[N2O3] Initial Rate
0.1 M r(t)=0.66 M/s
0.2 M r(t)=1.32 M/s
0.3 M r(t)=1.98 M/s
We can have the relationship:
([N2O3]/[N2O3]0) = 2
Also, we assume m=1 which is the order of the reaction.
Thus, the relationship is simplified to,
r(t)/r0(t) = 2
0.66 M/s=k×0.1 M
The value of k is .
Order of reaction:
It is sum of exponents of concentration of each of the reactants that are present in chemical reaction. In other words, order indicates power dependence of order of reaction on reactant concentration.
Given reaction is as follows:
When concentration of changes from 0.1 M to 0.2 M, rate of reaction becomes doubled. This implies order of reaction with respect to is one.
The expression for rate of given reaction is as follows:
Where, k is the rate constant of reaction.
Rearrange equation (1) to calculate the value of k.
Any of the three given concentrations of can be considered for calculation of rate constant along with the respective initial rates of reaction.
Here, we take concentration of as 0.1 M so rate corresponding to this is 0.66 M/s.
Substitute 0.1 M for and 0.66 M/s for rate in equation (2).
Therefore the value of rate constant (k) for the given reaction comes out to be .
1. What is the half-life of the reaction? 2. Rate of chemical reaction:
Grade: Senior School
Chapter: Chemical Kinetics
Keywords: k, rate, order, N2O3, NO, NO2, 0.1 M, 0.66, 0.2 M, 1.32, 0.3 M, 1.98, rate constant, sum, exponents.
rate2/rate1 = k[NO2 2]^a[O3 2]^b / k[NO2 1]^a[O3 1]^b
K and O3 cancel
4.4 / 2.6 = (1.1 / 0.65)^a
1.7 = 1.7^aa = 1
rate 3 / rate 2 = k[NO2 3][O3 3]^b / k[NO2 2][O3 2]^b
since you know the order of the rxn with respect to NO2, plug in the values and solve the equation. the [O3]^b is a separate value so reduce all numbers to a final value and then solve for b
12.32 / 4.4 = (1.76)(1.4)^b / (1.1)(0.8)^b1.76/1.1 = 1.61.4 / 0.8 = 1.75
2.8 = 1.6 x (1.75)^b
1.75 = (1.75)^bb = 1
rate = k[NO2][O3]
2.6x10^4 / (0.65)(0.8) = k
k = 5x10^4