What is the rate law for the overall process?
(1) NO(g) + #O_3# (g) #-># #NO_2# (g) + #O_2# (g) (slow)
(2) O(g) + #NO_2# (g) #-># NO(g) + #O_2# (fast)
(1) NO(g) +
(2) O(g) +
1 Answer
The mechanism is flawed, because it does not allow one to write a rate law in terms of ONLY reactants.
The best answer is:
#r(t) = k_1["NO"]["O"_3]#
but it shows no explicit dependency on
#r(t) = k_2["NO"_2]["O"]#
which shows no explicit dependency on
First off, the overall reaction is:
#"NO"(g) + "O"_3(g) stackrel(k_1" ")(->) "NO"_2(g) + "O"_2(g)# (slow)
#ul("O"(g) + "NO"_2(g) stackrel(k_2" ")(->) "NO"(g) + "O"_2(g))# (fast)
#"O"(g) + "O"_3(g) stackrel(k_"obs")(->) 2"O"_2(g)#
So, the rate law ought to contain
Using the slow step, we should be able to write a preliminary rate law based on the coefficients, only because it is the rate-determining step:
#color(blue)(r(t) = k_1["NO"]["O"_3])#
Now, it can be seen that
Since the first step is slow and the second is fast, we could try to assume that
#(d["NO"_2])/(dt) ~~ 0 = k_1["NO"]["O"_3] - k_2["O"]["NO"_2]#
Solving for
#["NO"] = k_2/k_1 (["NO"_2]["O"])/(["O"_3])#
So far we would have:
#r(t) = cancel(k_1)k_2/cancel(k_1) (["NO"_2]["O"])/cancel(["O"_3])cancel(["O"_3])#
#= k_2["NO"_2]["O"]#
But we see that although the approximation is valid, we get no new information from this... if we found an expression for
The mechanism proposed is therefore not good, because it does not show the dependency of the rate law on