Problem 32 - Chemistry: Principles and Reactions

Question

Consider the reaction $$\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \longrightarrow \mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{SCN}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}$$ The following data were obtained: $$\begin{array}{ccc}\hline\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{\underline{\phantom{xx}}}^{3+}\right] & {\left[\mathrm{SCN}^{-}\right]} & \text {Initial Rate }(\mathrm{mol} /\mathrm{L} \cdot \mathrm{min}) \\ \hline 0.025 & 0.060 & 6.5 \times 10^{-4} \\\0.025 & 0.077 & 8.4 \times 10^{-4} \\\0.042 & 0.077 & 1.4 \times 10^{-3} \\ 0.042 & 0.100 & 1.8 \times 10^{-3} \\\\\hline\end{array}$$ (a) Write the rate expression for the reaction. (b) Calculate \(k\). (c) What is the rate of the reaction when \(15 \mathrm{mg}\) of \(\mathrm{KSCN}\) is added to \(1.50 \mathrm{~L}\) of a solution \(0.0500 \mathrm{M}\) in \(\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{\underline{\phantom{xx}}}^{3+}\) ?

Step-by-Step Solution

Answer

Question: Determine the rate of the reaction involving Chromium hexaaqua ion and thiocyanate ion when 15 mg of KSCN is added to 1.50 L of a 0.0500 M Cr(H₂O)₆³⁺ solution. Answer: The rate of the reaction under the given conditions is approximately 2.22 × 10⁻³ M/min.

Step 1: (Step 1: Examine the experimental data)

Observe the concentration of each reactant and the corresponding initial reaction rate in each experiment. We will be using this data to find the reaction order with respect to each reactant, including chromium hexaaqua and thiocyanate ions.

Step 2: (Step 2: Determine the reaction order for each reactant)

To determine the reaction order for each reactant in the rate expression, we can look at how the rate changes when the concentration of the individual reactants is changed, keeping the other reactant constant. First, compare Experiment 1 and 3 where the thiocyanate concentration ([SCN⁻]) is constant at 0.077M: - From Experiment 1(Exp1) to 3(Exp3): - Concentration of Cr(H₂O)₆³⁺ doubles (0.025M to 0.042M) - The initial rate doubles (6.5 × 10⁻⁴ to 1.4 × 10⁻³); This is a first-order reaction with respect to Cr(H₂O)₆³⁺. Now, let's compare the experiments where the Chromium hexaaqua concentration ([Cr(H₂O)₆³⁺]) is constant at 0.025M. Compare Experiment 1 and 2: - From Exp1 to Exp2: - Concentration of SCN⁻ increases from 0.060M to 0.077M (1.283 times) - The initial rate increases from 6.5 × 10⁻⁴ to 8.4 × 10⁻⁴ (1.292 times); This indicates that the reaction is also first-order with respect to SCN⁻.

Step 3: (Step 3: Write the rate expression)

Since the reaction is first-order with respect to both reactants, we can write the rate expression as: Rate = k[Cr(H₂O)₆³⁺][SCN⁻].

Step 4: (Step 4: Calculate the rate constant k)

We can use the data from any experiment to calculate the rate constant (k). Let's choose Experiment 1: Rate = 6.5 × 10⁻⁴ M/min [Cr(H₂O)₆³⁺] = 0.025 M [SCN⁻] = 0.060 M Plugging these into the rate equation: 6.5 × 10⁻⁴ = k(0.025)(0.060). Solve for k: k ≈ 0.433 min⁻¹.

Step 5: (Step 5: Calculate the rate of the reaction under given conditions)

We are asked to determine the reaction rate when 15 mg of KSCN is added to 1.50 L of a 0.0500 M Cr(H₂O)₆³⁺ solution. First, we need to find the concentration of SCN⁻ in the solution: Molarity of KSCN = (15 mg KSCN) x (1 mol KSCN / 97.181g) x (1 L/ 1500 mL) ≈ 0.0103 M. Now we have the concentrations of reactants for the reaction: [Cr(H₂O)₆³⁺] = 0.0500 M [SCN⁻] = 0.0103 M Use the rate equation with the calculated rate constant to find the reaction rate: Rate = (0.433 min⁻¹)(0.0500)(0.0103) ≈ 2.22 × 10⁻³ M/min.

Key Concepts

Rate Expression

The rate expression is a mathematical equation that describes how the rate of a chemical reaction depends on the concentration of the reactants. For the reaction given in the problem, the rate expression can be deduced by examining the experimental data provided. By analyzing changes in concentration and the corresponding reaction rates, you can determine how each reactant concentration affects the overall rate.

• The reaction: \( \text{Cr(H}_2\text{O)}_6^{3+} + \text{SCN}^- \longrightarrow \text{Cr(H}_2\text{O)}_5\text{SCN}^{2+} + \text{H}_2\text{O} \)
• Rate expression: \( \text{Rate} = k[\text{Cr(H}_2\text{O)}_6^{3+}][\text{SCN}^-] \)

This is a simple model where the rate is directly proportional to the concentration of each reactant, leading to a first-order dependence for each.

Reaction Order

The reaction order is an essential concept in reaction kinetics, as it defines how the rate of a reaction is affected by the concentration of reactants. It is determined by analyzing experimental data where concentrations vary while others remain constant. In this particular reaction:

• The rate doubles as the concentration of \( \text{Cr(H}_2\text{O)}_6^{3+} \) doubles, indicating first-order dependence.
• Similarly, holding the concentration of \( \text{Cr(H}_2\text{O)}_6^{3+} \) constant and changing \([\text{SCN}^-]\) reveals that an increase in SCN⁻ also leads to a proportional rate increase, confirming it is first-order with respect to \( \text{SCN}^- \).

This means the overall reaction is second-order – first-order with respect to \( \text{Cr(H}_2\text{O)}_6^{3+} \) and first-order with respect to \( \text{SCN}^- \).

Rate Constant

The rate constant, represented by \(k\), is a critical factor in the rate expression and reflects how fast the reaction progresses under specific conditions. To find \(k\), you need to substitute known values from the experiment into the rate expression. For example, using the data from one of the trials:

• Using Experiment 1, Rate = \(6.5 \times 10^{-4} \) M/min, \([\text{Cr(H}_2\text{O)}_6^{3+}] = 0.025 \) M, and \([\text{SCN}^-] = 0.060 \) M.
• Insert values into the rate expression: \(6.5 \times 10^{-4} = k \cdot 0.025 \cdot 0.060\)
• Solve for \(k\) to discover: \(k ≈ 0.433 \text{ min}^{-1} \).

This value of \(k\) is constant for this reaction given the temperature and conditions specified.

Concentration Calculation

Accurate concentration calculation of reactants is essential for predicting reaction rates. When additional amounts of a reactant are involved, precise calculation ensures that you can use the rate expression correctly. In the example problem, when adding 15 mg of \(\text{KSCN}\) to a solution, you calculate the concentration like this:

• Convert mass of KSCN to moles using its molar mass: \(15 \text{ mg} \times \frac{1 \text{ mol}}{97.181 \text{ g}}\).
• Convert volume to liters: \(1.50 \text{ L}\).
• Calculate resulting molarity: \( \approx 0.0103 \text{ M} \).

This concentration joins the concentration of \( \text{Cr(H}_2\text{O)}_6^{3+} \), allowing you to substitute both into the rate expression to find the new rate of reaction. Understanding these calculations permits you to handle varied scenarios in reaction kinetics accurately.