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.)
1 pts
Tries 0/5 |
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.
For No catalyzedK_(N0)=Ae^(-(11.9 xx10^(3))/(RT))for al cataly zedK_(al)=Ae^(-(2*1xx10^(3))/(RT))A is same for both reaction{:[T rarr-52^(@)C=273-52=221K],[=e^(((9.8 xx10^(3))/(8.314 xx21)))],[=207.19],[(Kal_(KNO))/()=207.2]:}(2)Since frequency factor is same for catalyzed and uncatalyzed reaction therefore ratio of rate constant will depend only on activation energy and Temperature. for details solution please refer Image   Calculation of ratio of rate constant of catalyzed to ... See the full answer