Question

The rate constant of the first-order reaction A→3B is 0.291 s−1. The concentration of A at t=5 seconds is 0.081 mol/L. What was the initial concentration of A?

Answer 1

Answer : The initial concentration of A is, 0.347 mol/L

Explanation :

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  = $0.291s^{-1}$

t = time passed by the sample  = 5 s

a = initial amount of the reactant  = ?

a - x = amount left after decay process = 0.081 mol/L

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

$5=\frac{2.303}{0.291}\log\frac{a}{0.081}$

$a=0.347mol/L$

Therefore, the initial concentration of A is, 0.347 mol/L