Question

The data in the table below were obtained for the reaction: 3A + 2B ? C | Experiment Number | [A] (M) | [B] (M) | Initial Rate (M/s) | | :--- | :--- | :--- | :--- | | 1 | 0.052 | 2.36 | 4.25 | | 2 | 0.104 | 2.36 | 8.50 | | 3 | 0.052 | 4.72 | 8.50 | The rate law for this reaction is rate = ________. Select one: a.k[A][B] b.k[A]² c.k[A]²[B] d.k[B] e.k[A]³[B]²

          The data in the table below were obtained for the reaction:

3A + 2B ? C

| Experiment Number | [A] (M) | [B] (M) | Initial Rate (M/s) |
| :--- | :--- | :--- | :--- |
| 1 | 0.052 | 2.36 | 4.25 |
| 2 | 0.104 | 2.36 | 8.50 |
| 3 | 0.052 | 4.72 | 8.50 |

The rate law for this reaction is rate = ________.

Select one:
a.k[A][B]
b.k[A]²
c.k[A]²[B]
d.k[B]
e.k[A]³[B]²

Show more…

The data in the table below were obtained for the reaction:

3A + 2B ? C

| Experiment Number | [A] (M) | [B] (M) | Initial Rate (M/s) |
| :— | :— | :— | :— |
| 1 | 0.052 | 2.36 | 4.25 |
| 2 | 0.104 | 2.36 | 8.50 |
| 3 | 0.052 | 4.72 | 8.50 |

The rate law for this reaction is rate = .

Select one:
a.k[A][B]
b.k[A]²
c.k[A]²[B]
d.k[B]
e.k[A]³[B]²

Added by Burak Y.

Physical Chemistry for IIT JEE

Atul Singha 1st Edition

Chapter 7

Instant Answer

Solved by Expert Susan Hallstrom

06/11/2023

Step 1

Compare experiments 1 and 2: The concentration of A is doubled, while the concentration of B remains the same. The initial rate doubles as well. This suggests that the reaction is first order with respect to A. Show more…

Show all steps

lock

Thanks for your feedback!

The data in the table below were obtained for the reaction: 3A + 2B → C The rate law for this reaction is rate = ________. Select one: a. k[A][B] b. k[A]² c. k[A]²[B] d. k[B] e. k[A]³[B]²

Play audio

Feedback

Powered by NumerAI

Susan Hallstrom and 76 other subject Chemistry 101 educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Recommended Videos

Recommended Textbooks

Transcript

00:01 In this problem, we'll be discussing a rate law, and we're given the following data.

00:16 And we have our experiment number, our concentration of a, our concentration of b, and our initial rate.

00:31 And that is in units of molarity per second.

00:40 So our data are 0 .052, 0 .052, and 0 .104.

00:49 For b, we've got 2 .36, 2 .36, and 4 .72.

00:56 Our initial rates are 4 .25, 8 .50, and 8 .50.

01:04 My fives aren't looking very good here.

01:15 There we go.

01:17 So we're gonna do this.

01:19 Let's first do a.

01:21 And for a, we are going to use trials 1 and 2, because b is constant 1 and 2.

01:31 So i'll put a star here.

01:34 These are the two trials we're using for the blue a.

01:37 So we'll take our larger rate.

01:39 I always start with my larger on top.

01:44 And this will equal, for the first one, i'll put down what we're going to cross out.

01:48 And a is 0 .104 to the x power and 0 .052 to the x power.

02:03 My y is 2 .36 to the y power and 2 .36 to the y power.

02:15 So you can see that these, oops, i was going to just change colors on those.

02:19 These cross out...

Ace is your personal tutor. It breaks down any question with clear steps so you can learn.

Start Using Ace

Continue solving this problem

Create a free account to:

View full step-by-step solution
Ask follow-up questions with Ace AI
Save progress and study later