Respuesta :
Answer:
e) 1, 2, 1
Explanation:
The given kinetic data is:
A + B + C → Products
[A]₀ [B]₀ [C]₀ Initial Rate (mol/L.s)
1. 0.4 0.4 0.2 0.0160
2. 0.2 0.4 0.4 0.0080
3. 0.6 0.1 0.2 0.0015
4. 0.2 0.1 0.2 0.0005
5. 0.2 0.2 0.4 0.0020
The rate of the above reaction is :
[tex]Rate = k[A]^{x}[B]^{y}[C]^{z}[/tex]
where x, y and z is the order with respect to A, B and C respectively.
k = rate constant
[A], [B], [C] are the concentrations
In the method of initial rates involves running a given reaction multiple times and measuring the rate in each case. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.
Order w.r.t A : Use trials 3 and 4
[tex]\frac{Rate3}{Rate4}= [\frac{[A(3)]}{[A(4)]}]^{x}[/tex]
[tex]\frac{0.0015}{0.0005}= [\frac{[0.6]}{[0.2]}]^{x}[/tex]
[tex]3 = 3^{x} \\\\x =1[/tex]
Order w.r.t B : Use trials 2 and 5
[tex]\frac{Rate2}{Rate5}= [\frac{[B(2)]}{[B(5)]}]^{y}[/tex]
[tex]\frac{0.0080}{0.0020}= [\frac{[0.4]}{[0.2]}]^{y}[/tex]
[tex]4 = 2^{y} \\\\y =2[/tex]
Order w.r.t C : Use trials 1 and 2
Since the order wrt A and B are known, any of the two trails where the two trials can be used to deduce order wrt C
[tex]\frac{Rate1}{Rate2}= [\frac{[A(1)]}{[A(2)]}]^{x}[\frac{[B(1)]}{[B(2)]}]^{y}[\frac{[C(1)]}{[C(2)]}]^{z}[/tex]
we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:
[tex]\frac{0.0160}{0.0080}= [\frac{[0.4]}{[0.2]}]^{1}[\frac{[0.4]}{[0.4]}]^{2}[\frac{[0.2]}{[0.4]}]^{z}[/tex]
[tex]1 = (0.5)^{z}[/tex]
z = 1
Therefore:
Order wrt A = 1
Order wrt B = 2
Order wrt C = 1
