Math / Data & Statistics

QuestionProblem 3: Using the tabular data, determine the order of the reaction and the reaction rate. Graph the data to support your answer. The reaction is peroxide to water and oxygen as shown in stoichiometric equilibrium as below. $2 \mathrm{H}_{2} \mathrm{O}_{2} \Leftrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}$
The initial concentration of the hydrogen peroxide is 1.0 M and decreases according to the collected data shown in the table: \begin{tabular}{|c|c|} \hline Time $(\mathbf{s})$ & { $\left[\mathrm{H}_{2} \mathrm{O}_{2}\right] \mathbf{( M )}$ } \\ \hline 0 & 1.0 \\ \hline 10 & 0.8 \\ \hline 20 & 0.6 \\ \hline 30 & 0.4 \\ \hline 40 & 0.2 \\ \hline \end{tabular}
This dataset shows how the concentration of hydrogen peroxide decreases linearly over time in a zero-order reaction. If you have any questions or need further details, feel free to ask!

Studdy Solution

STEP 1

What is this asking? We need to find out how fast hydrogen peroxide breaks down into water and oxygen, and what kind of reaction it is, using the given data. Watch out! It's easy to assume reactions are more complicated than they are!
Look closely at the data before jumping to conclusions.

STEP 2

1. Analyze the Data
2. Determine the Order
3. Calculate the Rate Constant
4. Express the Rate Law

STEP 3

Let's look at how the concentration of hydrogen peroxide, $\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]$ , changes over time.
We're given a handy table!
Notice how the concentration decreases by the same amount, $\text{0.2 M}$ , every 10 seconds.
That's a big clue!

STEP 4

Since the concentration changes linearly with time, this tells us it's a zero-order reaction.
In other words, the rate of the reaction doesn't depend on how much hydrogen peroxide is present.
Wild, right?!

STEP 5

For a zero-order reaction, the rate law is simply: $\text{Rate} = k$ where $k$ is the rate constant.
We can find $k$ using the formula: $\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]_t = \left[\mathrm{H}_{2} \mathrm{O}_{2}\right]_0 - kt$ where $\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]_t$ is the concentration at time $t$ , and $\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]_0$ is the initial concentration.

STEP 6

Let's pick two points from our table.
How about the first two rows?
At $t = 0$ seconds, $\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]_0 = \text{1.0 M}$ .
At $t = 10$ seconds, $\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]_{10} = \text{0.8 M}$ . Plugging these values into our equation: $\text{0.8 M} = \text{1.0 M} - k \cdot \text{10 s}$

STEP 7

Now, let's solve for $k$ . Subtract $\text{1.0 M}$ from both sides: $\text{-0.2 M} = -k \cdot \text{10 s}$ Divide both sides by $\text{-10 s}$ : $k = \frac{\text{-0.2 M}}{\text{-10 s}} = \text{0.02 M/s}$ So, our rate constant, $k$ , is $\text{0.02 M/s}$ !

STEP 8

Now we can write our rate law: $\text{Rate} = \text{0.02 M/s}$ See? Nice and clean!

STEP 9

The reaction is zero-order, and the rate constant is $\text{0.02 M/s}$ .
The rate law is simply $\text{Rate} = \text{0.02 M/s}$ .

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Studdy Solution

STEP 1

What is this asking? We need to find out how fast hydrogen peroxide breaks down into water and oxygen, and what kind of reaction it is, using the given data. Watch out! It's easy to assume reactions are more complicated than they are!
Look closely at the data before jumping to conclusions.

STEP 2

1. Analyze the Data
2. Determine the Order
3. Calculate the Rate Constant
4. Express the Rate Law

STEP 3

Let's look at how the concentration of hydrogen peroxide, $\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]$ , changes over time.
We're given a handy table!
Notice how the concentration decreases by the same amount, $\text{0.2 M}$ , every 10 seconds.
That's a big clue!

STEP 4

Since the concentration changes linearly with time, this tells us it's a zero-order reaction.
In other words, the rate of the reaction doesn't depend on how much hydrogen peroxide is present.
Wild, right?!

STEP 5

STEP 6

STEP 7

Now, let's solve for $k$ . Subtract $\text{1.0 M}$ from both sides: $\text{-0.2 M} = -k \cdot \text{10 s}$ Divide both sides by $\text{-10 s}$ : $k = \frac{\text{-0.2 M}}{\text{-10 s}} = \text{0.02 M/s}$ So, our rate constant, $k$ , is $\text{0.02 M/s}$ !

STEP 8

Now we can write our rate law: $\text{Rate} = \text{0.02 M/s}$ See? Nice and clean!

STEP 9

The reaction is zero-order, and the rate constant is $\text{0.02 M/s}$ .
The rate law is simply $\text{Rate} = \text{0.02 M/s}$ .

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Studdy Solution

STEP 1

What is this asking? We need to find out how fast hydrogen peroxide breaks down into water and oxygen, and what kind of reaction it is, using the given data. Watch out! It's easy to assume reactions are more complicated than they are!Look closely at the data before jumping to conclusions.

STEP 2

1. Analyze the Data2. Determine the Order3. Calculate the Rate Constant4. Express the Rate Law

STEP 3

Let's **look** at how the concentration of hydrogen peroxide, [H2O2]\left[\mathrm{H}_{2} \mathrm{O}_{2}\right][H2​O2​], changes over time.We're given a handy table!Notice how the concentration decreases by the **same amount**, 0.2 M\text{0.2 M}0.2 M, every **10 seconds**.That's a big clue!

STEP 4

Since the concentration changes linearly with time, this tells us it's a **zero-order** reaction.In other words, the rate of the reaction *doesn't* depend on how much hydrogen peroxide is present.Wild, right?!

STEP 5

STEP 6

STEP 7

STEP 8

Now we can **write** our rate law: Rate=0.02 M/s \text{Rate} = \text{0.02 M/s} Rate=0.02 M/s See? Nice and clean!

STEP 9

The reaction is **zero-order**, and the rate constant is 0.02 M/s\text{0.02 M/s}0.02 M/s.The rate law is simply Rate=0.02 M/s\text{Rate} = \text{0.02 M/s}Rate=0.02 M/s.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? We need to find out how fast hydrogen peroxide breaks down into water and oxygen, and what kind of reaction it is, using the given data. Watch out! It's easy to assume reactions are more complicated than they are!
Look closely at the data before jumping to conclusions.

1. Analyze the Data
2. Determine the Order
3. Calculate the Rate Constant
4. Express the Rate Law

Let's look at how the concentration of hydrogen peroxide, $\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]$ , changes over time.
We're given a handy table!
Notice how the concentration decreases by the same amount, $\text{0.2 M}$ , every 10 seconds.
That's a big clue!

Since the concentration changes linearly with time, this tells us it's a zero-order reaction.
In other words, the rate of the reaction doesn't depend on how much hydrogen peroxide is present.
Wild, right?!

Now we can write our rate law: $\text{Rate} = \text{0.02 M/s}$ See? Nice and clean!

The reaction is zero-order, and the rate constant is $\text{0.02 M/s}$ .
The rate law is simply $\text{Rate} = \text{0.02 M/s}$ .