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Question: Decomposition of \({H_2}{O_2}\) follows a first order reaction. In fifty minutes the concentration o...

Decomposition of H2O2{H_2}{O_2} follows a first order reaction. In fifty minutes the concentration of H2O2{H_2}{O_2} decreases from 0.5M0.5M to 0.125M0.125M in one such decomposition. When the concentration of H2O2{H_2}{O_2} reaches 0.05M0.05M, the rate of formation of O2{O_2}, will be:
A. 2.78×10−4molmin⁡−12.78 \times {10^{ - 4}}mol{\min ^{ - 1}}
B. 2.66Lmin⁡−12.66L{\min ^{ - 1}} at STP
C. 1.34×10−2molmin⁡−11.34 \times {10^{ - 2}}mol{\min ^{ - 1}}
D. 6.93×10−4molmin⁡−16.93 \times {10^{ - 4}}mol{\min ^{ - 1}}

Explanation

Solution

The differential equation describing first – order kinetics is given as-
Rate- =−d[A]dt=k[A]1=k[A] = - \dfrac{{d[A]}}{{dt}} = k{[A]^1} = k[A]
The integral equation for first order is ln⁡[A]=−kt+ln⁡[A]0\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- dAdt=−kt\dfrac{{dA}}{{dt}} = - kt. Here dAdt\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=2.303tlog⁡aa−xk = \dfrac{{2.303}}{t}\log \dfrac{a}{{a - x}}
Given
t=50min⁡ a=0.5M a−x=0.125M t = 50\min \\\ a = 0.5M \\\ a - x = 0.125M
Putting those value in the first order equation we get,
⇒k=2.30350log⁡0.50.125=0.0277min⁡−1\Rightarrow k = \dfrac{{2.303}}{{50}}\log \dfrac{{0.5}}{{0.125}} = 0.0277{\min ^{ - 1}}
Now, the balanced decomposition reaction of H2O2{H_2}{O_2} is given as
2H2O2→2H2O+O22{H_2}{O_2} \to 2{H_2}O + {O_2}
Rate expression for this reaction will be
⇒−12d[H2O2]dt=12d[H2O]dt=d[O2]dt\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, −d[H2O2]dt=k[H2O2] - \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,
⇒d[O2]dt=−12d[H2O2]dt=12k[H2O2]\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 H2O2{H_2}{O_2} reach 0.05M0.05M, d[O2]dt=12×0.0277×0.05\dfrac{{d[{O_2}]}}{{dt}} = \dfrac{1}{2} \times 0.0277 \times 0.05
After calculation we will have d[O2]dt=6.93×10−4\dfrac{{d[{O_2}]}}{{dt}} = 6.93 \times {10^{ - 4}} molmin⁡−1mol{\min ^{ - 1}}

**So the correct option will be D. 6.93×10−4molmin⁡−16.93 \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 2t(1/2)=50min⁡,t(1/2)=25min⁡2{t_{(1/2)}} = 50\min ,{t_{(1/2)}} = 25\min .