06 Question (2 points) e See page 552 Atmospheric chemistry involves highly reac
ID: 589305 • Letter: 0
Question
06 Question (2 points) e See page 552 Atmospheric chemistry involves highly reactive, odd-electron molecules such as the hydroperoxyl radical HO2, which decomposes into H2O2 and O2 The following data was obtained at 298 K. me (us) 0.00 0.60 1.00 1.40 1.80 2.40 8.50 5.10 3.60 2.60 1.80 1.10 8th attempt Part 1 (1 point) Determine the rate law for the reaction. Do not add multiplication symbols to your answer x x -, +, log, cos. . Rate- K NO Part 2 (1 point) Q See Hint Determine the value of the rate constant at 298 K. 256 usExplanation / Answer
By Hit and trial method
Zero order law :
At = Ao - kt
1st order rate law
ln At = -kt + ln Ao
At = concentration after time t
Ao = intial concetration
t = time
k =rate constant
check "k" value for different At values if it is constant it is that order
I am checking with first order
ln 5.1 = ln 8.5 - k*0.6
k = 0.85
ln 3.6 = ln 8.5 - k*1
k = 0.859
K is constant here,So it is first order reaction
Rate = k [O2H]^1
Rate constant = 0.853 us (take average of all k values from the rate law )
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.