**Understanding Reaction Rates: Example Problem****Problem Statement:**A rate is equal to 0.0200 M/s. If [A] = 0.100 M and the rate equation is given by rate = k[A]^2, what is the new rate if the concentration of [A] is increased to 0.200 M?**Possible Answers:**- ☐ 0.0200 M/s- ☐ 0.100 M/s- ☐ 0.0400 M/s- ☐ 0.0600 M/s- ☐ 0.0800 M/s**Explanation:**To solve this problem, you need to understand how the rate of a reaction changes with the concentration of a reactant. The given rate law is rate = k[A]^2, indicating that the rate of reaction is directly proportional to the square of the concentration of [A].Given:- Initial rate = 0.0200 M/s- Initial concentration, [A] = 0.100 M- Rate = k[A]^2Firstly, determine the rate constant (k) using the provided data:\[ 0.0200 \text{ M/s} = k \times (0.100 \text{ M})^2 \]\[0.0200 \text{ M/s} = k \times 0.0100 \text{ M}^2\]\[ k = \frac{0.0200 \text{ M/s}}{0.0100 \text{ M}^2} \]\[ k = 2 \text{ M}^{-1}\text{s}^{-1} \]Now, use this rate constant to find the new rate when [A] = 0.200 M:\[ \text{New rate} = k \times (0.200 \text{ M})^2 \]\[ \text{New rate} = 2 \text{ M}^{-1}\text{s}^{-1} \times (0.200 \text{ M})^2 \]\[ \text{New rate} = 2 \text{ M}^{-1}\text{s}^{-1} \times 0.0400 \text{ M}^2 \] \[ \text{New rate} = 0.080

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Answered: A rate is equal to 0.0200 M/s. If [A] =…

A rate is equal to 0.0200 M/s. If [A] = 0.100M and rate = k[A]<, what is the new rate if the concentration of [A] is increased to 0.200 M?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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