physical chemistry

Question

At high temperature, a compound decomposes with a rate constant of 90 s-1. (i) Deduce the order of the decomposition explaining your reasonings. (ii) What is the concentration of the compound after 0.05 s if the initial concentration is 5.00 mol/dm3. Derive the equations to be used

Answer 1

(i). The units of the rate constant are in reciprocal seconds (s-1). This tells us the reaction is FIRST ORDER.

(ii). The rate law for a 1st order reaction is ...

-d[A] / dt = k[A]

The integrated rate law for a 1st order reaction is ...

[A] = [A]_oe^-kt

[A] = ?

[A]_o = 5.00 M

k = 90 s^-1

t = 0.05 s

[A] = [5.00]e^-(90)(0.05) = (5)e^-4.5

[A] = 0.055 M

This can be solved another way. For a 1st order reaction, the t_1/2 = ln 2 / k, so for this reaction

t_1/2 = 0.693 / 90 s = 0.0077 sec

Also for a first order reaction, the fraction remaining (FR) = 0.5ⁿ where n = number of half lives elapsed.

n = 0.05 s / 0.0077 s = 6.5 half lives

FR = 0.5^6.5 = 0.011 this is the fraction that remains

Since we began with 5.00 M, the amount remaining = 5.00 x 0.011 = 0.055 M