The reaction of CO(g) + NO(g) is second-order in NO2 and zero-order in CO at temperatures less than 500K.
(a) Write the rate law for the reaction.
(b) How will the reaction rate change if the NO2 concentration is halved?
(c) How will the reaction rate change is the concentration of CO is doubled?
typo.
rate = (NO)2(CO)o
(a) The rate law for the reaction can be expressed as follows:
rate = k [NO]^2 [CO]^0
Since the exponent of [CO] is zero, the concentration of CO does not affect the rate of the reaction.
(b) If the NO2 concentration is halved, the new rate law can be written as:
rate' = k [NO']^2 [CO]^0
Since the concentration of CO does not affect the rate, we can simplify the equation to:
rate' = k [NO']^2
This means that if the NO2 concentration is halved, the rate of the reaction will also be halved.
(c) If the concentration of CO is doubled, the new rate law can be written as:
rate' = k [NO]^2 [CO']^0
Since the exponent of [CO'] is zero, the concentration of CO does not affect the rate of the reaction. Therefore, the rate of the reaction will remain the same if the concentration of CO is doubled.
To answer these questions, we need to first understand how to write the rate law for the reaction. The rate law is an expression that relates the rate of the reaction to the concentrations of the reactants.
(a) Writing the rate law:
From the information given, we know that the reaction is second-order in NO2 and zero-order in CO. This means that the rate of the reaction is directly proportional to the square of the concentration of NO2 and is not affected by the concentration of CO.
Therefore, the rate law for this reaction can be written as:
Rate = k[NO2]^2
Where:
- Rate is the rate of the reaction
- k is the rate constant
- [NO2] is the concentration of NO2
(b) Effect of halving the NO2 concentration:
If the NO2 concentration is halved, it means that the new concentration is now half of the original concentration. Let's call the original concentration [NO2]0 and the new concentration [NO2]1.
Using the rate law, we can compare the rates before and after halving the NO2 concentration:
Rate before = k[NO2]0^2
Rate after = k[NO2]1^2
Since [NO2]1 = (1/2)[NO2]0, we can substitute this into the rate after equation:
Rate after = k[(1/2)[NO2]0]^2
= k(1/4)[NO2]0^2
= (1/4) * (k[NO2]0^2)
Therefore, the reaction rate will decrease by a factor of 1/4 (or 25%) if the NO2 concentration is halved.
(c) Effect of doubling the CO concentration:
According to the rate law, the reaction rate is not affected by the concentration of CO. Therefore, doubling the CO concentration will not change the reaction rate. The rate will remain the same.
In summary:
(a) The rate law for the given reaction is Rate = k[NO2]^2.
(b) If the NO2 concentration is halved, the reaction rate will decrease by a factor of 1/4 (or 25%).
(c) Doubling the CO concentration will not change the reaction rate.
a). rate = k(NO2)^2
b). (1/2)^2 = ??
c). Where is CO in the rate equation? I don't see it although I could have written the answer to a as rate = (NO2)2(CO)o>/>