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Hydrogen peroxide decomposes to water and oxygen according to the reaction below: $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{~g}) $$ In the presence of large excesses of \(\mathrm{I}^{-}\)ion, the following set of data is obtained. What is the average rate of disappearance of \(\mathrm{H}_{2} \mathrm{O}_{2}\) (aq) in \(\mathrm{M} / \mathrm{s}\) in the first \(45.0\) seconds of the reaction if \(1.00\) litre of \(\mathrm{H}_{2} \mathrm{O}_{2}\) reacts at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) pressure? \begin{tabular}{ll} Time, \(s\) & \(\mathrm{O}_{2}(\mathrm{~g})\) collected, \(\mathrm{ml}\) \\ \(0.0\) & \(0.0\) \\ \(45.0\) & \(2.00\) \\ \(90.0\) & \(4.00\) \\ \(135.0\) & \(6.00\) \\ \hline \end{tabular} 26 \(2.63 \times 10^{-4} \mathrm{M} / \mathrm{s}\) a. \(.2 .63 \times 10^{-4} \mathrm{M} / \mathrm{s}\) \(6.33 \times 10^{-6} \mathrm{M} / \mathrm{s}\) b. \(6.33 \times 10^{-6} \mathrm{M} / \mathrm{s}\) \(3.63 \times 10^{-6} \mathrm{M} / \mathrm{s}\) c. d. \(1.36 \times 10^{-3} \mathrm{M} / \mathrm{s}\)

Step by step solution

01

Understand the Reaction

We are given the decomposition reaction of hydrogen peroxide: \( 2\text{H}_2\text{O}_2(\text{aq}) \rightarrow 2\text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g}) \). We need to calculate the average rate of disappearance of \( \text{H}_2\text{O}_2 \) over the first 45 seconds.

02

Find the Volume of Gas Collected

We have the volumes of \( \text{O}_2 \) collected at different times. In the first 45 seconds, the volume of \( \text{O}_2 \) collected is 2.00 mL.

03

Use the Ideal Gas Law

Convert the volume of \( \text{O}_2 \) from mL to L: 2.00 mL is 0.002 L. Using the ideal gas law \( PV = nRT \) to find moles \( n \), where \( P = 1.00 \, \text{atm} \), \( V = 0.002 \, \text{L} \), \( R = 0.0821 \, \text{L} \cdot \text{atm/mol} \cdot \text{K} \), and \( T = 298 \, \text{K} \).

04

Calculate Moles of Oxygen Formed

Solving for \( n \), we have: \( n = \frac{PV}{RT} = \frac{(1.00)(0.002)}{(0.0821)(298)} \). Calculate this to find \( n \approx 8.16 \times 10^{-5} \) moles of \( \text{O}_2 \).

05

Use Stoichiometry to Find Moles of \( \text{H}_2\text{O}_2 \)

According to the reaction, 1 mole of \( \text{O}_2 \) is formed from 2 moles of \( \text{H}_2\text{O}_2 \). Calculate the moles of \( \text{H}_2\text{O}_2 \) used: \( 8.16 \times 10^{-5} \times 2 = 1.63 \times 10^{-4} \).

06

Calculate Concentration Change

The change in concentration of \( \text{H}_2\text{O}_2 \) is the moles used divided by the total volume (1 L): \( \Delta\text{[H}_2\text{O}_2] = 1.63 \times 10^{-4} \, \text{M} \).

07

Find Rate of Reaction

The average rate of disappearance of \( \text{H}_2\text{O}_2 \) is:\[ \text{Rate} = -\frac{\Delta [\text{H}_2\text{O}_2]}{\Delta t} = -\frac{1.63 \times 10^{-4} \, \text{M}}{45 \, \text{s}} \].

08

Calculate Rate

Solving the equation gives: \( \text{Rate} = 3.63 \times 10^{-6} \, \text{M/s} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate of Reaction

When we talk about the rate of a chemical reaction, we are discussing how quickly or slowly a reaction occurs. The rate of reaction is usually defined as the change in concentration of a reactant or product per unit time. For the decomposition of hydrogen peroxide, we focus on how fast hydrogen peroxide disappears as oxygen gas is formed in the reaction.

To calculate the rate, we use the formula: \[\text{Rate} = -\frac{\Delta [\text{H}_2\text{O}_2]}{\Delta t}\] Here, - \(\Delta [\text{H}_2\text{O}_2]\) represents the change in concentration of hydrogen peroxide.- \(\Delta t\) is the time interval.

For this particular reaction, the negative sign indicates that the concentration of hydrogen peroxide decreases over time. It's useful because it reminds us that this is a disappearance rate rather than an appearance rate. The rate of reaction can differ greatly depending on factors like temperature, pressure, and concentration.

Hydrogen Peroxide Decomposition

Hydrogen peroxide (H_2O_2) commonly decomposes into water (H_2O) and oxygen (O_2). This process can be spontaneous, but it's generally quite slow. However, the presence of a catalyst, such as the I^{-} ion mentioned in the experimental setup, dramatically accelerates the decomposition.

The reaction is as follows: \[2 \text{H}_2\text{O}_2(\text{aq}) \rightarrow 2 \text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g})\]This balanced reaction shows that two molecules of hydrogen peroxide break down into two molecules of water and one molecule of oxygen gas.

In a lab setting, it's not uncommon to monitor how much O_2 gas is produced to understand the behavior of the reaction over time. Observing the rate at which H_2O_2 decomposes helps chemists in fields ranging from microbiology to environmental science, especially in situations where H_2O_2 is used as a disinfectant or clean-up agent.

Stoichiometry

Stoichiometry is the bridge between reactants and products in a chemical reaction. It allows us to use the balanced equation to predict how much product can be made from a given amount of reactant, or vice versa. In this reaction, stoichiometry tells us that every mole of O_2 produced corresponds to two moles of H_2O_2 that have decomposed.

This relationship is crucial because it allows us to calculate how much H_2O_2 disappears based on the volume of O_2 gas collected. Using the stoichiometric coefficients, we multiply the moles of O_2 by 2 to find the moles of H_2O_2 that have reacted.

Understanding stoichiometry is essential for solving any chemical problem because it helps translate between the "world" of actual chemical substances and the abstract world of numbers and equations chemists work with.

Ideal Gas Law

The ideal gas law is a pivotal equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. It is expressed as: \[ PV = nRT \]where:

• \(P\) represents pressure (in atm)
• \(V\) is the volume (in L)
• \(n\) is the number of moles
• \(R\) is the ideal gas constant (0.0821 L·atm/mol·K), a constant that makes the units compatible
• \(T\) is the temperature (in Kelvin)

For our experiment, we use the ideal gas law to convert the volume of O_2 gas collected to moles, which gives us insight into how many moles of H_2O_2 were used. Ensuring that your units are consistent is crucial when applying this approach. Typically, measurements must be in liters, atm, and Kelvin to use standard values for R effectively.

This law is instrumental in reactions like H_2O_2 decomposition where gas is a product, as it links measurable quantities to the amounts of substances reacting or being produced.