Answer:
1) Overall reaction is
2H₂(g) + 2NO(g) → N₂(g) + 2H₂O(g)
2) The catalyst cannot be determined from the given information about this reaction. None of the species in the elementary reactions can pass as a catalyst for the reaction.
3) The only intermediate for this reaction is N₂O(g).
4) Rate = K [H₂] [NO]²
Comparing this with
Rate = K [A]ᵐ [B]ⁿ
A = H₂
B = NO
m = 1
n = 2
Explanation:
1) The overall reaction is obtained by adding all of the elementary reactions up.
Step 1 (slow step)
H₂(g) + 2 NO(g) → N₂O(g) + H₂O(g)
Step 2 (fast step)
N₂O(g) + H₂(g) → N₂(g) + H₂O(g)
Summing up, we obtain,
H₂(g) + 2 NO(g) + N₂O(g) + H₂(g) → N₂O(g) + H₂O(g) + N₂(g) + H₂O(g)
We then eliminate the species that appear on both sides of this
2H₂(g) + 2NO(g) → N₂(g) + 2H₂O(g)
2) The catalyst cannot be determined from the given information about this reaction.
The catalyst doesn't participate in the reaction, it just affects the rate of the reaction. So, none of the species in the elementary reactions can pass as a catalyst for the reaction.
3) The reaction intermediates are the species that appear in the elementary reactions but do not appear in the overall reaction. They are formed and disappear all in the process of the reaction.
From combining the elementary reactions in (1), it is evident that the only intermediate for this reaction is N₂O(g).
4) The rate law is the one that gives the rate of the overall reaction. It is obtained from the slow step of the elementary reactions. And the intermediates that appear in it are substituted using the other steps in the elementary reactions.
For this reaction, the slow step is
H₂(g) + 2 NO(g) → N₂O(g) + H₂O(g)
Rate = K [H₂] [NO]²
Since no intermediates appear in the rate law given by the slow step, there is no need for any substitution.
The rate of the overall reaction is
Rate = K [H₂] [NO]²
Comparing this with
Rate = K [A]ᵐ [B]ⁿ
A = H₂
B = NO
m = 1
n = 2
Hope this Helps!!!
