Question

In a solution at a constant \(\mathrm{H}^{+}\) concentration, iodide ions react with hydrogen peroxide to produce iodine. $$ \mathrm{H}^{+}(a q)+\mathrm{I}^{-}(a q)+\frac{1}{2} \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \frac{1}{2} \mathrm{I}_{2}(a q)+\mathrm{H}_{2} \mathrm{O} $$ The reaction rate can be followed by monitoring the appearance of \(\mathrm{I}_{2}\). The following data are obtained: $$ \begin{array}{ccc} \hline\left[\mathrm{I}^{-}\right] & {\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]} & \text { Initial Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{min}) \\ \hline 0.015 & 0.030 & 0.0022 \\ 0.035 & 0.030 & 0.0052 \\ 0.055 & 0.030 & 0.0082 \\ 0.035 & 0.050 & 0.0087 \\ \hline \end{array} $$ (a) Write the rate expression for the reaction. (b) Calculate \(k\). (c) What is the rate of the reaction when \(25.0 \mathrm{~mL}\) of a \(0.100 \mathrm{M}\) solution of \(\mathrm{KI}\) is added to \(25.0 \mathrm{~mL}\) of a \(10.0 \%\) by mass solution of \(\mathrm{H}_{2} \mathrm{O}_{2}(d=1.00 \mathrm{~g} / \mathrm{mL}) ?\) Assume volumes are additive.

Short answer

Question: Determine the rate expression, the value of the rate constant k, and the rate of reaction under specific conditions for a given iodide-hydrogen peroxide reaction. Answer: The rate expression is \(rate = k[\mathrm{I}^{-}][\mathrm{H}_{2} \mathrm{O}_{2}]\), the rate constant k is \(4.89 \mathrm{~L.mol^{-1}.min^{-1}}\), and the rate of reaction under the given conditions is \(0.0357\, \mathrm{mol/L.min}\).

Step-by-step solution

(a) Determine the rate expression

To find the rate expression for the reaction, we need to determine the order of the reaction with respect to each reactant. This can be done by comparing the initial rates for different concentrations of the reactants. Let the rate expression be: \(rate = k[\mathrm{I}^{-}]^m[\mathrm{H}_{2} \mathrm{O}_{2}]^n\) Comparing trials 1 and 2: \(\frac{0.0052}{0.0022} = \frac{k[\mathrm{I}^{-}]^m[\mathrm{H}_{2} \mathrm{O}_{2}]^n}{k[\mathrm{I}^{-}]^m[\mathrm{H}_{2} \mathrm{O}_{2}]^n} = \frac{(0.035)^m}{(0.015)^m}\) Since the concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is the same for both trials, it can be cancelled out. Solving for m: \(2.364 = (0.035/0.015)^m \Rightarrow m = 1\) Comparing trials 2 and 4: \(\frac{0.0087}{0.0052} = \frac{k[\mathrm{I}^{-}]^m[\mathrm{H}_{2} \mathrm{O}_{2}]^n}{k[\mathrm{I}^{-}]^m[\mathrm{H}_{2} \mathrm{O}_{2}]^n} = \frac{(0.05)^n}{(0.03)^n}\) Since the concentration of \(\mathrm{I}^{-}\) is the same for both trials, it can be cancelled out. Solving for n: \(1.673 = (0.05/0.03)^n \Rightarrow n = 1\) Given the order of the reaction with respect to each reactant, the rate expression is: \(rate = k[\mathrm{I}^{-}][\mathrm{H}_{2} \mathrm{O}_{2}]\)

(b) Calculate k

Now that we have the rate expression, we can use data from any trial to calculate the rate constant k. For instance, using trial 1: \(0.0022 = k(0.015)(0.03)\) Solving for k: \(k = \frac{0.0022}{(0.015)(0.03)} = 4.89 \mathrm{~L.mol^{-1}.min^{-1}}\)

(c) Calculate the rate of reaction

We are asked to calculate the reaction rate when \(25.0 \mathrm{~mL}\) of a \(0.100 \mathrm{~M}\) solution of \(\mathrm{KI}\) is mixed with \(25.0 \mathrm{~mL}\) of a \(10\%\) by mass solution of \(\mathrm{H}_{2} \mathrm{O}_{2}\), with a density of \(1.00 \mathrm{~g/mL}\). First, we need to find the new concentrations of \(\mathrm{I}^{-}\) and \(\mathrm{H}_{2} \mathrm{O}_{2}\) after mixing: For \(\mathrm{I}^{-}\): \(\mathrm{I}^{-}\; concentration = \frac{0.100\, mol/L \times 25.0\, mL}{50.0\, mL} = 0.050\, M\) For \(\mathrm{H}_{2} \mathrm{O}_{2}\): First, we need to find the concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) in the initial \(10\%\) by mass solution: \(10\%\) by mass means \(10\,g\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\) in \(100\,g\) solution. Since \(d = 1.00 \mathrm{~g/mL}\), the volume of \(100\,g\) solution is \(100\,mL\). \(\mathrm{H}_{2} \mathrm{O}_{2}\; initial\; concentration = \frac{10\,g \, \times \, \frac{1\,mol}{34.01\,g}}{100\,mL} = 0.293\,\mathrm{M}\) Now, we can find the new concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) after mixing: \(\mathrm{H}_{2} \mathrm{O}_{2}\; concentration = \frac{0.293\, mol/L \times 25.0\, mL}{50.0\, mL} = 0.146\, M\) Now, we can use the rate expression and the calculated k to find the rate of reaction: \(rate = k[\mathrm{I}^{-}][\mathrm{H}_{2} \mathrm{ O}_{2}] = 4.89 \mathrm{~L.mol^{-1}.min^{-1}} \times 0.050\, M \times 0.146\, M = 0.0357\, \mathrm{mol/L.min}\) The rate of the reaction is \(0.0357\, \mathrm{mol/L.min}\).