Reaction rate in |\text{mol/s}\!:|
-
Identify the known values: ||\begin{align}r &= 2.0\times10^{-2}\ \text{g/s}\\ M_{\text{Mg}}&= 24.31\ \text{g/mol}\\ r &=\ ?\ \text{mol/s} \end{align}||
-
Use the molar mass formula to convert the mass of |\text{Mg}| that is consumed per second into the amount of moles |(n)\!:| ||\begin{align} M = \dfrac{m}{n}\Rightarrow n &=\dfrac{m}{M}\\\\ n &= \dfrac{2.0\times10^{-2}\ \cancel{\text{g}}}{24.31\ \cancel{\text{g}}\text{/mol}}\\\\ n &\approx 0.00082\ \text{mol} = 8.2\times10^{-4}\ \text{mol} \end{align}||
-
Express the rate in |\text{mol/s}\!:| ||r \approx 8.2\times10^{-4}\ \text{mol/s}||
During a chemical reaction conducted at STP, |50.0\ \text{mL}| of gas forms over the course of |80.0\ \text{s}.| Determine the average reaction rate with respect to this gas in |\text{mol/s}.|
First, let’s determine the reaction rate in |\text{mL/s}\!:|
-
Identify the known values: ||\begin{align} V_{i(\text{Product})}&= 0.0\ \text{mL}\\ V_{f(\text{Product})}&= 50.0\ \text{mL}\\ \Delta t &= 80\ \text{s}\\ r &=\ ?\ \text{mL/s} \end{align}||
-
Write the reaction rate formula that can be used in this case: ||r = \dfrac{\vert\Delta V_{\text{Product}}\vert}{\Delta t}||
Note that: ||\Delta V_{\text{Product}}=V_{f(\text{Product})}- V_{i(\text{Product})}|| -
Plug in the known values and calculate the answer: ||\begin{align}r &= \dfrac{\vert50.0\ \text{mL} - 0.0\ \text{mL}\vert}{80.0\ \text{s}}\\ r &= 0.625\ \text{mL/s} \end{align}||
Now, convert the reaction rate into |\text{mol/s}| using Avogadro’s law. Keep in mind that the molar volume at STP is |(1\ \text{mol} = 22.4\ \text{L})\!:|
-
Identify the known values. Note that the volume must be expressed in |\text{L}\!:| ||\begin{align} n_1 &= 1\ \text{mol}\\ V_1 &= 22.4\ \text{L}\\ n_2 &=\ ?\ \text{mol}\\ V_2 & = 0.625\ \cancel{\text{mL}}\times\dfrac{1\ \text{L}}{1000\ \cancel{\text{mL}}} = 6.25\times10^{-4}\ \text{L} \end{align}||
-
Use Avogadro’s law to convert the volume of the gas that is formed per second into the amount of moles |(n)\!:| ||\begin{align} \dfrac{n_1}{V_1} = \dfrac{n_2}{V_2}\Rightarrow n_2 &= \dfrac{n_1\times V_2}{V_1}\\\\ n_2 &= \dfrac{1\ \text{mol}\times 6.25\times10^{-4}\ \cancel{\text{L}}}{22.4\ \cancel{\text{L}}}\\\\ n_2 &\approx 0.00002790 = 2.79\times10^{-5}\ \text{mol} \end{align}||
-
Express the rate in |\text{mol/s}\!:| ||r \approx 2.79\times10^{-5}\ \text{mol/s}||
During a chemical reaction conducted at |15.00^\circ\text{C}| and |102\ \text{kPa},| |105.0\ \text{mL}| of |\text{H}_2,| gas forms over the course of |6\ \text{min}.| Determine the average reaction rate with respect to this gas in |\text{mol/L}\cdot\text{s}.|
First, let’s use the ideal gas law to convert the volume of the gas that is formed per second into the amount of moles |(n).| Pay close attention to the units:
-
Identify the known values: ||\begin{align} P &= 102\ \text{kPa}\\ \Delta V_{\text{H}_2} &=105.0\ \cancel{\text{mL}}\times \dfrac{1\ \text{L}}{1000\ \cancel{\text{mL}}}=0.1050\ \text{L}\\ T &= 15.00^\circ\text{C} + 273.15 = 288.15\ \text{K}\\ R &= 8.314\ \text{L}\cdot\text{kPa}\text{/mol}\cdot{\text{K}}\\ n &=\ ?\ \text{mol}\end{align}||
-
Isolate |n| from the ideal gas law and plug in the known values: ||\begin{align} PV = nRT \Rightarrow n &= \dfrac{PV}{RT}\\\\ n &= \dfrac{102\ \cancel{\text{kPa}}\times0.1050\ \cancel{\text{L}}}{8.314\ \cancel{\text{L}}\cdot\cancel{\text{kPa}}\text{/mol}\cdot\cancel{{\text{K}}}\times288.15\ \cancel{\text{K}}}\\\\ n &\approx 0.00447\ \text{mol} = 4.47\times10^{-3}\ \text{mol}\\ \end{align}||
Then, determine |[\text{H}_2]| in |\text{mol/L}\!:|
-
Identify the known values: ||\begin{align}n_{\text{H}_2} &= 4.47\times10^{-3}\ \text{mol}\\ V_{\text{H}_2} &= 0.1050\ \text{L}\end{align}||
-
Plug the values into the molar concentration formula: ||\begin{align}[\text{H}_2] &=\dfrac{n}{V}\\\\ [\text{H}_2]&= \dfrac{4.47\times10^{-3}\ \text{mol}}{0.1050\ \text{L}}\\\\ [\text{H}_2] &\approx 0.0426\ \text{mol/L}\end{align}||
Finally, determine the reaction rate in |\text{mol/L}\cdot\text{s}\!:|
-
Identify the known values: ||\begin{align} \Delta[\text{H}_2] &\approx 0.0426\ \text{mol/L}\\\\ \Delta t &= 6\ \cancel{\text{min}}\times\dfrac{60\ \text{s}}{1\ \cancel{\text{min}}} = 360\ \text{s} \end{align}||
-
Plug the values into the rate formula: ||\begin{align}r&=\dfrac{\vert\Delta[\text{H}_2]\vert}{\Delta t}\\\\ r&=\dfrac{0.0426\ \text{mol/L}}{360\ \text{s}}\\\\ r &\approx 0.000118\ \text{mol/L}\cdot\text{s} = 1.18\times10^{-4}\ \text{mol/L}\cdot\text{s}\end{align}||