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

The new rate of the reaction is 9.53 x 10^-2 mol/L·s.

Step-by-step solution

Find the rate constant (k) using the initially given data.

The rate equation is given by: \(\mathrm{Rate} = k[\mathrm{ClO}_{2}]^2[\mathrm{OH}^{-}]\) Using the given initial data, we can solve for k: $$ k = \frac{\mathrm{Rate}}{[\mathrm{ClO}_{2}]^2[\mathrm{OH}^{-}]} = \frac{6.00 \times 10^{-4}}{(0.010)^2 \times (0.030)} = 20.00 \mathrm{s}^{-1}$$ We now have the rate constant, \(k = 20.00 \mathrm{s}^{-1}\).

Calculate the initial concentrations of \(\mathrm{ClO}_{2}\) and \(\mathrm{OH}^{-}\) in the new mixture.

First, we need to find the moles of \(\mathrm{ClO}_{2}\) and \(\mathrm{OH}^{-}\) using the given volumes and concentrations: Moles of \(\mathrm{ClO}_{2} = \mathrm{Volume} \times \mathrm{Concentration} = 0.050 \, \mathrm L \times 0.200 \, \mathrm{M} = 0.010 \, \mathrm {mol}\) Moles of \(\mathrm{OH}^{-} = \mathrm{Volume} \times \mathrm{Concentration} = 0.095 \, \mathrm L \times 0.155 \, \mathrm{M} = 0.014725 \, \mathrm {mol}\) To find the new concentrations, we must divide moles by the total volume of the new mixture: Total volume = Volume of \(\mathrm{ClO}_{2}\) + Volume of \(\mathrm{OH}^{-} = 0.050 \, \mathrm L + 0.095 \, \mathrm L = 0.145 \, \mathrm L\) New concentration of \(\mathrm{ClO}_{2} = \frac{0.010 \, \mathrm {mol}}{0.145 \, \mathrm L} = 0.06897 \, \mathrm M\) New concentration of \(\mathrm{OH}^{-} = \frac{0.014725 \, \mathrm {mol}}{0.145 \, \mathrm L} = 0.10155 \, \mathrm M\)

Use the rate constant and the new initial concentrations to find the new reaction rate.

Now, we will use the rate equation to find the new reaction rate: $\mathrm{Rate} = k[\mathrm{ClO}_{2}]^2[\mathrm{OH}^{-}] = 20.00 \mathrm{s}^{-1}\times (0.06897)^2\times 0.10155\\ = 0.09530 \, \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}$ So the new rate of the reaction is \(9.53 \times 10^{-2} \mathrm{mol}/\mathrm{L} \cdot \mathrm{s}\).

Key concepts

Understanding Reaction Rate

The term reaction rate is fundamental to the study of chemical kinetics, a branch of physical chemistry. It refers to the speed at which reactants are converted into products in a chemical reaction. The rate can be expressed as the change in concentration of a reactant or product per unit time. For instance, in the provided exercise, the reaction rate initially was given as \(6.00 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\), representing how fast the reactants \(\mathrm{ClO}_{2}\) and \(\mathrm{OH}^{-}\) are turning into products.

To better understand reaction rates, imagine watching a video of a chemical reaction in process. A high reaction rate would be akin to a fast-forwarded video where the reactants quickly disappear, and the products form almost instantaneously. Conversely, a low reaction rate would suggest that the video is playing in slow motion, with reactants leisurely transitioning into products. Factors that can influence reaction rates include the concentration of reactants, temperature, and the presence of a catalyst.

Determining the Rate Constant

In chemical kinetics, the rate constant is a proportionality factor in the rate equation that provides a connection between the reaction rate and the concentrations of reactants. It is usually denoted as \(k\) and varies with temperature and the presence of a catalyst. The rate constant is a measure of the intrinsic speed at which a reaction proceeds.

In the exercise, solving for the rate constant required initial concentration values and the initial rate. By rearranging the rate equation \(\mathrm{Rate} = k[\mathrm{ClO}_{2}]^2[\mathrm{OH}^{-}]\), you were able to deduce that \(k = 20.00 \mathrm{s}^{-1}\). The dimensions of \(k\) depend on the reaction order, which in this case, is three (first-order with respect to \(\mathrm{OH}^{-}\) and second-order with respect to \(\mathrm{ClO}_{2}\)). The rate constant is critical because it remains constant at a given temperature, allowing you to predict the rate of the reaction with different concentrations of reactants.

Assessing Reaction Order

The reaction order is a telling aspect of how the rate is affected by the concentration of each reactant used in the reaction. It is essentially a sum of powers of the concentration terms in the rate equation. In the exercise, the reaction is described as first-order with respect to \(\mathrm{OH}^{-}\) and second-order with respect to \(\mathrm{ClO}_{2}\), totalling to a third-order reaction. This implies that doubling the concentration of \(\mathrm{OH}^{-}\) will double the reaction rate, whereas doubling \(\mathrm{ClO}_{2}\) would quadruple it.

To conceptualize this, think of reaction order as a recipe for a dish. The amount of each ingredient (reactant) affects the outcome of the dish (reaction rate). Some ingredients have more influence on the taste (rate), much like how the concentration of certain reactants affects the reaction more prominently depending on their order in the rate equation.