Answer:
e.It would decrease by a factor of 4.
Explanation:
first order in [NO2]
first order in [F2]
The rate is then given as;
Rate = k [NO2][F2]
where k = rate constant = And is constant for a reaction.
Let's insert some dummy values (Any values work, just be consistent);
[NO2] = 2
[F2] = 2
K = 3
Rate = 3*2*2
Rate = 12
What would happen to the reaction rate if the concentrations of both reactants were halved with everything else held constant?
[NO2] = 2 / 2 = 1
[F2] = 2 / 2 = 1
K = 3
Rate = 3*1*1
Rate = 3
Comparing both rates (12 and 3); the correct option is;
e.It would decrease by a factor of 4.
