Question

Rate constant of a first order reaction is $1.15\times {10}^{-5}$. The percentage of initial concentration that remained after 1 hour is?

• A. $89.15\%$
• B. $95.96\%$
• C. $55.86\%$
• D. $83.45\%$

Answer 1

The correct option is B $95.96\%$

Given: $I^{st}$ order reaction,

$k=1.15\times {10}^{-5}$

$t=1hour=60\times 60=3600seconds$

We have :

$k=\frac{2.303}{t}\frac{Initialconc}{finalConc}$

On putting all given values:

$1.15\times {10}^{-5}=\frac{2.303}{3600}\frac{I_c}{F_c}$

$I_c$= initial concenmtration, $f_c$= final concenmtration

On solving we will get $F_c=95.96$%

Option B is correct.