Question

Consider the following reaction: $$ \mathrm{CH}_{3} \mathrm{Br}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Br}^{-}(a q) $$ The rate law for this reaction is first order in \(\mathrm{CH}_{3} \mathrm{Br}\) and first order in \(\mathrm{OH}^{-}\). When \(\left[\mathrm{CH}_{3} \mathrm{Br}\right]\) is \(5.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.050 \mathrm{M}\), the reaction rate at \(298 \mathrm{~K}\) is \(0.0432 \mathrm{M} / \mathrm{s}\). (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) What would happen to the rate if the concentration of \(\mathrm{OH}^{-}\) were tripled?

Short answer

(a) The value of the rate constant is \(k = 173.6 \,\mathrm{s^{-1}}\). (b) The units of the rate constant are \(\mathrm{s^{-1}}\). (c) The new reaction rate when the concentration of \(\mathrm{OH}^{-}\) is tripled would be \(0.130\, \mathrm{M/s}\).

Step-by-step solution

Write the rate law expression

The rate law for the reaction is given as first order in \(\mathrm{CH}_3\mathrm{Br}\) and first order in \(\mathrm{OH}^{-}\). So, the rate law expression can be written as: $$ Rate = k\left[\mathrm{CH}_3\mathrm{Br}\right]\left[\mathrm{OH}^{-}\right] $$ where \(Rate\) is the reaction rate, \(k\) is the rate constant, and the concentrations are in moles per liter (M).

Calculate the rate constant

Using the given concentrations and reaction rate, plug the values into the rate law expression: $$ 0.0432 \, \mathrm{M/s} = k(5.0\times10^{-3}\,\mathrm{M})(0.050\,\mathrm{M}) $$ Solve for \(k\): $$ k = \frac{0.0432\,\mathrm{M/s}}{(5.0\times10^{-3}\,\mathrm{M})(0.050\,\mathrm{M})}= 173.6\,\mathrm{s^{-1}} $$ So, the value of the rate constant \(k=173.6 \, \mathrm{s^{-1}}\).

Determine the units of the rate constant

The units of the rate constant in the reaction can be determined using the rate expression: $$ \mathrm{M/s} = k\,(\mathrm{M})(\mathrm{M}) $$ Solving for the units of \(k\): $$ \frac{\mathrm{M}}{\mathrm{s}}\times\frac{1}{\mathrm{M}\times\mathrm{M}} = k $$ So, the units of the rate constant are \(\mathrm{s^{-1}}\).

Calculate the new reaction rate

Now we need to find the new reaction rate when the concentration of \(\mathrm{OH}^{-}\) is tripled. This means the new concentration of \(\mathrm{OH}^{-}\) will be \(3\times0.050\,\mathrm{M}=0.150\,\mathrm{M}\). Using the rate law expression and the calculated value of \(k\): $$ Rate_\text{new} = (173.6\,\mathrm{s^{-1}})(5.0\times10^{-3}\,\mathrm{M})(0.150\,\mathrm{M}) = 0.130\, \mathrm{M/s} $$ The new reaction rate when the concentration of \(\mathrm{OH}^{-}\) is tripled would be \(0.130\, \mathrm{M/s}\).

Key concepts

Rate Law

The rate law is a mathematical equation that relates the rate of a chemical reaction to the concentration of its reactants. It allows us to predict how changes in concentration affect the reaction rate. The form of the rate law for a reaction is determined experimentally and cannot be deduced from the balanced chemical equation alone. For instance, in our exercise, the rate law is given by the expression:
\[ Rate = k[\mathrm{CH}_3\mathrm{Br}][\mathrm{OH}^-] \]
where \(k\) is the rate constant and the concentrations of the reactants (\( [\mathrm{CH}_3\mathrm{Br}] \) and \( [\mathrm{OH}^-] \)) are raised to some power. These powers are known as the 'order' of the reaction with respect to each reactant—in this case, first order for both, indicating that the rate is directly proportional to the concentration of both \(\mathrm{CH}_3\mathrm{Br}\) and \(\mathrm{OH}^-\).

Rate Constant

The rate constant, denoted as \(k\), is a proportionality factor in the rate law that is specific to a particular reaction at a given temperature. It has a fixed value at a constant temperature but can change with temperature. The magnitude of \(k\) reflects how quickly a reaction proceeds; a larger rate constant means a faster reaction. In our exercise, we find that the rate constant for the reaction at \(298\ K\) is:
\[ k = 173.6\, \mathrm{s^{-1}} \]
This means that at \(298\ K\), the reaction rate will increase by \(173.6\) times for every one-molar increase in the product of the concentrations of \(\mathrm{CH}_3\mathrm{Br}\) and \(\mathrm{OH}^-\). The units of the rate constant depend upon the overall order of the reaction, which, for a first order reaction with respect to both reactants, will be \(\mathrm{s^{-1}}\) in this case.

Reaction Rate

The reaction rate indicates how quickly the concentration of a reactant or product changes over time. In chemical kinetics, it is typically expressed in moles per liter per second (\(\mathrm{M/s}\)). The rate can be influenced by several factors, including the reactant concentrations, temperature, and the presence of a catalyst. In the example provided, the reaction rate is measured and given as :
\[ 0.0432\, \mathrm{M/s} \]
This value indicates that the concentration of products is increasing (or reactants decreasing) by \(0.0432\) molarity every second under the specified conditions. If we alter concentrations or other conditions, the reaction rate will adjust accordingly, as demonstrated by tripling the concentration of \(\mathrm{OH}^-\), which increases the reaction rate.

Concentration Effect

The concentration effect, outlined by the rate law, describes how the reaction rate changes with varying concentrations of reactants. A key principle is that, for most reactions, an increase in the concentration of reactants will lead to an increase in the reaction rate. This relationship is because more reactant molecules allow for more frequent effective collisions per unit time. In our exercise, when the concentration of \(\mathrm{OH}^-\) is tripled, the reaction rate increases significantly. The calculation based on the rate law shows:
\[ Rate_{\text{new}} = (173.6\,\mathrm{s^{-1}})(5.0\times10^{-3}\,\mathrm{M})(0.150\,\mathrm{M}) = 0.130\, \mathrm{M/s} \]
This illustrates the direct relationship between reactant concentration and rate—the rate is directly proportional to the concentration of \(\mathrm{OH}^-\) when other conditions are held constant. Understanding this concept helps predict and control the speed of chemical processes.