Question

The half life for the first order conversion of A to B is 56.6 hours. How long does it take for the concentration of A to decrease 10.0% of its original concentration

Answer 1

Answer:

It will take 188.06 hours for the concentration of A to decrease 10.0% of its original concentration.

Explanation:

A → B

Initial concentration of the reactant = x

Final concentration of reactant = 10% of x = 0.1 x

Time taken by the sample, t = ?

Formula used :

$A=A_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

where,

$A_o$ = initial concentration of reactant

A = concentration of reactant left after the time, (t)

$t_{\frac{1}{2}}$ = half life of the first order conversion  = 56.6 hour

$\lambda$ = rate constant

$A=A_o\times e^{-(\frac{0.693}{t_{1/2}})\times t}$

Now put all the given values in this formula, we get

$0.1x=x\times e^{-(\frac{0.693}{56.6 hour})\times t}$

t = 188.06 hour

It will take 188.06 hours for the concentration of A to decrease 10.0% of its original concentration.