Consider the following reaction:4 HBr(g) + O2(g) 2 H2O(g) + 2 Br2(g)(a) The rate law for this reaction is first order in HBr(g) and first order in O2(g). What is the rate law for this reaction?Rate = k [HBr(g)] [O2(g)]Rate = k [HBr(g)]2 [O2(g)] Rate = k [HBr(g)] [O2(g)]2Rate = k [HBr(g)]2 [O2(g)]2Rate = k [HBr(g)] [O2(g)]3Rate = k [HBr(g)]4 [O2(g)](b) If the rate constant for this reaction at a certain temperature is 8860, what is the reaction rate when [HBr(g)] = 0.00558 M and [O2(g)] = 0.00825 M?Rate = _____M/s.(c) What is the reaction rate when the concentration of HBr(g) is doubled, to 0.0112 M while the concentration of O2(g) is 0.00825 M?Rate =____M/s

Answered: Image /qna-images/answer/4a2f6757-f4bf-4c26-ac10-7d5ad4c8cc3f.jpg

Answered: Consider the following reaction: 4…

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...

See similar textbooks

Related questions

Question

Consider the following reaction:

4 HBr(g) + O₂(g) 2 H₂O(g) + 2 Br₂(g)

(a) The rate law for this reaction is first order in HBr(g) and first order in O₂(g). What is the rate law for this reaction?

Rate = k [HBr(g)] [O₂(g)]

Rate = k [HBr(g)]² [O₂(g)]

Rate = k [HBr(g)] [O₂(g)]²

Rate = k [HBr(g)]² [O₂(g)]²

Rate = k [HBr(g)] [O₂(g)]³

Rate = k [HBr(g)]⁴ [O₂(g)]

(b) If the rate constant for this reaction at a certain temperature is 8860, what is the reaction rate when [HBr(g)] = 0.00558 M and [O₂(g)] = 0.00825 M?

Rate = _____M/s.

(c) What is the reaction rate when the concentration of HBr(g) is doubled, to 0.0112 M while the concentration of O₂(g) is 0.00825 M?

Rate =____M/s

Expert Solution

Step by step

Solved in 4 steps with 4 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions