Respuesta :
Answer:
a) The reaction is first order, that is, order 1. Option C is correct.
b) The half life of the reaction is 23 minutes. Option B is correct
c) The initial rate of production of NO2 for this reaction is approximately = (3.7 × 10⁻⁴) M/min. Option has been cut off.
Explanation:
First of, we try to obtain the order of the reaction from the data provided.
t (minutes) [N2O5] (mol/L)
0 1.24x10-2
10 0.92x10-2
20 0.68x10-2
30 0.50x10-2
40 0.37x10-2
50 0.28x10-2
70 0.15x10-2
Using a trial and error mode, we try to obtain the order of the reaction. But let's define some terms.
C₀ = Initial concentration of the reactant
C = concentration of the reactant at any time.
k = rate constant
t = time since the reaction started
T(1/2) = half life
We Start from the first guess of zero order.
For a zero order reaction, the general equation is
C₀ - C = kt
k = (C₀ - C)/t
If the reaction is indeed a zero order reaction, the value of k we will obtain will be the same all through the set of data provided.
C₀ = 0.0124 M
At t = 10 minutes, C = 0.0092 M
k = (0.0124 - 0.0092)/10 = 0.00032 M/min
At t = 20 minutes, C = 0.0068 M
k = (0.0124 - 0.0068)/20 = 0.00028 M/min
At t = 30 minutes, C = 0.0050 M
k = (0.0124 - 0.005)/30 = 0.00024 M/min
It's evident the value of k isn't the same for the first 3 trials, hence, the reaction isn't a zero order reaction.
We try first order next, for first order reaction
In (C₀/C) = kt
k = [In (C₀/C)]/t
C₀ = 0.0124 M
At t = 10 minutes, C = 0.0092 M
k = [In (0.0124/0.0092)]/10 = 0.0298 /min
At t = 20 minutes, C = 0.0068 M
k = 0.030 /min
At t = 30 minutes, C = 0.0050 M
k = 0.0303
At t = 40 minutes
k = 0.0302 /min
At t = 50 minutes,
k = 0.0298 /min
At t = 60 minutes,
k = 0.031 /min
This shows that the reaction is indeed first order because all the answers obtained hover around the same value.
The rate constant to be taken will be the average of them all.
Average k = 0.0302 /min.
b) The half life of a first order reaction is related to the rate constant through this relation
T(1/2) = (In 2)/k
T(1/2) = (In 2)/0.0302
T(1/2) = 22.95 minutes = 23 minutes.
c) The initial rate of production of the product at the start of the reaction
Rate = kC (first order)
At the start of the reaction C = C₀ = 0.0124M and k = 0.0302 /min
Rate = 0.0302 × 0.0124 = 0.000374 M/min = (3.74 × 10⁻⁴) M/min
The initial rate of production of NO2 is 3.72 x10-4 mol/L min-1.
We need to determine the order of reaction from the data presented. The reaction can not be a zero order reaction. Hence, we test the first order reaction equation;
In (C₀/C) = kt
k = [In (C₀/C)]/t
Where;
Co = Initial concentration
C = concentration at time t
t = Time taken
Let us try two entries at t = 10 min and t = 20 min
At t = 10;
k = [ln(1.24x10-2/0.92x10-2)]/10
k = 0.0300 min-1
At t = 20 min
k = [ln(1.24x10-2/0.68x10-2)/20
k = 0.0300 min-1
Hence, the reaction is first order
The half life of a first order reaction is;
T(1/2) = (In 2)/k
Where k = rate constant
T(1/2) = (In 2)/0.0300 min-1
T(1/2) = 23 minutes
Recall that for a first order reaction;
Rate = k[A]
Where;
[A] = concentration of the specie
Initial rate of production of NO2 = 0.0300 min-1 × 1.24x10-2 = 3.72 x10-4 mol/L min-1
Learn more: https://brainly.com/question/23184115
