Problem 67 - Chemistry The Science in Context

Question

Hydrogen peroxide decomposes spontaneously into water and oxygen gas via a first-order reaction: $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g) $$ but in the absence of catalysts this reaction proceeds very slowly. If a small amount of a salt containing the \(\mathrm{Fe}^{3+}\) ion is added to a \(0.437 M\) solution of \(\mathrm{H}_{2} \mathrm{O}_{2}\) in water, the reaction proceeds with a half-life of 17.3 min. What is the concentration of the solution after 10.0 min under these conditions?

Step-by-Step Solution

Answer

Answer: The concentration of the hydrogen peroxide solution after 10.0 minutes is 0.260 M.

Step 1: Identify relevant formulas and given values

For a first-order reaction, the formula that relates the initial concentration, final concentration, rate constant, and time is $$ \ln\frac{[A]_0}{[A]} = kt $$ where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \(t\), \(k\) is the rate constant, and \(t\) is the time elapsed. Additionally, for a first-order reaction, the half-life (\(t_{1/2}\)) is related to the rate constant as $$ t_{1/2} = \frac{\ln 2}{k} $$ We are given the initial concentration \([A]_0 = 0.437 M\), the half-life \(t_{1/2} = 17.3\) min, and we want to find the concentration at \(t = 10.0\) min.

Step 2: Calculate the rate constant from the half-life

Using the half-life formula, we can find the rate constant: $$ k = \frac{\ln 2}{t_{1/2}} = \frac{\ln 2}{17.3} = 0.0401 \textrm{ min}^{-1} $$

Step 3: Use the first-order reaction formula to find the concentration at 10.0 min

Now, using the first-order reaction formula and the calculated value of \(k\), we will calculate the concentration of hydrogen peroxide at \(t = 10.0\) min. Plug in the given values: $$ \ln\frac{0.437}{[A]} = 0.0401 \times 10.0 $$ Solve for \([A]\): $$ \frac{0.437}{[A]} = \exp(0.401) \\ [A] = \frac{0.437}{\exp(0.401)} \\ [A] = 0.260\,\text{M} $$

Step 4: State the final answer

The concentration of the hydrogen peroxide solution after 10.0 minutes under these conditions is 0.260 M.

Key Concepts

Reaction Kinetics

Chemical reactions occur at different rates, and understanding these speeds is essential in chemistry. Reaction kinetics studies these rates and the factors that influence them. For first-order reactions, the rate depends linearly on the concentration of a single reactant. This means if you double the concentration of the reactant, the reaction rate also doubles.
Key factors affecting reaction rates include:

• Concentration of reactants
• Temperature
• Presence of a catalyst
• Surface area available for reaction

Understanding how these factors work together helps chemists control and predict the outcomes of reactions. In the case of hydrogen peroxide's decomposition, adding a catalyst speeds up the process, offering a practical application of reaction kinetics.

Rate Constant Calculation

The rate constant, denoted as \( k \), is a crucial parameter in the kinetics of chemical reactions. It provides information about the speed of a reaction at a given temperature. The value of the rate constant depends on the type of reaction and the conditions under which it occurs.
For first-order reactions, the relationship between the rate constant and half-life is given by the equation:\[t_{1/2} = \frac{\ln(2)}{k}\]This formula allows us to calculate \( k \) when the half-life is known. In our example, the half-life of hydrogen peroxide is 17.3 minutes, so:\[k = \frac{\ln(2)}{17.3} = 0.0401 \text{ min}^{-1}\]Once you have the rate constant, you can use it to find the concentration of a reactant at any time through the first-order reaction equation. This illustrates the importance of \( k \) in predicting reaction behavior.

Half-Life of Reactions

The concept of half-life is commonly used in the study of first-order reactions. The half-life of a reaction is the time taken for the concentration of a reactant to decrease to half its initial value. It is an important metric because it provides a simple way to describe the speed of a reaction.
For first-order reactions, the half-life is independent of the initial concentration, which means no matter how much reactant you start with, the time it takes to reduce to half will always be the same.
With the formula:\[t_{1/2} = \frac{\ln(2)}{k}\]the half-life can be easily calculated if the rate constant \( k \) is known. This property makes half-life particularly useful for reactions that proceed at a measurable rate. It helps chemists and industry professionals predict how long a chemical, such as a drug or a pollutant, will remain active or at necessary levels in a solution or environment. For hydrogen peroxide, this information aids in understanding how long it will last before it decomposes significantly.