For the following reaction, a+b+c rightarrow d+e the initial reaction rate was m

Question

For the following reaction, a+b+c rightarrow d+e the initial reaction rate was measured for various initial concentrations of reactants. The following data were collected: What is the reaction order with respect to A? What is the reaction order with respect to B? What is the reaction order with respect toC? Calculate the value of the rate constant for the reaction. Be certain to include the correct units with your rate constant! After 50.0 minutes. 19.0% of a compound has decomposed. What b the half-life of this reac tion assuming first order kinetics?

Explanation / Answer

4. let the rate equation be,

rate = k[A]^x.[B]^y.[C]^z

with,

k being rate constant

x, y, and z be the order with respect to [A], [B] and [C], respectrively

a. Trial 1 and Trial 3

[B] and [C] are constant

rate1/rate3 = 3 x 10^-5/1.20 x 10^-4 = (0.1/0.2)^x

taking log of both sides and solving for x,

x = 2

Order with respect to [A] is 2

b. Trial 3 and trial 4

rate3/rate4 = 1.2 x 10^-4/1.2 x 10^-4 = (0.1/0.2)^y

taking log of both sides and solving for y,

y = 0

So, order is 0 with respect to [B]

d. Trial 1 and trial 2

[A] and [B] are constant

rate1/rate2 = 3 x 10^-5/9 x 10^-5 = (0.1/0.3)^z

taking log of both sides and solving for z

z = 1

So order is 1 with respect to [C]

e. Rate constant k = rate/[A]^2.[C] = 3 x 10^-5/(0.1)^2.(0.1) = 0.03 M^-2.s^-1

5. Agfter t = 50 min, 19% decomposed,

So amount remaining, [A]t = 100 - 19 = 81

For a first order reaction,

ln[Ao] - ln[A] = kt

ln(100) - ln(81) = k x 50 x 60

k = 7.024 x 10^-5 s-1

So, t1/2 = 0.693/k = 0.693/7.024 x 10^-5 = 9866.125 s

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