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? (\mathbf{b} )What are the units of the rate constant? (c) What would happen to the rate if the concentration of OH \(^{-}\) were tripled? (d) What would happen to the rate if the concentration of both reactants were tripled?

Short answer

The rate constant value (k) is 17.28 s⁻¹. The units of the rate constant (k) are s⁻¹(M⁻¹). The rate will triple when the concentration of OH⁻ is tripled. The rate will be nine times faster when the concentrations of both reactants are tripled.

Step-by-step solution

Find the rate constant (k) value

To find the rate constant, we can use the given rate law and the reaction rate at specific concentrations of the reactants. The rate law is given as: Rate = k [CH3Br] [OH⁻] We can plug in the given values: 0.0432 M/s = k (5.0 × 10⁻³ M)(0.050 M) Now, we can solve for k: k = 0.0432 M/s / [(5.0 × 10⁻³ M)(0.050 M)] k = 17.28 s⁻¹ Therefore, the rate constant value (k) is 17.28 s⁻¹.

Find the units of the rate constant (k)

The units of the rate constant can be derived from the rate law equation: Rate = k [CH3Br] [OH⁻] Since the rate itself has units of M/s: k = Rate / ([CH3Br] [OH⁻]) k will have the units of (M/s) / (M²), which simplifies to: Units of k = s⁻¹ (M⁻¹) The units of the rate constant (k) are s⁻¹ (M⁻¹).

Effect on the rate when the concentration of OH⁻ is tripled

If the concentration of OH⁻ is tripled, the new reaction rate can be calculated using the rate law: New Rate = k [CH3Br] [3 × OH⁻] New Rate = k [CH3Br] (3 [OH⁻]) Since the reaction is first order in OH⁻, the new rate will be three times the initial rate: New Rate = 3 × (k [CH3Br] [OH⁻]) Therefore, the rate will triple when the concentration of OH⁻ is tripled.

Effect on the rate when the concentrations of both reactants are tripled

If both reactant concentrations are tripled, the new reaction rate can be calculated using the rate law: New Rate = k [3 × CH3Br] [3 × OH⁻] New Rate = k (3 [CH3Br])(3 [OH⁻]) Since the reaction is first order in both CH3Br and OH⁻, the new rate will be nine times the initial rate: New Rate = 9 × (k [CH3Br] [OH⁻]) Therefore, the rate will be nine times faster when the concentrations of both reactants are tripled.

Key concepts

Rate Law

In chemical kinetics, the rate law is an equation that links the reaction rate with the concentrations of reactants. For the reaction \[\mathrm{CH}_{3} \mathrm{Br}(aq)+\mathrm{OH}^{-}(aq) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(aq)+\mathrm{Br}^{-}(aq)\], the rate law can be expressed as \[\text{Rate} = k [\mathrm{CH}_{3} \mathrm{Br}][\mathrm{OH}^{-}] \],where \(k\) is the rate constant and \([\mathrm{CH}_{3} \mathrm{Br}]\) and \([\mathrm{OH}^{-}]\) are the concentrations of each reactant.
This particular rate law is first order in both \(\mathrm{CH}_{3} \mathrm{Br}\) and \(\mathrm{OH}^{-}\), meaning that if you double the concentration of any one of the reactants, the reaction rate will also double.

• First order in a reactant means that the rate depends linearly on its concentration.
• Understanding the rate law is crucial as it provides insights into which reactants affect the speed of a reaction and by what degree.

Remember, the coefficients in the rate law are determined experimentally and may not match the coefficients in the balanced chemical equation.

Reaction Rate

The reaction rate is an important concept in understanding how quickly a reaction proceeds. In the given reaction, the initial rate of formation of products is 0.0432 \(\text{M/s}\). The reaction rate can vary based on factors such as the concentration of reactants, temperature, and the presence of a catalyst.
By applying the rate law, \[\text{Rate} = k [\mathrm{CH}_{3} \mathrm{Br}][\mathrm{OH}^{-}]\], we can determine the reaction rate at any specific moment by using the concentration values of \(\mathrm{CH}_{3} \mathrm{Br}\) and \(\mathrm{OH}^{-}\).
In practical terms:

• Increasing the concentration of reactants usually increases the rate of reaction.
• Tripling the concentration of a reactant will triple the rate if the reaction is first order with respect to that reactant.
• The rate will increase ninefold if both reactants are tripled, demonstrating the multiplicative effect in reactions following the given rate law.

This predictability helps scientists and engineers control reaction velocities for optimal efficiency in industrial and laboratory settings.

Rate Constant

The rate constant, denoted as \(k\), is a crucial component of the rate law. It quantifies how rapidly a reaction proceeds under certain conditions. In our example, the rate constant is calculated as 17.28 \(\text{s}^{-1} \text{M}^{-1}\). Its value is determined from the known reaction rate and the concentrations of the reactants using the equation \[k = \frac{\text{Rate}}{[\mathrm{CH}_{3} \mathrm{Br}][\mathrm{OH}^{-}]}\].
The units of \(k\) depend on the overall order of the reaction. Since this is a second-order reaction (the sum of exponents in the rate law is 2), the units are \(\text{s}^{-1} \text{M}^{-1}\).
Key points regarding rate constants include:

• Rate constants are unique to each reaction and vary with temperature.
• A larger value of \(k\) indicates a faster reaction.
• The units of \(k\) help verify the correctness of the derived rate law order.

When the temperature increases, usually \(k\) will too, meaning that reactions tend to proceed faster at higher temperatures.