In a solution at a constant H^+ concentration, iodide ions react with hydrogen peroxide to produ

In a solution at a constant H^+ concentration, iodide ions...

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.

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.

Key Concepts

Dilution and Concentration Calculations

When solutions are mixed, their volumes change, and the concentrations of the solutes must be recalculated accordingly. This is important in kinetics as the rate depends on the concentration of the reactants. The process involves using the initial amounts (moles) of reactants and the new total volume to determine the resultant concentrations for further use in rate calculations.

Rate Constant Calculation

The rate constant, k, is a proportionality factor in the rate law that is independent of the reactant concentrations but can depend on temperature and other conditions. Calculating k involves substituting experimental values for the rate and reactant concentrations into the rate law. It provides a quantitative measure of the reaction speed under given conditions.

Initial Rate Method

The initial rate method involves measuring the reaction rate immediately after the reactants are mixed, before significant product formation occurs. This method is useful for kinetic studies because it allows for the determination of rate laws by minimizing complications from reverse reactions or changing concentrations during the course of the reaction.

Rate Law

The rate law in chemical kinetics is an equation that expresses the rate of a chemical reaction in terms of the concentrations of the reactants, each raised to an exponent that represents the order of reaction with respect to that reactant. It provides insights into how changes in concentration affect the speed of the reaction and is foundational for understanding reaction mechanisms.

Reaction Order Determination

Determining the reaction order involves analyzing experimental data by seeing how the rate changes as the concentration of one reactant is varied while others are held constant. The exponents in the rate law (reaction orders) indicate the dependency of the rate on the concentration of each reactant, and they are crucial for understanding the kinetics and mechanism of the reaction.

Transcript

00:02 Let's go over how to determine the rate law for the reaction in this problem.

00:11 We need to find the experiments in which one of the reacting concentrations remains the same and one of them changes.

00:26 So if we look at reaction 1 and 3, we can see that hydrogen peroxide concentration remains the same, and the iodide concentration changes.

00:41 So let's take a look at how the iodide concentration changes and how this changes the rate of the reaction.

00:54 So if we divide the concentrations of iodide then we see that the concentration goes up by a of four.

01:15 Now we're going to do the same with the rates.

01:28 So we're going to take the rate from reaction three divided by the rate from reaction one and we also get about four.

01:51 So from this we can see since the concentration and the reaction rate go up by the same factor, this reaction is first order in the iodide ion.

02:09 Now we're going to look at experiment two and four.

02:13 So in these two experiments you can see that the iodide ion concentration stays the same but the hydrogen peroxide concentration is changing.

02:22 So let's determine how the hydrogen and peroxide concentration changes.

02:28 So if we divide the concentrations, we see that the concentration goes up by a factor of two and then we're going to do the same for the rates and we see that's also a factor of two.

02:56 So because the concentration and the rate is going to do the rate is going to up by the same factor.

03:04 It's also a first order in hydrogen peroxide.

03:07 So this is going to be the expression for the rate law.

03:12 Now we are asked to calculate the value of k.

03:17 To get the value of k, we're going to plug in the numbers from any experiment.

03:24 So i'm just going to do experiment one.

03:41 So now we're going to to plug in the numbers and now solve for k.

03:50 So this is the value that we get for k...