physical chemistry

Question

The rate constant is 7.5 × 10-3 dm3mol–1s –1. Given that the initial number of moles of A is 0.05 moles in a volume of 2 litres.

(i) Derive the relationship showing the variation of the concentration of A with time t. [7 marks]

(ii) Calculate the time (s) taken for [A] to be reduced by 20 %.

(iii) Determine the half-life of A.

Answer 1

You asked this question several days ago, and I answered it. You have added some additional parts so I'll try to address those. If you aren't happy with my previous answer, it would help to know why, so maybe I can improve it. If you are just asking again so we can answer the additional parts, then fine.

(i). For a 2nd order reaction (which this is, based on the units of k), we can derive the integrated rate law, but I'm not sure that's what you're really asking as it requires calculus. But here it is in an abbreviated form:

for a 2nd order reaction, -d[A] / dt = k[A]²

d[A] / [A]² = -kdt

integrate this to obtain ...

1 /[A] - 1 / [A]_o = kt and rearrange this to...

1 / [A] = 1 / [A]_o + kt ... Integrate Rate Law for 2nd order reaction

(ii). The integrated rate law for a 2nd order reaction is 1/[A] = 1/[A]₀ + kt (as derived above)

[A]₀ = 0.05 mol/2 L = 0.025 M

[A] = 0.20 x 0.025 = 0.005 M

k = 7.5x10^-3 M^-1s^-1

t = ?

Substituting and solving for t, we have...

1/0.005 = 1/0.025 + 7.5x10^-3 t

200 = 40 + 7.5x10^-3 t

t = 160 / 7.5x10^-3

t = 21,333 s = 356 min = 5.9 hours

(iii). The half life for a 2nd order reaction is

t_1/2 = 1 / k[A]_o

t_1/2 = 1 / (7.5x10^-3 M^-1s^-1)(0.025 M)

t_1/2 = = 1 / 1.875x10^-4 s^-1

t_1/2 = 5333 s = 88.9 min = 1.48 hrs