Respuesta :
Answer : The initial rate for a reaction will be [tex]3.4\times 10^{-3}Ms^{-1}[/tex]
Explanation :
Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
For the given chemical equation:
[tex]A+B+C\rightarrow D+E[/tex]
Rate law expression for the reaction:
[tex]\text{Rate}=k[A]^a[B]^b[C]^c[/tex]
where,
a = order with respect to A
b = order with respect to B
c = order with respect to C
Expression for rate law for first observation:
[tex]6.0\times 10^{-5}=k(0.20)^a(0.20)^b(0.20)^c[/tex] ....(1)
Expression for rate law for second observation:
[tex]1.8\times 10^{-4}=k(0.20)^a(0.20)^b(0.60)^c[/tex] ....(2)
Expression for rate law for third observation:
[tex]2.4\times 10^{-4}=k(0.40)^a(0.20)^b(0.20)^c[/tex] ....(3)
Expression for rate law for fourth observation:
[tex]2.4\times 10^{-4}=k(0.40)^a(0.40)^b(0.20)^c[/tex] ....(4)
Dividing 1 from 2, we get:
[tex]\frac{1.8\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1[/tex]
Dividing 1 from 3, we get:
[tex]\frac{2.4\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2[/tex]
Dividing 3 from 4, we get:
[tex]\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0[/tex]
Thus, the rate law becomes:
[tex]\text{Rate}=k[A]^2[B]^0[C]^1[/tex]
Now, calculating the value of 'k' by using any expression.
Putting values in equation 1, we get:
[tex]6.0\times 10^{-5}=k(0.20)^2(0.20)^0(0.20)^1[/tex]
[tex]k=7.5\times 10^{-3}M^{-2}s^{-1}[/tex]
Now we have to calculate the initial rate for a reaction that starts with 0.75 M of reagent A and 0.90 M of reagents B and C.
[tex]\text{Rate}=k[A]^2[B]^0[C]^1[/tex]
[tex]\text{Rate}=(7.5\times 10^{-3})\times (0.75)^2(0.90)^0(0.90)^1[/tex]
[tex]\text{Rate}=3.4\times 10^{-3}Ms^{-1}[/tex]
Therefore, the initial rate for a reaction will be [tex]3.4\times 10^{-3}Ms^{-1}[/tex]
The rate of reaction can be given as 0.00379 M/s.
The given reaction has been:
A + B +C D + E
The rate of reaction can be given by:
Rate = [tex]\rm k\;[A]^a\;[B]^b\;[C]^c[/tex]
The value of the coefficients can be calculated from the given trials, When the concentration for two reactants is the same, the coefficient for the third reactant can be calculated from the ratio of the two reactions.
The reaction rate and concentrations have been provided from the image attached.
- For calculating a:
Ratio of trial 1 to trial 3
[tex]\rm \dfrac{Rate\;3}{Rate\;1}[/tex] = [tex]\rm \dfrac{[A]^a\;[B]^b\;[C]^c}{[A]^a\;[B]^b\;[C]^c}[/tex]
[tex]\rm \dfrac{2.4\;\times\;10^-^4}{6.0\;\times\;10^-^5}[/tex] = [tex]\rm \dfrac{[0.4]^a\;[0.2]^b\;[0.2]^c}{[0.2]^a\;[0.2]^b\;[0.2]^c}[/tex]
4 = [tex]\rm (2)^a[/tex]
[tex]\rm (2)^2\;=\;(2)^a[/tex]
2 = a
- For calculating b:
By changing the concentrations of B in 3rd and 4th trial, the rate has not been changed. Thus the rate has been independent of the concentration of B.
- For calculating c:
Ratio of trials 2 and 1.
[tex]\rm \dfrac{Rate\;2}{Rate\;1}[/tex] = [tex]\rm \dfrac{[A]^a\;[B]^b\;[C]^c}{[A]^a\;[B]^b\;[C]^c}[/tex]
[tex]\rm \dfrac{1.8\;\times\;10^-^4}{6.0\;\times\;10^-^5}[/tex] = [tex]\rm \dfrac{[0.4]^a\;[0.4]^b\;[0.2]^c}{[0.2]^a\;[0.2]^b\;[0.2]^c}[/tex]
3 = [tex]\rm (3)^c[/tex]
1 = c
The rate of reaction can be given by:
Rate = [tex]\rm k\;[A]^a\;[C]^c[/tex]
From the trial 1:
[tex]\rm 6.0\;\times\;10^-^5[/tex] = k [tex]\rm (0.2)^2\;(0.2)^1[/tex]
k = 7.5 [tex]\rm \times\;10^-^3\;M^-^2.s^-^1[/tex]
For the reaction with [A] = 0.75 M, and [C] = 0.90 M.
Rate = 7.5 [tex]\rm \times\;10^-^3\;M^-^2.s^-^1[/tex] [tex]\times\;[0.75]^2\;[0.90]^1[/tex]
Rate = 0.00379 M/s
The rate of reaction can be given as 0.00379 M/s.
For more information about the rate of reaction, refer to the link:
https://brainly.com/question/8592296
