4A + 2B + C --> D + 2E a) Write the rate expression for this reaction. What is t

Question

4A + 2B + C --> D + 2E

a) Write the rate expression for this reaction. What is the rate appearance of D if the rate of dissappearance of A is 4.80x10^-6 M/sec?

b) using collision theory, explain why this reaction does not occur in one step.

c) derive the rate law for this reaction

[A] .300 .600 .300 .300
[B] .0303 .0303 .0909 .0303
[C] .100 .100 .100 .200
Rate (M/s) 4.8E-6 1.92E-5 1.44E-5 9.60E-6

d) what is the overall order of this reaction?

e) what is the constant k for this reaction?

f) what is the rate of the reaction if the initial concentrations of [A] = .400M, [B]= .0500M and [C]= .075M?

Explanation / Answer

Let change in concentration of A per unit time be A

Let change in concentration of B per unit time be B

Let change in concentration of C per unit time be C

Let change in concentration of D per unit time be D

Let change in concentration of E per unit time be E

then,

a) Rate of reaction = (-1/4)A = (-1/2)B = -C = +D = (+1/2)E = 4.8x10^-6 M/sec

b) Collision theory states that for a reaction in which reactants react to give products, it is essential for the reactants to colloide in correct orientation with required energy such that products can be formed. In this reaction we have three chemical species namely A, B and C which react to give two products D and E. In nature tri-molecular reactions are very unlikely as the energy barrier predicted by collision theory is very high. Hence it is nit possible for all three reactants to react and give the product in one step.

c) Let the rate law be

rate, r = k[A]a[B]b[C]c ,

where a,b and c are orders w.r.t A,B and C respectively

in trials 1 and 2, conc of B and C is constant

take ratios of rates of trial 1 and 2

so,

(1.92e-5/4.8e-6) = (0.6/0.3)a

solving, a = 2

in trials 1 and 3, conc of A and C is constant

take ratios of rates of trial 1 and 3

so,

(1.44e-5/4.8e-6) = (0.0909/0.0303)b

solving, b = 1

in trials 1 and 4, conc of A and B is constant

take ratios of rates of trial 1 and 4

so,

(9.6e-6/4.8e-6) = (0.2/0.1)c

solving, c = 1

so the rate law is, r = k[A]2[B]1[C]1

d) overall order = a+b+c = 2+1+1 = 4

e) for finding k, use values of any trial

take trial 1

rate = 4.8e-6 M/sec

[A] = 0.3 M

[B] = 0.0303 M

[C] = 0.1 M

put in rate law,

4.8e-6 = r = k[0.3]2[0.0303]1[0.1]1

solving

k = 0.0176 sec/M3

f) again put values in rate law,

rate = 0.0176(0.4*0.4*0.05*0.075) = 1.056x10^-5 M/sec

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