Can you please include your work so I can gain a full understanding. Thank you!
ID: 999505 • Letter: C
Question
Can you please include your work so I can gain a full understanding. Thank you!
A study of the gas phase oxidation of nitrogen monoxide at 25 degrees C gave the following results: 2NO (g) + 2H2 (g) N2 (g) + 2H20 (g) NO] 4.5 x 10 2.2 x 102 0.1475 4.5 x 102 6.6 x 102 0.4425 H2] Rate (M/s) Experiment 1 Experiment 2 Experiment 3 1.35 x 101 6.6 x 10 3.9825 (a.) Determine the rate law for this reaction (b.) Determine the rate constant for this reaction (include units) (c.) Determine which of the following proposed mechanism(s)/is/are valid.(give proof of why or why not for each mechanism) H2 + 2NO H2O + N2O (slow) N2O + H2 N2 + H2O (fast) Mechanism 1: 2NON,O (fast equilibrium) N,02 + H2 N2O + H2O (slow) N20+ H2 N2 + H2O (fast) Mechanism 2:Explanation / Answer
Let the rate law(r) = K[NO]a [H2]b
Where Kis rate constant , a and b are orders with respect to a and b respectively
From Experiment-1
K[4.5*10-2]a [2.2*10-2]b= 0.1475 (1)
From experiment 2
K[4.5*10-2]a [6.6*10-2]b= 0.4425 (2)
From experiment 3
K[1.35*10-1]a [6.6*10-2]b= 3.9825 (3)
Eq.2/Eq.1 gives (6.6/2.2)b= (0.4425/0.1475)
Which gives b=1
Eq.3/Eq.2 gives
(13.5/4.5)a= (3.9825/0.4425) =9
Hence a= 2
The rate becomes r= K[NO]2 [H2]1
Eq.1 becomes
K[4.5*10-2]2 [2.2*10-2]1= 0.1475 (1)
K= 3311/M2.s
The rate law becomes
Rate ,r= 3311*[NO]2 [H2]1
The slowest step will be the rate limiting step and this is the first step of Mechanism-1.
Rate for this = K[ H2] [NO]2