Question

ubstance A undergoes a first order reaction A®B with a half-life, t½, of 20 min at 25 °C. If the initial concentration of A in a sample is 1.6 M, what will be the concentration of A after 80 min?

Answer 1

Answer : The concentration of A after 80 min is, 0.100 M

Explanation :

Half-life = 20 min

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{20\text{ min}}$

$k=3.465\times 10^{-2}\text{ min}^{-1}$

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant  = $3.465\times 10^{-2}\text{ min}^{-1}$

t = time passed by the sample  = 80 min

a = initial amount of the reactant  = 1.6 M

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

$80=\frac{2.303}{3.465\times 10^{-2}}\log\frac{1.6}{a-x}$

$a-x=0.100M$

Therefore, the concentration of A after 80 min is, 0.100 M