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1.3 Calculating Reaction Rates

  • Page ID
    32216
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    Let's try an example of calculating a reaction rate. Consider the following reaction:

    \(\ce{A \rightarrow B}\)

    The following data were obtained for how the concentration of these substances changed during the experiment

    Time
    (min)
    [A]
    (mol ·L-1)
    [B]
    (mol ·L-1)
    0.0
    1.000
    0.000
    3.0
    0.400
    0.600
    6.0
    0.250
    0.750

    We could measure the rate of the reaction either by measuring how the concentration of reactant A changes or how the concentration of product B changes. Let's measure A's average rate of change first:

    \(\mathrm{rate}={\dfrac{\Delta[A]}{\Delta{time}}} = \dfrac{(.250-1.00)}{(6.00-0.00)} = -0.125\: mol \cdot L^{-1} \cdot min^{-1}\)

    Compare this rate to the rate of just the first three minutes of the reaction:

    \(\mathrm{rate}={\dfrac{\Delta[A]}{\Delta{time}}} = \dfrac{(.400-1.00)}{(3.00-0.00)} = -0.200\: mol \cdot L^{-1} \cdot min^{-1}\)

    We can see that, as was mentioned in the last section, the rate did slow down as the overall rate is slower than the rate of the first three minutes of the reaction.

    If we calculate the average rate based on the production of product B:

    \(\mathrm{rate}={\dfrac{\Delta[B]}{\Delta{time}}} = \dfrac{(.750-0.00)}{(6.00-0.00)} = 0.125\: mol \cdot L^{-1} \cdot min^{-1}\)

    Notice that we must compare rates measured during the same time period.

    You will see that we got the same rate, except for the sign, when we calculated the overall average rate for the disappearance of A as for the formation of B. This is only true, however, because our balanced equation shows us that one mole of A produces one mole of B.

    We would not find the same rates if we did not have a 1:1 relationship between reaction participants. For example if we examine the following reaction:

    \(\ce{2{H_2O_2} \rightarrow 2{H_2O} + {O_2}} \)

    we find that only one mole of oxygen forms for every two moles of hydrogen peroxide that decomposes.

    Therefore, we could make the following relationships:

    \(\ce{2H_2O_2}\)

    \(\ce{ \rightarrow}\)

    \(\ce{2H_2O}\)

    \(\ce{+}\)

    \(\ce{O_2}\)


    \(\ce{2H_2O_2}\)

    then

    1 mole of \(\ce{O_2}\) is formed


    if the rate of decomposition of \(\ce{2H_2O_2}\) is \(\ce{4.00\: mol \cdot L^{-1} \cdot min^{-1}}\)

    then

    the rate of formation of \(\ce{O_2}\) is ½ × 4.00 or 2.00 mol · L-1· min -1

    Some Comments about Units for Concentration

    Concentration, as we will see in depth in our unit on Solutions, may be measured in several ways, but we will often use the unit mol ·L-1 for concentration. Be comfortable seeing this unit written in any of the following ways:

    \(\mathrm{mol \cdot L^{-1}}= {mol / L}= M = \dfrac{mol}{L}\)


    1.3 Calculating Reaction Rates is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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