Nitric oxide (NO) reacts with H2 to form N2O and H2O. It is known that there are two elementarysteps to this reaction:Step I: NO(g) + NO(g) ---> N2O2(g)Step II: N2O2(g) + H2(g) ---> N2O(g) + H2O(g)The measured rate law for this overall reaction is:Rate = k[NO]2 [H2]What can we conclude about the above elementary reaction steps based on this rate law?(A) Steps I & II are about the same rate.(B) Step I is much faster than step II.(C) Step II is much faster than step I.(D) Not enough information to tell. How do you rule out answer B? I think that if step II is the slow step, then the rate law would be Rate = k[N2O2] [H2]. Since N2O2 is an intermediate made from [NO]2, then couldn't the rate law still be Rate = k[NO]2 [H2] when step II is the slow step?
Nitric oxide (NO) reacts with H2 to form N2O and H2O. It is known that there are two elementary
steps to this reaction:
Step I: NO(g) + NO(g) ---> N2O2(g)
Step II: N2O2(g) + H2(g) ---> N2O(g) + H2O(g)
The measured rate law for this overall reaction is:
Rate = k[NO]2 [H2]
What can we conclude about the above elementary reaction steps based on this rate law?
(A) Steps I & II are about the same rate.
(B) Step I is much faster than step II.
(C) Step II is much faster than step I.
(D) Not enough information to tell.
How do you rule out answer B?
I think that if step II is the slow step, then the rate law would be Rate = k[N2O2] [H2]. Since N2O2 is an intermediate made from [NO]2, then couldn't the rate law still be Rate = k[NO]2 [H2] when step II is the slow step?
Step by step
Solved in 2 steps with 1 images
