Question

A first order reaction has specific reactions rate of 1/100 per sec. How much time it take for 20g of the reactant to reduce to 5g?

Answer 1

Answer:

138.6 secs

Explanation:

Integrated rate law equation for first order reaction is

$A=A_0e^{-k.t}$

Rearranging the equation we obtained

$\frac{A}{A_0}=-k.t\\or \\taking ln on both side we get \\ln\frac{A}{A_0}=-k.t$

taking    negative    on   both   sides   we get

$ln\frac{A_0}{A}= k.t$  (1)

where A is the concentration left and A0 is the initial concentration.

Concentration is expressed as

$A=\frac{mass of reactant}{Molecular mass of reactant*Volume}$

where mass/molar mass become number of moles.

Lets take the volume = 1

Now when we apply this assumption to our rate law equation 1

We get

$ln\frac{mass taken initial}{mass left}=k.t$

We have the rate constant or specific reaction rate as 1/100 per sec.

substituting the data in the equation

$ln\frac{20}{5}=\frac{1}{100}*t$

so  

 $t= 100*ln 4 \\t=100*1.3859\\=138.6 sec$