Chemistry, 22.06.2019 09:40 loveoneonly9153

Consider this initial-rate data at a certain temperature for the reaction described by

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Answer from: wsdafvbhjkl

N2O3(g)→NO(g)+NO2(g)

[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)^m=r(t)/r0(t)
However,
([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

r(t)=k[N2O3]

0.66 M/s=k×0.1 M

k=6.6 s−1

Answer from: brookekolmetz

The value of k is $\boxed{\text{6.6 s}^{-1}}$ .

Further Explanation:

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:

$\text{N}_2\text{O}_3(\text{g})\rightarrow\text{NO}(\text{g})+\text{NO}_2(\text{g})$

When concentration of $\text{N}_2\text{O}_3$ changes from 0.1 M to 0.2 M, rate of reaction becomes doubled. This implies order of reaction with respect to $\text{N}_2\text{O}_3$ is one.

The expression for rate of given reaction is as follows:

$\text{Rate}=\text{k}[\text{N}_2\text{O}_3]$ ...... (1)

Where, k is the rate constant of reaction.

Rearrange equation (1) to calculate the value of k.

$\text{k}=\dfrac{\text{Rate}}{[\text{N}_2\text{O}_3]}$ ...... (2)

Any of the three given concentrations of $\text{N}_2\text{O}_3$ can be considered for calculation of rate constant along with the respective initial rates of reaction.

Here, we take concentration of $\text{N}_2\text{O}_3$ as 0.1 M so rate corresponding to this is 0.66 M/s.

Substitute 0.1 M for $[\text{N}_2\text{O}_3]$ and 0.66 M/s for rate in equation (2).

$\begin{aligned}{\text{k}&=\dfrac{\text{0.66 M/s}}{\text{0.1 M}}\\&=\text{6.6 s}^{-1}}\end{aligned}$

Therefore the value of rate constant (k) for the given reaction comes out to be $\text{6.6 s}^{-1}}$ .

Learn More:

1. What is the half-life of the reaction? 2. Rate of chemical reaction:

Answer Details:

Grade: Senior School

Subject: Chemistry

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.

Answer from: Teedollasign

NO2O3rate
0.650.82.6x10^4
1.10.84.4x10^4
1.761.412.32x10^4

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
k cancels
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

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