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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 \(1.00\) atm pressure?\begin{aligned} &\text { Time, } \mathrm{s} \quad \mathrm{O}_{2}(\mathrm{~g}) \text { collected, } \mathrm{ml} \\ &0.0 & 0.0 \\ &45.0 & 2.00 \\ &90.0 & 4.00 \\ &135.0 & 6.00 \\ &\text { a. } 2.63 \times 10^{-4} \mathrm{M} / \mathrm{s} \\ &\text { b. } 6.33 \times 10^{-6} \mathrm{M} / \mathrm{s} \\ &\text { c. } 3.63 \times 10^{-6} \mathrm{M} / \mathrm{s} \\ &\text { d. } 1.36 \times 10^{-3} \mathrm{M} / \mathrm{s} \end{aligned}

Step by step solution

01

Understanding the Reaction

The reaction given is \(2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{~g})\). For every 2 moles of \(\mathrm{H}_{2} \mathrm{O}_{2}\) that decompose, 1 mole of \(\mathrm{O}_2\) gas is produced.

02

Calculate Moles of \(\mathrm{O}_2\) Produced

First, convert the volume of \(\mathrm{O}_2\) gas collected at 45 seconds from ml to liters: \(2.00 \text{ ml} = 0.002 \text{ L}\). At STP, the molar volume of a gas is 22.4 L/mol. So, the moles of \(\mathrm{O}_2\) can be calculated using: \(\frac{0.002 \text{ L}}{22.4 \text{ L/mol}}\).

03

Calculate Moles of \(\mathrm{H}_2\mathrm{O}_2\) Decomposed

Using the stoichiometry of the reaction, 1 mole of \(\mathrm{O}_2\) comes from 2 moles of \(\mathrm{H}_{2} \mathrm{O}_{2}\). Thus, the moles of \(\mathrm{H}_{2} \mathrm{O}_{2}\) are twice the moles of \(\mathrm{O}_2\) calculated.

04

Calculate Concentration Change

Calculate the concentration change of \(\mathrm{H}_2\mathrm{O}_2\). The moles of \(\mathrm{H}_2\mathrm{O}_2\) decomposed divide by the 1 L volume of the solution gives the concentration change in moles per liter (M).

05

Calculate Average Rate of Disappearance

The average rate of disappearance of \(\mathrm{H}_2\mathrm{O}_2\) over the first 45 seconds is determined by dividing the concentration change by the time interval (45 seconds).

06

Select Closest Option

Compare the calculated average rate with the provided options to select the closest match.

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Key Concepts

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

Reaction Stoichiometry

Understanding the reaction stoichiometry is like solving a detailed chemical puzzle. In our original exercise, the decomposition reaction of hydrogen peroxide can be shown as:
\[ 2 \text{H}_2\text{O}_2(\text{aq}) \rightarrow 2 \text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g}) \]
This equation tells us that for every 2 moles of hydrogen peroxide, we get 1 mole of oxygen gas. Stoichiometry allows us to keep track of the number of moles consumed and produced in a reaction.
This is essential for understanding how much reactant is needed and how much product can be expected, which is key for balancing reactions.
In our problem, stoichiometry helps us link the moles of hydrogen peroxide decomposed to the moles of oxygen produced.

Rate of Reaction

The **rate of reaction** gives insight into how quickly a reaction occurs. In chemical kinetics, it’s calculated by how much concentration changes over time.
For the decomposition of hydrogen peroxide, the **average rate** of disappearance might be what you're interested in:

• First, identify the concentration change of hydrogen peroxide by looking at how many moles are decomposed.
• Next, divide this change by the time interval to find the average rate: \[ \text{Rate} = \frac{\Delta [\text{H}_2\text{O}_2(\text{aq})]}{\Delta \text{time}(s)} \]

In our example, the rate is determined over the first 45 seconds. Calculations like this are useful for understanding the efficiency and speed of chemical processes.

Gas Volume Conversion

Converting gas volume is essential when working with reactions involving gases. When **oxygen gas** is collected in laboratory conditions, its volume often needs to be converted to standard conditions for easier calculations.
Here's how you can do it:

• Convert the volume collected in milliliters to liters by dividing by 1000 (since 1 L = 1000 mL).
• Use the molar volume of gas at standard temperature and pressure (STP), which is typically 22.4 L/mol, to convert from volume to moles. \[\text{Moles of } \text{O}_2 = \frac{\text{Volume of } \text{O}_2 (L)}{22.4 \text{ L/mol}}\]

Such conversions allow us to standardize the volumes for comparison and estimation in reactions.

Molar Volume of Gas

The molar volume of gas is a fundamental concept that tells us the space one mole of gas occupies at standard conditions. This value, often taken as **22.4 liters per mole** at 0°C and 1 atm, simplifies calculations involving gases:
In the lab setting or certain problems, this lets you directly convert gas volumes to moles, crucial when calculating reactant or product amounts.
For our specific context, this meant when we collected oxygen gas, we needed its volume in liters. Applying this concept enables time-saving conversions between volume and moles without much hassle.