Question

It takes $1h$ for a first order reaction to go to $50$% completion. The total time required for the same reaction to reach $87.5$% completion will be:

• A. $1.75h$
• B. $6.00h$
• C. $3.50h$
• D. $3.00h$

Answer 1

The correct option is D $3.00h$

For first order reaction we have:-

$K=\frac{1}{t}ln\left(\frac{a}{a-x}\right)$ $-\left(i\right)$

where, $K$= rate constant

$t$=time

$a$= initial concentration of the reactant

$x$= amount of reactant reacted in time $t$

Case I:- $t=1$ hour & $x=0.5a$

$(a-x)=0.5a$

Putting the values in $\left(i\right)$:-

$K=\frac{1}{1}ln\left(\frac{1}{0.5}\right)=ln2$ $-\left(ii\right)$

Case II:- $t=?$ & $x=0.875a$

$(a-x)=0.125a$

Putting the value in $\left(ii\right)$:-

$K=\frac{1}{t}ln\left(\frac{1}{0.125}\right)=\frac{1}{t}ln\left(8\right)$ $-\left(iii\right)$

From, $\left(ii\right)$ and $\left(iii\right)$ $\Rightarrow ln2=\frac{1}{t}ln8$

$\Rightarrow t=\frac{ln8}{ln2}=3$ hours $[\because ln(2^3)=3ln2]$