Question

The following set of data was obtained by the method of initial rates for the reaction: $$ \begin{aligned} &\mathrm{S}_{2} \mathrm{O}_{8}^{2-}(\mathrm{aq})+3 \mathrm{I}^{-}(\mathrm{aq}) \rightarrow \\ &2 \mathrm{SO}_{4}^{2-}(\mathrm{aq})+\mathrm{I}_{3}-(\mathrm{aq}) \end{aligned} $$ What is the rate law for the reaction? $$ \begin{array}{lll} \hline\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right], \mathrm{M} & {[\mathrm{I}-], \mathrm{M}} & \text { Initial rate, } \mathrm{M} \mathrm{s}^{-1} \\ \hline 0.25 & 0.10 & 9.00 \times 10^{-3} \\ 0.10 & 0.10 & 3.60 \times 10^{-3} \\ 0.20 & 0.30 & 2.16 \times 10^{-2} \\ \hline \end{array} $$ a. Rate \(=\mathrm{k}\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right]\left[\mathrm{I}^{-}\right]^{2}\) b. Rate \(=\mathrm{k}\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right]^{2}\left[\mathrm{I}^{-}\right]\) c. Rate \(=\mathrm{k}\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right]\left[\mathrm{I}^{-}\right]\) d. Rate \(=\mathrm{k}\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right]\left[\mathrm{I}^{-}\right]^{5}\)

Short answer

The rate law is \( \text{Rate} = \text{k} [\mathrm{S}_{2} \mathrm{O}_{8}^{2-}] [\mathrm{I}^{-}] \).

Step-by-step solution

Understanding the Rate Law

The rate law for a reaction is determined by the relation \( ext{rate} = ext{k}[ ext{reactant}_1]^{x}[ ext{reactant}_2]^{y} \), where k is the rate constant, and x and y are the reaction orders with respect to each reactant. Our task is to find these orders using the given experimental data.

Analyzing Data to Determine Order of \([\mathrm{S}_{2} \mathrm{O}_{8}^{2-}]\)

Compare experiments where the concentration of \([\mathrm{I}^{-}]\) remains constant, i.e., the first and second experiments: \(\begin{align*} \text{Experiment 1:} & [\mathrm{S}_{2} \mathrm{O}_{8}^{2-}] = 0.25, \text{rate} = 9.00 \times 10^{-3} \ \text{Experiment 2:} & [\mathrm{S}_{2} \mathrm{O}_{8}^{2-}] = 0.10, \text{rate} = 3.60 \times 10^{-3} \end{align*}\).Calculate the rate ratio:\, \(\frac{9.00 \times 10^{-3}}{3.60 \times 10^{-3}} = 2.5\). Calculate the concentration ratio:\, \(\frac{0.25}{0.10} = 2.5\). Since \(2.5 = 2.5^{1}\), the order with respect to \([\mathrm{S}_{2} \mathrm{O}_{8}^{2-}]\) is 1.

Analyzing Data to Determine Order of \([\mathrm{I}^{-}]\)

Now, compare experiments where the concentration of \([\mathrm{S}_{2} \mathrm{O}_{8}^{2-}]\) remains constant, i.e., the second and third experiments: \(\begin{align*} \text{Experiment 2:} & [\mathrm{I}^{-}] = 0.10, \text{rate} = 3.60 \times 10^{-3} \ \text{Experiment 3:} & [\mathrm{I}^{-}] = 0.30, \text{rate} = 2.16 \times 10^{-2} \end{align*}\). Calculate the rate ratio:\, \(\frac{2.16 \times 10^{-2}}{3.60 \times 10^{-3}} = 6.0\). Calculate the concentration ratio:\, \(\frac{0.30}{0.10} = 3.0\). Since \(6.0 = 3.0^{1}\), the order with respect to \([\mathrm{I}^{-}]\) is 1.

Formulating the Rate Law

With the reaction orders for both \([\mathrm{S}_{2} \mathrm{O}_{8}^{2-}]\) and \([\mathrm{I}^{-}]\) as 1, the rate law is:\( \text{Rate} = \text{k} [\mathrm{S}_{2} \mathrm{O}_{8}^{2-}]^{1} [\mathrm{I}^{-}]^{1} \). Simplifying, we get \( \text{Rate} = \text{k} [\mathrm{S}_{2} \mathrm{O}_{8}^{2-}] [\mathrm{I}^{-}] \).

Key concepts

Reaction Order

The concept of reaction order is crucial in understanding how the concentration of reactants influences the rate of a chemical reaction. In a rate law, reaction order refers to the power to which the concentration of a reactant is raised. For example, if we have a rate law expressed as \( \text{Rate} = \text{k}[\text{A}]^x[\text{B}]^y \), then \( x \) and \( y \) are the orders of the reaction with respect to reactants A and B respectively.

Reaction order cannot be deduced from the balanced equation but rather must be determined experimentally. It tells how the rate is affected by the concentration of the reactants.- **Zero-order**: The rate is independent of the concentration of that reactant.- **First-order**: The rate is directly proportional to the concentration of that reactant.- **Second-order**: The rate is proportional to the square of the concentration of that reactant.

In the given exercise, the reaction order with respect to \([\mathrm{S}_{2}\mathrm{O}_{8}^{2-}]\) was determined to be 1 by comparing experiments with constant \([\mathrm{I}^{-}]\) concentration. Similarly, the reaction order with respect to \([\mathrm{I}^{-}]\) was also found to be 1, indicating that the rate depends linearly on both reactants.

Method of Initial Rates

The method of initial rates is a common experimental approach used to determine the rate law of a chemical reaction. By conducting experiments with different initial concentrations of reactants, one can observe how the initial rate of reaction varies. This method is particularly useful because it allows chemists to deduce reaction order from practical data without having to fully understand the mechanism of the reaction.

To apply this method, experiments are conducted multiple times with varying initial concentrations of one reactant while keeping the others constant. The initial rate is measured for each setup.- **Comparing Rates and Concentrations**: By comparing the ratio of initial rates to the ratio of concentrations, one can determine the reaction order for each reactant.- **Example from Exercise**: When the concentration of \([\mathrm{S}_{2}\mathrm{O}_{8}^{2-}]\) was changed, the initial rate provided evidence for first-order dependence, as seen from the matching rate and concentration ratios.

This approach is elegant for its simplicity and provides a clear insight into how changes in concentrations affect the reaction speed.

Rate Constant

The rate constant, denoted as \( k \), is a crucial component of the rate law. It provides the proportionality factor that links the rate of reaction with the concentrations of reactants raised to their respective powers, as determined by the reaction order. The rate constant is unique for each reaction and changes with temperature, so it reflects the specific conditions under which the reaction occurs.

Key traits of the rate constant:- **Units**: The units of \( k \) depend on the overall order of the reaction. For a first-order reaction, \( k \) might have units of \( s^{-1} \), while for a second-order reaction, it could be \( M^{-1} \cdot s^{-1} \).- **Effect of Temperature**: As temperature increases, typically, the rate constant increases, reflecting a higher rate of reaction.

In the exercise provided, while specific values for \( k \) are not given, it represents a constant that would be experimentally determined given the correct initial conditions and can vary with different experimental setups.