Question

The time for half change in a first order decomposition of a substance A is 60 seconds. Calculate the rate constant. How much of A will be left after 180 seconds?

Answer 1

1. Order of a reaction = 1

`"t"_(1/2)` = 60 seconds,

k = ?

We know that, k = `0.6932/"t"_(1/2)`

k = `2.303/60` = 0.01155 s⁻¹

2. [A₀] = 100%

t = 180 s

k = 0.01155 seconds⁻¹

[A] = ?

For the first order reaction k = `2.303/"t" log  (["A"_0])/(["A"])`

0.01155 = `2.303/180 log (100/(["A"]))`

`(0.01155  xx 180)/2.303 = log  (100/(["A"]))`

0.9207 = log 100 – log [A]

log [A] = log 100 – 0.9207

log [A] = 2 – 0.9207

log[A] = 1.0973

[A] = antilog of (1.0973)

[A] = 12.5%