91Ó°ÊÓ

Consider the reaction described by \\[ \mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(\mathrm{aq})+\mathrm{SCN}^{-}(\mathrm{aq}) \longrightarrow \mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{SCN})^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\] for which the following initial rate data were obtained at \(298.15 \mathrm{K}\) \\[ \begin{array}{ccc} {\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\right]_{0} / \mathrm{mol} \cdot \mathrm{dm}^{-3}} & {\left[\mathrm{SCN}^{-}\right]_{0} / \mathrm{mol} \cdot \mathrm{dm}^{-3}} & v_{0} / \mathrm{mol} \cdot \mathrm{dm}^{-3} \cdot \mathrm{s}^{-1} \\ \hline 1.21 \times 10^{-4} & 1.05 \times 10^{-5} & 2.11 \times 10^{-11} \\ 1.46 \times 10^{-4} & 2.28 \times 10^{-5} & 5.53 \times 10^{-11} \\ 1.66 \times 10^{-4} & 1.02 \times 10^{-5} & 2.82 \times 10^{-11} \\ 1.83 \times 10^{-4} & 3.11 \times 10^{-5} & 9.44 \times 10^{-11} \end{array} \\] Determine the rate law for the reaction and the rate constant at \(298.15 \mathrm{K}\). Assume the urders are integers.

Step by step solution

01

Write the General Rate Law Equation

The general rate law for the given reaction can be written as \( v = k [\mathrm{Cr(H_2O)}_6^{3+}]^m [\mathrm{SCN}^-]^n \), where \( v \) is the rate, \( k \) is the rate constant, and \( m \) and \( n \) are the orders of reaction with respect to \( \mathrm{Cr(H_2O)}_6^{3+} \) and \( \mathrm{SCN}^- \) respectively.

02

Determining the Orders of Reaction

We compare experiments where only one concentration changes at a time. Between experiments 1 and 3, \([\mathrm{Cr(H_2O)}_6^{3+}]\) increases and \([\mathrm{SCN}^-]\) remains nearly constant, leading to the ratio \( \frac{2.82 \times 10^{-11}}{2.11 \times 10^{-11}} = (\frac{1.66 \times 10^{-4}}{1.21 \times 10^{-4}})^m \). Solving gives \( m \approx 1 \). Next, comparing experiments 2 and 4, where \([\mathrm{SCN}^-]\) changes, gives \( \frac{9.44 \times 10^{-11}}{5.53 \times 10^{-11}} = (\frac{3.11 \times 10^{-5}}{2.28 \times 10^{-5}})^n \). Solving, \( n \approx 1 \). Therefore, both orders are 1.

03

Write the Full Rate Law

Since both orders were found to be 1, the rate law is \( v = k [\mathrm{Cr(H_2O)}_6^{3+}] [\mathrm{SCN}^-] \).

04

Calculate the Rate Constant

Using any of the experiment data and the rate law, solve for \( k \). Using experiment 1: \( 2.11 \times 10^{-11} = k (1.21 \times 10^{-4})(1.05 \times 10^{-5}) \), solve for \( k \), resulting in \( k = \frac{2.11 \times 10^{-11}}{(1.21 \times 10^{-4})(1.05 \times 10^{-5})} \approx 1.67 \times 10^{4} \, \text{mol}^{-1} \cdot \text{dm}^3 \cdot \text{s}^{-1} \).

05

Verify Consistency

Confirm the consistency of \( k \) with the other experiments. Calculate \( k \) using data from experiments 2, 3, and 4 as done in Step 4, and ensure the values of \( k \) are approximately the same to validate the reaction orders and rate constant calculations.

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Key Concepts

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

Reaction Orders

In chemical kinetics, reaction orders play a pivotal role in understanding how varying concentrations of reactants influence the rate of a reaction. For the given reaction, the rate law equation is presented as \( v = k [\mathrm{Cr(H_2O)}_6^{3+}]^m [\mathrm{SCN}^-]^n \). Here, \( m \) and \( n \) represent the reaction orders with respect to each reactant.

To determine these orders, only one reactant's concentration should change while keeping the others constant. By analyzing experimental data, reaction order \( m \) is found by comparing experiments 1 and 3, resulting in \( m \approx 1 \). Similarly, \( n \) is discovered by comparing experiments 2 and 4, where \( n \approx 1 \).

The conclusion is that the reaction is first order with respect to both \( \mathrm{Cr(H_2O)}_6^{3+} \) and \( \mathrm{SCN}^- \). Understanding these orders allows predictions about how changes in concentration will impact the reaction speed.

Rate Constant Calculation

Once the reaction orders are known, the next step is to calculate the rate constant \( k \). The rate constant is a crucial parameter that quantifies the reaction speed at a specific temperature, independent of reactant concentration.

Using the rate law, \( v = k [\mathrm{Cr(H_2O)}_6^{3+}] [\mathrm{SCN}^-] \), the rate constant can be found from the initial rate data of any experiment. For instance, using experiment 1's data:

• Initial rate \( v_0 = 2.11 \times 10^{-11} \)
• \( [\mathrm{Cr(H_2O)}_6^{3+}] = 1.21 \times 10^{-4} \)
• \( [\mathrm{SCN}^-] = 1.05 \times 10^{-5} \)

Substitute these values into the rate law and solve for \( k \), resulting in approximately \( k \approx 1.67 \times 10^{4} \, \text{mol}^{-1} \cdot \text{dm}^3 \cdot \text{s}^{-1} \).

Ensuring the values are consistent across different experiments confirms the reliability of the calculated \( k \). The rate constant is vital for understanding how the reaction progresses.

Chemical Kinetics

Chemical kinetics is the study of reaction rates and the factors affecting them. It's central to predicting how changes in conditions affect the speed of chemical reactions. For any reaction, knowing the rate and the mechanism provides insights into the reactivity of the substances involved.

In our example, the reaction between \( \mathrm{Cr(H_2O)}_6^{3+} \) and \( \mathrm{SCN}^- \) is explored at a molecular level. By identifying the reaction's mechanism and measuring the rate under varied conditions, kinetics aids in identifying the role each reactant plays in the process.

Utilizing this knowledge, chemical kinetics provides crucial information for developing new reactions or optimizing existing processes in industrial and research settings.

Experiment Analysis

Analyzing experimental data is crucial for deriving meaningful insights into chemical reaction behavior. In this exercise, by carefully selecting and comparing different sets of experiments, one can validate theoretical predictions about reaction rates and mechanisms.

To determine reaction orders and the rate constant, comparisons of specific experiments allowed isolation of each reactant's effect. This method ensured accuracy in defining how each concentration change impacts the reaction rate.

Thorough experiment analysis allows chemists to back up their findings, ensuring all experimental variables are considered. This approach enhances the robustness of the conclusions drawn about the reaction's behavior, forming the foundation for further research and application.