7.11

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805373590
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Joined: Wed Sep 11, 2019 12:17 am

7.11

Postby 805373590 » Tue Mar 10, 2020 1:13 pm

The rate law of the reaction 2 NO(g) 1 2 H2(g) S N2(g) 1 2 H2O(g) is Rate 5 kr[NO]2[H2], and the mechanism that has been proposed is
21 21 21
, the rate of the reaction is 1.21 mmol?L s . The
Potential energy
Step 1 NO1NO¡N2O2
Step 2 N2O2 1H2 ¡N2O1H2O A Step 3 N2O1H2 ¡N2 1H2O
(a) Whichstepinthemechanismislikelytoberatedetermining? Explain your answer. (b) Sketch a reaction profile for the overall
Can someone explain this question please

Katherine Wu 1H
Posts: 104
Joined: Fri Aug 30, 2019 12:15 am
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Re: 7.11

Postby Katherine Wu 1H » Tue Mar 10, 2020 10:29 pm

a. The objective is to reproduce the observed rate law. If step 2 is the slow step, if step 1 is a rapid equilibrium, and if step 3 is fast also, then our proposed rate law will be rate=k2[N2O2][H2]. Consider the equilibrium of step 1:
k1[NO]^2=k prime[N2O2]
[N2O2]=k1/k prime [NO]^2
Substituting in our proposed rate law, we have rate= k2(k1/k prime)[NO]^2[H2]=k[NO]^2[H2] where k=k2(k1/k prime)
The assumptions made above reproduce the observed rate law; therefore, step 2 is the slow step.
b. On a graph, the progress of the rxn would peak three times.
Note: The dips that represent the formation of the intermediate N2O2 and N2O will not be at the same energy, but we have no info to determine which should be lower.


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