Question

The initial concentration of reactant in a first-order reaction is 0.27 m. the rate constant for the reaction is 0.75 s-1. what is the concentration (mol/l) of reactant after 1.5 s?

Answer 1

The concentration in  mol/l  of reactant  after  1.5 s  is calculated as  follows

from  first order  integrated  equation

 In (A)t  = - Kt + In (A)o  where
  At = final  concentration =?
A)o =initial  concentration =0.27 M
K =  constant=0.75
T=temperature =  1.5 s

in( A)t = -0.75(1.5) + in(0.27)

In (A)t =-1.125  +-1.31 =-2.435
In(A)t =-2.435

In (A)t  =  e
find the e  value 

 ( A)t is therefore =0.0876  mol/l

Answer 2

Answer: 0.087 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  = $0.75s^{-1}$

t = time taken for the reaction  = 1.5 sec

a = initial amount of the reactant  = 0.27 M

a - x = amount left  = ?

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

$1.5=\frac{2.303}{0.75}\log\frac{0.27}{a-x}$

$(a-x)=0.087M$

Thus the concentration of reactant after 1.5 s is 0.087 mol/L