Question

Question #8cdfb

Answer 1

You were given the mechanism, which is (fixing the typo):

`2"N"_2"O"_5 stackrel(k_1" ")(=>) 4"NO"_2 + color(red)("O"_2)`
`2"NO"_2 + 1/2"O"_2 stackrel(k_2" ")(=>) "N"_2"O"_5`

The overall reaction therefore is:

`\mathbf("N"_2"O"_5 stackrel(k_(obs))(->) 2"NO"_2 + 1/2"O"_2)`

Notice how:

`2"N"_2"O"_5 stackrel(k_1" ")(=>) 4"NO"_2 + color(red)("O"_2)`
`2(2"NO"_2 + 1/2"O"_2 stackrel(k_2" ")(=>) "N"_2"O"_5)`
`"-----------------------------------------------"`
`"Nothing"`

The second step is half the scale of the first. Hence, the second elementary step's activation energy is half as large. Or:

`color(blue)(E_2 = 1/2E_1)`

Correct me if I'm wrong, but I don't think it's c, because that looks backwards. In that case, I would say `E_1 > E_2`.