If tripling the initial concentration of a reactant triples the initial rate of
ID: 1018646 • Letter: I
Question
If tripling the initial concentration of a reactant triples the initial rate of reaction, what is the order of the reaction with respect to the reactant? If the rate increases by a factor of nine, what is the order? If the rate remains the same, what is the order? The decomposition of N_2O_5 at 45 degree C is 2 N_2O_5 rightarrow 4 NO_2 + O_2 and is found experimentally to follow the rate law expression rate = k[N_2O_5] The value of the rate constant is found to be 6.2 times 10^-4 s^-1. What is the overall order of the reaction? What would be the rate of the reaction when [N_2O_5] = 1.20 M? What would be the rate of the reaction when [N_2O_5] = 0.55 M? What concentration of N_2O_5 would be necessary if we wish to have the reaction rate be 1.3 times 10^-4 Ms^-1? Express the rate of the reaction in terms of the change in concentration of each of the reactants and products. 2D rightarrow 3E + 5F When E is increasing at 0.25 Ms^-1, how fast is F increasing? When D is decreasing at 0.1 Ms^-1, how fast is E increasing?
Explanation / Answer
(2) Let the rate be , r = k[A]n
Where [A] = concentration of A
n = order of the reaction
k = rate constant
On tripling the concentration of A the rate of the reaction becomes triples then
the new rate will be r' = k[3A]n = 3r
This is possible iff n=1
So the order of the reaction will be 1 ----(a)
On tripling the concentration of A by nine times the rate of the reaction becomes nine times
So r' = k[3A]n = 9r
So n = 2
Therefore the order of the reaction will be 2 ----(b)
On tripling the concentration of A by nine times the rate of the reaction becomes the same
So r' = k[3A]n = r
So n = 0
Therefore the order of the reaction will be zero ----(c)
(3) Given rate , r = k[N2O5]
So the order of the reaction = sum of the powers of the concentrations of the reactants in the rate law
= 1
Therefore the order of the reaction is 1 -----(a)
Given rate constant , k = 6.2x10-4 s-1
[[N2O5] = 1.20 M
Plug the values in above rate law we get
r = 6.2x10-4 s-1x 1.20 M
= 7.44x10-4 Ms-1 ----(b)
Given rate constant , k = 6.2x10-4 s-1
[[N2O5] = 0.55 M
Plug the values in above rate law we get
r = 6.2x10-4 s-1x 0.55 M
= 3.41x10-4 Ms-1 ----(c)
[N2O5] = r / k
= (1.3x10-4 Ms-1) / (6.2x10-4 s-1)
= 0.21 M -----(d)
(4) Given reaction is 2D ----> 3E + 5F
Rate of the reaction = rate of disaapearance of reactants = rate of appearance of products
r = -(1/2) x d[D] / dt = +(1/3) x d[E]/dt = +(1/5) x d[F] / dt
Given d[E]/dt = 0.25 Ms-1
So From the above relation d[D] / dt = (2/3)xd[E] / dt
= (2/3) x 0.25
= 0.17 Ms-1 ---(a)
Given d[D]/dt = 0.1 Ms-1
So From the above relation d[E] / dt = (3/22/3)xd[D] / dt
= (3/2) x 0.1
= 0.15 Ms-1 ---(b)