ng.cengage.com MindTap - Cengage Learning Freelance Brochure Design Services Onl
ID: 1083633 • Letter: N
Question
ng.cengage.com MindTap - Cengage Learning Freelance Brochure Design Services Online | Fiverr MINDTAP Braulio Cunha -L 14 Challenge Due Tomorrow at 8 AM CST Late Submission until Feb 14 at 8 AM CST is allowed, with a one time 10% penalty to subrmitted score. Question 1 Use the References to access important values if needed for this question. NO2 reacts with CO in the gas phase according to the following chemical equation: NO2g) (g) +NO(g) Question 4 It is observed that, when the concentration of CO is reduced to 1/4 of its initial value, the rate of the reaction is does not change. When the concentration of NO2 is multiplied by 6.63, the ratc of the reaction incrcases by a factor of 44.0. Question 5 (a) Write the rate expression for this reaction, and give the units of the rate constant k, assuming concentration is expressed as mol L- and time is in seconds. rate = | k[NO2][CO] kunts Ms-1 Question 7 A-Z (b) If [NO21 were multiplied by 4.89 and [CO] by 3.24, what change in the rate would be observed? The rate would increase by a factor of 15.84 Question 10 Submit Answer 1 question attempt remaining Question 11 Current score: 1012 pts (83.33 %) Autosaved at 2:23 PMExplanation / Answer
Ans :
a) since the rate of reaction does not depend on the concentration of CO , the order of reaction with respect to CO will be zero.
And now the rate of reaction becomes 44 times , when the conecntration of NO is increased 6.63 times, so the order of reactio with respect to NO is 2.
The rate law will be given as :
Rate = k [NO2]2[CO]0
b) When concentration of NO2 is multiplied by 4.89 , and CO by 3.24 , the rate of reaction will be :
Rate = k[4.89 NO2]2 [3.24CO]0
the rate will increase by a factor of 23.9
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.