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Consider the base-catalyzed reaction \(\mathrm{OCl}^{-}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq}) \longrightarrow \mathrm{OI}^{-}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\) Use the following initial-rate data to determine the rate law and the corresponding rate constant for the reaction. $$\begin{array}{cccc} {\left[\mathrm{OCl}^{-}\right] / \mathrm{mol} \cdot \mathrm{dm}^{-3}} & {\left[\mathrm{I} \mathrm{I} / \mathrm{mol} \cdot \mathrm{dm}^{-3}\right.} & {\left[\mathrm{OH}^{-}\right] / \mathrm{mol} \cdot \mathrm{dm}^{-3}} & v_{0} / \mathrm{mol} \cdot \mathrm{dm}^{-3} \cdot \mathrm{s}^{-1} \\ \hline 1.62 \times 10^{-3} & 1.62 \times 10^{-3} & 0.52 & 3.06 \times 10^{-4} \\\ 1.62 \times 10^{-3} & 2.88 \times 10^{-3} & 0.52 & 5.44 \times 10^{-4} \\ 2.71 \times 10^{-3} & 1.62 \times 10^{-3} & 0.84 & 3.16 \times 10^{-4} \\ 1.62 \times 10^{-3} & 2.88 \times 10^{-3} & 0.91 & 3.11 \times 10^{-1} \end{array}$$

Step by step solution

01

Identifying the Rate Law Form

For reactions in a solution, the general form of the rate law is \( v_0 = k [\text{OCl}^-]^a [\text{I}^-]^b [\text{OH}^-]^c \). We need to determine the exponents \(a\), \(b\), and \(c\) to find the actual rate law.

02

Determining Order with Respect to OCl鈦�

Compare experiments 1 and 3 where the concentration of \( [\text{I}^-] \) and \( [\text{OH}^-] \) are held constant:\[\frac{v_3}{v_1} = \frac{k [\text{OCl}^-]_3^a}{k [\text{OCl}^-]_1^a} = \frac{3.16 \times 10^{-4}}{3.06 \times 10^{-4}} = 1.03\]This ratio implies \( [\text{OCl}^-]^a \) has minimal impact, suggesting \( a \approx 0 \), making it zero-order with respect to \( \text{OCl}^- \).

03

Determining Order with Respect to I鈦�

Compare experiments 1 and 2 where \( [\text{OCl}^-] \) and \( [\text{OH}^-] \) are constant:\[\frac{v_2}{v_1} = \frac{k [\text{I}^-]_2^b}{k [\text{I}^-]_1^b} = \frac{5.44 \times 10^{-4}}{3.06 \times 10^{-4}} \approx 1.78\]\[\frac{1.78}{1} = \frac{[2.88 \times 10^{-3}]^b}{[1.62 \times 10^{-3}]^b} \approx \left(\frac{2.88}{1.62}\right)^b = 1.78\]Solving gives \( b = 1 \). Thus, the reaction is first order in \( \text{I}^- \).

04

Determining Order with Respect to OH鈦�

Compare experiments 2 and 4 where \( [\text{OCl}^-] \) and \( [\text{I}^-] \) are constant:\[\frac{v_4}{v_2} = \frac{k [\text{OH}^-]_4^c}{k [\text{OH}^-]_2^c} = \frac{3.11 \times 10^{-1}}{5.44 \times 10^{-4}} \approx 572\]\[572 = \left(\frac{0.91}{0.52}\right)^c\]Solving for \(c\) gives \( c = 5 \). Thus, the reaction is fifth order in \( \text{OH}^- \).

05

Writing the Rate Law

The determined rate orders allow us to write the rate law as:\[v_0 = k [\text{I}^-]^1 [\text{OH}^-]^5\]

06

Calculating the Rate Constant

Using any data row, e.g., experiment 1:\[3.06 \times 10^{-4} = k [1.62 \times 10^{-3}]^1 [0.52]^5\]Solving for \(k\):\[k = \frac{3.06 \times 10^{-4}}{1.62 \times 10^{-3} \times (0.52)^5} \approx 5.0 \times 10^{3} \text{ mol}^{-4} \text{dm}^{12} \text{s}^{-1}\]

07

Conclusion

The rate law is \( v_0 = k [\text{I}^-][\text{OH}^-]^5 \) with \( k = 5.0 \times 10^{3} \text{ mol}^{-4} \text{dm}^{12} \text{s}^{-1} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate Law

Understanding the rate law is key to grasping chemical kinetics. It involves a mathematical expression that relates the rate of a reaction to the concentration of its reactants. In the case of a base-catalyzed reaction like \(\mathrm{OCl}^{-} + \mathrm{I}^{-} \longrightarrow \mathrm{OI}^{-} + \mathrm{Cl}^{-}\), the rate law can be written as \( v_0 = k [\mathrm{OCl}^{-}]^a [\mathrm{I}^{-}]^b [\mathrm{OH}^{-}]^c \). In this expression:

• \(v_0\) represents the initial rate of reaction.
• \(k\) is the rate constant, a specific value for each reaction at a given temperature.
• \([\mathrm{OCl}^{-}], [\mathrm{I}^{-}], [\mathrm{OH}^{-}]\) are concentrations of the reactants.
• \(a, b, c\) are the orders of the reaction with respect to each reactant.

These terms are crucial as they guide how we change the amount of each substance to control the rate of the reaction.

Reaction Order

The reaction order is perhaps the most enlightening concept as it indicates the influence of each reactant on the rate of reaction. It is determined experimentally, often by holding concentrations constant and observing the rate changes. For the example reaction:- **Zero Order**: If a reactant shows zero order, like \(\mathrm{OCl}^{-}\) in this case, changing its concentration has little to no effect on the reaction rate.- **First Order**: When a reactant has a first order, such as \(\mathrm{I}^{-}\), the reaction rate linearly increases with its concentration.- **Fifth Order**: Impressively, \(\mathrm{OH}^{-}\) has a fifth order, meaning even minor changes in its concentration will dramatically affect the reaction rate. Understanding these orders helps manipulate conditions to achieve desired speeds in reactions.

Rate Constant

The rate constant \(k\) is a crucial part of the rate law that often puzzles students. It quantifies the relationship between reactant concentrations and reaction rate. Unlike concentration orders, \(k\) has a fixed value for a reaction at a constant temperature. For the reaction studied, \(k\) is determined by plugging values into the rate law equation.In our original exercise, solving this involved using initial concentrations from a specific dataset:\[k = \frac{3.06 \times 10^{-4}}{1.62 \times 10^{-3} \times (0.52)^5}\]This calculation yielded \(k = 5.0 \times 10^{3} \text{ mol}^{-4} \text{dm}^{12} \text{s}^{-1}\). Understanding how to derive and apply \(k\) enables predictions on how the reaction behaves under various conditions.

Base-Catalyzed Reaction

Base-catalyzed reactions, like the one involving \(\mathrm{OCl}^{-}\) and \(\mathrm{I}^{-}\), are intriguing due to their dependence on the presence of a base, such as \(\mathrm{OH}^{-}\). In these reactions, the base increases the reaction rate without itself being consumed.In this particular reaction:

• \(\mathrm{OH}^{-}\) acts as a catalyst, enhancing the formation of \(\mathrm{OI}^{-}\) and \(\mathrm{Cl}^{-}\).
• The fifth order dependence on \(\mathrm{OH}^{-}\) signifies its powerful role in accelerating the reaction.

Catalysis is a key concept in chemical kinetics, as it can efficiently lower the activation energy needed, allowing reactions to occur swiftly and efficiently under suitable conditions.