Question

The rate constant for a first order reaction with a single reactant is 7.54x10-4 s-1. If the initial reactant concentration is 1.18 M, what will the reactant concentration be after 8.30 minutes?

Answer 1

Answer : The reactant concentration be after 8.30 minutes is, 0.812 M

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  = $7.54\times 10^{-4}s^{-1}$

t = time passed by the sample  = 8.30 min = 498 s

a = initial amount of the reactant  = 1.18 M

a - x = amount left after decay process = ?

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

$498=\frac{2.303}{7.54\times 10^{-4}}\log\frac{1.18}{a-x}$

$a-x=0.812M$

Therefore, the reactant concentration be after 8.30 minutes is, 0.812 M