Question

When a base is added to an aqueous solution of chlorine dioxide gas, the following reaction occurs: \(2 \mathrm{ClO}_{2}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{ClO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}\) The reaction is first-order in \(\mathrm{OH}^{-}\) and second-order for \(\mathrm{ClO}_{2}\). Initially, when \(\left[\mathrm{ClO}_{2}\right]=0.010 \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]=0.030 \mathrm{M}\), the rate of the reaction is \(6.00 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} .\) What is the rate of the reaction when \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{ClO}_{2}\) and \(95.0 \mathrm{~mL}\) of \(0.155 \mathrm{M} \mathrm{NaOH}\) are added?

Short answer

Question: Calculate the rate of the reaction when 50.0 mL of 0.200 M ClO₂ solution is mixed with 95.0 mL of 0.155 M NaOH solution. The initial reaction rate when 0.010 M ClO₂ reacts with 0.030 M OH⁻ is 6.00 x 10⁻⁴ M/s. The reaction is first-order in OH⁻ and second-order in ClO₂. Answer: The new rate of the reaction is 0.00870 M/s.

Step-by-step solution

Write the rate law for the given reaction

We are given that the reaction is first-order in \(\mathrm{OH}^{-}\) and second-order in \(\mathrm{ClO}_{2}\). So the rate law for the reaction will be: Rate = \(k[\mathrm{ClO}_2]^2[\mathrm{OH}^-]\)

Calculate the rate constant, k, using the initial rate and initial concentrations

Plug in the initial concentrations and rate into the rate law equation to solve for k: \(6.00 \times 10^{-4} \mathrm{~M/s} = k(0.010 \mathrm{M})^2(0.030 \mathrm{M})\) Solve for k: \(k = \frac{6.00 \times 10^{-4} \mathrm{~M/s}}{(0.010 \mathrm{M})^2(0.030 \mathrm{M})}\) \(k = 2.00 \times 10^3\, \mathrm{M^{-2} s^{-1}}\)

Calculate the new concentrations of ClO\(_2\) and OH\(^-\) after mixing

New concentration of ClO\(_2\): \(\frac{(0.200\, \mathrm{M})(50.0\, \mathrm{mL})}{(50.0\, \mathrm{mL} + 95.0\, \mathrm{mL})}\) New concentration of ClO\(_2 = 0.0653\, \mathrm{M}\) New concentration of OH\(^-\): \(\frac{(0.155\, \mathrm{M})(95.0\, \mathrm{mL})}{(50.0\, \mathrm{mL} + 95.0\, \mathrm{mL})}\) New concentration of OH\(^- = 0.1007\, \mathrm{M}\)

Calculate the new rate using the calculated values of k, new concentration of ClO\(_2\), and new concentration of OH\(^-\)

Plug in the values of k, new concentration of ClO\(_2\), and new concentration of OH\(^-\) into the rate law equation to get the new rate: Rate = \((2.00 \times 10^{3}\, \mathrm{M^{-2} s^{-1}})(0.0653\, \mathrm{M})^2(0.1007\, \mathrm{M})\) New Rate = \(0.00870\, \mathrm{M/s}\)

Key concepts

Rate Law

Understanding the rate law is essential in chemical kinetics. It's an equation that relates the rate of a chemical reaction to the concentration of the reactants.

For a general reaction where reactants A and B form product C, the rate law might look something like this: Rate = k[A]ⁿ[B]^m, where [A] and [B] represent the molar concentrations of the reactants, n and m are the orders of the reaction with respect to A and B, and k is the rate constant. It's important to note that the values of n and m are not derived from the balanced chemical equation but are determined experimentally.

When solving problems involving the rate law, we first need to determine the value of the rate constant, k, which is unique for every reaction and depends on conditions such as temperature. With this constant, we can predict the rate of a reaction under different concentrations.

Reaction Order

The reaction order is the power to which the concentration of a reactant is raised in the rate law equation, and it gives important information about how sensitive the rate of reaction is to changes in concentration.

For instance, in the presented exercise, the reaction is first-order with respect to the hydroxide ion (OH^-), meaning that if the concentration of OH^- doubles, the rate of the reaction also doubles. Conversely, it's second-order with respect to chlorine dioxide (ClO₂), indicating that if the concentration of ClO₂ doubles, the rate increases by a factor of four (since 2² = 4).

The overall reaction order is the sum of the individual orders with respect to each reactant. In our example, the overall order is 1 (from OH^-) + 2 (from ClO₂) = 3. This tells us that the rate of the reaction is highly sensitive to changes in the concentrations of the reactants.

Rate Constant

The rate constant, k, is a proportionality factor that bridges the rate of a reaction with the concentrations of the reactants according to the rate law. It has a specific value for each reaction at a given temperature and is not affected by changes in reactant concentrations.

The units of the rate constant vary depending on the overall reaction order. For example, a second-order reaction has units of M^-1s^-1, as seen in our exercise, where k is calculated to be 2.00 x 10³M^-2s^-1. This constant is integral for calculating the rate of the reaction under new conditions, as it did not change when the reactant concentrations were altered.

Remember, if the temperature were to change, the value of the rate constant might also change, which is why temperature control is important in experiments regarding reaction rates.