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: $$\begin{array}{ccc}\hline\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}\right] & {\left[\mathrm{SCN}^{-}\right]} & \text {Initial Rate }(\mathrm{mol} /\mathrm{L} \cdot \mathrm{min}) \\ \hline 0.025 & 0.060 & 6.5 \times 10^{-4} \\\0.025 & 0.077 & 8.4 \times 10^{-4} \\\0.042 & 0.077 & 1.4 \times 10^{-3} \\ 0.042 & 0.100 & 1.8 \times 10^{-3} \\\\\hline\end{array}$$ (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 \(\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}\) ?

Step by step solution

01

(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.

02

(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⁻.

03

(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⁻].

04

(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.

05

(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.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!