just last question!
One pathway for the destruction of ozone in the upper atmosphere
is
O3(g) + NO(g) -> NO2(g) + O2(g) Slow |
NO2(g) + O(g) -> NO(g) + O2(g) Fast |
Overall reaction: O3(g) + O(g) -> 2O2(g) |
Identify the correct term to describe each of the following
species.
Catalyst: nitrogen monoxide
Product: molecular (di)oxygen
Intermediate: NO2(g)
Reactant: O3(g)
Reactant: O(g)
1 pts
You are correct. | Previous Tries |
Ea is 14.0 kJ/mol, for the uncatalyzed reaction:
O3(g) + O(g) -> 2O2(g) |
Ea for the same reaction, when catalyzed as
above, is 11.9 kJ/mol.
At -44°C, what is the ratio of the rate constant for the catalyzed
reaction to that for the uncatalyzed reaction?
(Assume the frequency factor A is the same for each
reaction.)
ans:3.01
One of the concerns about the use of freons is that they will
migrate to the upper atmosphere, where chlorine atoms can be
generated by the reaction
CCl2F2 -> CF2Cl + Cl (photolysis of Freon-12) |
Chlorine atoms can also act as a catalyst for the destruction of
ozone.
The activation energy for the reaction
Cl + O3 -> ClO + O2 |
is 2.1 kJ/mol.
Cl is regenerated by the fast reaction
ClO + O -> O2 + Cl |
If the frequency factors are the same and both catalysts are in equal concentration, how many times faster, at -52°C, does the Cl catalysed reaction proceed compared to the reaction catalysed as described in the previous question.
{:[K=Ae^(-Ea//RT)],[" If "E_(a)=14kJ∣mol],[k_(1)=Ae^(-14//RT)],[1fE_(a)=11*gkJ∣mol],[k_(2)=A(-11.9)/(RT)],[:.(K_(2))/(K_(1))=(Ae^(-(11.9)/(RT)))/(Atheta^(-14//RT))],[=3.01]:}2nd partReac ... See the full answer