Question

Decomposition of $H_2{O}_2$ is a first order reaction. A solution of $H_2{O}_2$ labelled as $20$ volumes was left open. Due to this some $H_2{O}_2$ decomposed. To determine the new volume strength after $6$ hours, $10mL$ of this solution was diluted to $100mL.10mL$ of this diluted solution was titrated against $25mL$ of $0.025MKMn{O}_4$ acidified solution. Calculate the rate constant for decomposition of $H_2{O}_2$.

Answer 1

After hour

Meq. of $H_2{O}_2$ in 10 mL dilute solution

$=25\times 0.025\times 5=3.125$

Meq. of $H_2{O}_2$ in 10 mL conc. solution

$=3.125\times 10=31.25$

Meq. of $H_2{O}_2$ in 10 mL of 20 volume $H_2{O}_2$ present initially

$=\frac{68\times 20\times 10}{22400\times 17}\times 1000=35.71$

$[\because 22400mL{O}_2=68g{H}_2{O}_2]$

$\therefore 20mL{O}_2=\frac{68\times 20}{22400}g{H}_2{O}_2$ in 1 mL

$=\frac{68\times 20\times 10}{22400}g{H}_2{O}_2$ in 10 mL

$\therefore K=\frac{2.303}{6}log\frac{35.71}{31.25}=0.022hr$

Hence, answer is 2.