Verified Answer :
The rate of formation of a product depends on the the concentrations of the reactants in a variable way.
If two products, call them A and B react together to form product C, a general equation for the formation of C has the form:
rate = k[A]^m [B]^n
Where the symbol [ ] is the concentration of each compound.
Then, plus the concentrations of compounds A and B you need k, m and n.
Normally you run controled trials in lab which permit to calculate k, m and n .
Here the data obtained in the lab are:
Trial [A] [B] Rate
(M)
(M) (M/s)
1 0.50
0.010 3.0×10−3
2 0.50 0.020 6.0×10−3
3 1.00
0 .010 1.2×10−2
Given that for trials 1 and 2 [A] is the same you can use those values to find n, in this way
rate 1 = 3.0 10^ -3 = k [A1]^m [B1]^n
rate 2 = 6.010^-3 = k [A2]^m [B2]^n
divide rate / rate 1 => 2 = [B1]^n / [B2]^n
[B1] = 0.010 and [B2] = 0.020 =>
6.0 / 3.0 =( 0.020 / 0.010)^n =>
2 = 2^n => n = 1
Given that for data 1 and 3 [B] is the same, you use those data to find m
rate 3 / rate 1 = 12 / 3.0 = (1.0)^m / (0.5)^m =>
4 = 2^m => m = 2
Now use any of the data to find k
With the first trial: rate = 310^-3 m/s = k (0.5)^2 (0.1) =>
k = 3.010^-3 m/s / 0.025 m^3 = 0.12 m^-2 s^-1
Now that you have k, m and n you can use the formula of the rate with the concentrations given
rate = k[A]^2[B] = 0.12 m^-2 s^-1 (0.50m)^2 (0.075m) = 0.0045 m/s = 4.510^=3 m/s
Answer: 4.5 10^-3 m/s
The initial rate for the formation of 'c' at 25\(\rm ^\circ C\) has been calculated to be 2.25 \(\rm \times\;10^-^2\) M/s.
The rate of formation of the product has been dependent on the concentration of reactants.
The given reaction has been:
A + 2B \(\rightarrow\) C
The rate of reaction can be given as:
Rate = k \(\rm [A]^a\;[B]^b\)
By the experimental data:
[A] [B] Rate
0.5 M 0.01 M 3 \(\rm \times\;10^-^3\) M/s
0.5 M 0.02 M 6 \(\rm \times\;10^-^3\) M/s
1 M 0.0.1 M 1.2 \(\rm \times\;10^-^2\) M/s
For the value of [A] = 0.5 M,
The possible rate can be: 3 \(\rm \times\;10^-^3\) M/s or 6 \(\rm \times\;10^-^3\) M/s
The ratio of two possible rates has been directly proportional to the concentration of b.
\(\rm \dfrac{Rate\;2}{Rate\;1}\;=\;\dfrac{[B2]^b}{[B1]^b}\)
\(\rm \dfrac{6.0\;\times\;10^-^3}{3.0\;\times\;10^-^3}\;=\;(\dfrac{0.02}{0.01})^b\)
2 = \(\rm 2^b\)
b = 1
Since rate 3 and rate 1 have the same concentration of [B], the possible value for a can be calculated with their ratio.
\(\rm \dfrac{Rate\;3}{Rate\;1}\;=\;\dfrac{[A3]^a}{[A1]^a}\)
\(\rm \dfrac{1.2\;\times\;10^-^2}{3.0\;\times\;10^-^3}\;=\;(\dfrac{1}{0.5})^a\)
4 = \(\rm 2^a\)
a = 2
The value of rate constant k can be calculated by substituting the values for the rate 1.
Rate = k \(\rm [A]^a\;[B]^b\)
3 \(\rm \times\;10^-^3\) M/s = k \(\rm [0.5]^2\;[0.01]^1\) M
3 \(\rm \times\;10^-^3\) = k (0.25) (0.01)
3 \(\rm \times\;10^-^3\) = 0.0025 k
k = 1.2 \(\rm M^-^2\;s^-^1\)
To calculate the initial rate for the formation of c at [A] = 0.5 M, and [B] = 0.075 M.
Rate = 1.2 \(\rm [0.5]^2\;[0.075]^1\)
Rate = 0.0225 M/s
Rate = 2.25 \(\rm \times\;10^-^2\) M/s
The initial rate for the formation of 'c' at 25\(\rm ^\circ C\) has been calculated to be 2.25 \(\rm \times\;10^-^2\) M/s.
For more information about the rate of reaction, refer to the link:
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