Question

Decomposition of ${H_2}{O_2}$ follows a first order reaction. In fifty minutes the concentration of ${H_2}{O_2}$ decreases from $0.5M$ to $0.125M$ in one such decomposition. When the concentration of ${H_2}{O_2}$ reaches $0.05M$, the rate of formation of ${O_2}$, will be:
A. $2.78 \times {10^{ - 4}}mol{\min ^{ - 1}}$
B. $2.66L{\min ^{ - 1}}$ at STP
C. $1.34 \times {10^{ - 2}}mol{\min ^{ - 1}}$
D. $6.93 \times {10^{ - 4}}mol{\min ^{ - 1}}$

Answer 1

Hint:The differential equation describing first – order kinetics is given as-
Rate- $ = - \dfrac{{d[A]}}{{dt}} = k{[A]^1} = k[A]$
The integral equation for first order is $\ln [A] = - kt + \ln {[A]_0}$

Complete answer:
Decaying at an exponential rate, such as radioactivity and some chemical reaction shows first order kinetics. In first order kinetics, the amount of material decaying in a given period of time is directly proportional to the amount of material remaining. This can be represented as a differential equation- $\dfrac{{dA}}{{dt}} = - kt$. Here $\dfrac{{dA}}{{dt}}$ is the rate per unit at which the quantity of material is increasing, t is time, k is constant. Here the minus sign indicates that the material remaining will be decreasing with time.
The formula for the first order kinetics: $k = \dfrac{{2.303}}{t}\log \dfrac{a}{{a - x}}$
Given
$
  t = 50\min \\
  a = 0.5M \\
  a - x = 0.125M
 $
Putting those value in the first order equation we get,
$\Rightarrow k = \dfrac{{2.303}}{{50}}\log \dfrac{{0.5}}{{0.125}} = 0.0277{\min ^{ - 1}}$
Now, the balanced decomposition reaction of ${H_2}{O_2}$ is given as
$2{H_2}{O_2} \to 2{H_2}O + {O_2}$
Rate expression for this reaction will be
$\Rightarrow - \dfrac{1}{2}\dfrac{{d[{H_2}{O_2}]}}{{dt}} = \dfrac{1}{2}\dfrac{{d[{H_2}O]}}{{dt}} = \dfrac{{d[{O_2}]}}{{dt}}$
So as per differential rate expression of first order kinetics we can write, $ - \dfrac{{d[{H_2}{O_2}]}}{{dt}} = k[{H_2}{O_2}]$
Since we need the amount of oxygen therefore we by combining both above equation we get,
$\Rightarrow\dfrac{{d[{O_2}]}}{{dt}} = - \dfrac{1}{2}\dfrac{{d[{H_2}{O_2}]}}{{dt}} = \dfrac{1}{2}k[{H_2}{O_2}]$
When the concentration of ${H_2}{O_2}$ reach $0.05M$, $\dfrac{{d[{O_2}]}}{{dt}} = \dfrac{1}{2} \times 0.0277 \times 0.05$
After calculation we will have $\dfrac{{d[{O_2}]}}{{dt}} = 6.93 \times {10^{ - 4}}$ $mol{\min ^{ - 1}}$

So the correct option will be D. $6.93 \times {10^{ - 4}}mol{\min ^{ - 1}}$

Note:
If in fifty minutes, concentration of hydrogen peroxide decreases from 0.5 to 0.125. In one half life concentration decreases from 0.5 to 0.25 . So in two half lives concentration will decrease from 0.5 to 0.125 or we can say $2{t_{(1/2)}} = 50\min ,{t_{(1/2)}} = 25\min $.