Question

From the following data, the decomposition of $H_2{O}_2$ in aqueous solution follows first order kinetics. Then calculate its rate constant (k) value.

Time (in minutes)	0	10	20
Volume (V) in $mL$	22	13	7.5

where V is the volume of potassium permanganate required to react with undecomposed hydrogen peroxide solution.

Take
$log\frac{22}{13}=0.23log\frac{22}{7.5}=0.46$

• A. $k=0.053mi{n}^{-1}$
• B. $k=0.102mi{n}^{-1}$
• C. $k=0.204mi{n}^{-1}$
• D. $k=0.40mi{n}^{-1}$

Answer 1

The correct option is A $k=0.053mi{n}^{-1}$

The equation is
$2H_2{O}_2\stackrel{\Delta }{⟶}2H_{2}O\left(l\right)+O_2\left(g\right)$

The volume of $KMn{O}_4$ used, evidently corresponds to the undecomposed hydrogen peroxide.
$\therefore$
Volume of $KMn{O}_4$ used at zero time $\left(V_0\right)$ corresponds to the initial concentration (a) of $H_2{O}_2$
$V_0\propto a$

Volume of $KMn{O}_4$ used after time 't' $V_t$ corresponds to the undecomposed hydrogen peroxide concentration $(a-x)$.
$V_t\propto a-x$

The integrated rate law for 1st order reaction
$k=\frac{2.303}{t}log\frac{a}{a-x}$

$k=\frac{2.303}{t}log\frac{V_0}{V_t}$

After 10 min,
$k_1=\frac{2.303}{10}log\frac{22}{13}=0.0526mi{n}^{-1}$

After 20 min,
$k_2=\frac{2.303}{20}log\frac{22}{7.5}=0.0538mi{n}^{-1}$

$k_{avg}=0.053mi{n}^{-1}$
The constancy of k shows that, the reaction follows first order kinetics.