Question

Decomposition of $H_2O_2$ follows a first order reaction. In fifty minutes the concentration of $H_2O_2$ decreases from $0.5$ to $0.125 \,M$ in one such decomposition. When the concentration of $H_2O_2$ reaches $0.05 \,M$, the rate of formation of $O_2$ will be :

• A. $6.93 \times 10^{-2} \,mol \, min^{-1}$
• B. $6.93 \times 10^{-4} \,mol \, min^{-1}$
• C. $2.66 \,L \, min^{-1} $ at $STP$
• D. $1.34 \times 10^{-2} mol \, min^{-1}$

Answer 1

The Correct Option is B

$t_{3/4} = 2 \times t_{1/2} = 50 \min$

i.e. $ t_{1/2} = 25 \min $
$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{25} \min^{-1} $

Rate of $H_2O_2$ decomposition $= k[H_{2}O_{2}] $
$= \frac{0.693}{25} \times0.05 = - \frac{d\left[H_{2}O_{2}\right]}{dt} $
$H _{2} O _{2} \longrightarrow H _{2} O +\frac{1}{2} O _{2}$
$ - \frac{d\left[H_{2}O_{2}\right]}{dt} = 2 \frac{d\left[O_{2}\right]}{dt}$
$ \Rightarrow \frac{d\left[O_{2}\right]}{dt} = 6.93 \times10^{-4} \text{mol} \min^{-1} $

Answer 2

The decaying at an exponential rate, such as the radioactivity and some of the chemical reactions shows first order kinetics. 

So, In first order kinetics, the amount of material decaying in the given period of time is directly proportional to amount of material remaining. 

Hence, this can be represented as a differential equation \(\frac{dA}{dt}=-kt\). 

So, here \(\frac{dA}{dt}\) is the rate per unit at which the quantity of material is increasing, t is the time, k is the constant and the minus sign indicates that the material remaining will be decreasing with time. 

So the formula for the first order kinetics is: k=\(\frac{2.303}{t}log\frac{a}{a-x}\) 

Here the given values are as follows: 

t = 50min 

a = 0.5M 

a−x = 0.125M 

After putting the given values in the first order equation we get, 

⇒\(\frac{2.303}{50}log\frac{0.5}{0.125}=0.0277min^{-1}\)

So now, the balanced decomposition reaction of H₂O₂ is given as follows: 2H₂O₂→2H₂O+O₂ 

Therefore, the Rate expression for this reaction will be 

⇒\(-\frac{1}{2}\frac{dH_{2}O_{2}}{dt}=\frac{1}{2}\frac{d[H_{2}O]}{dt}=\frac{d[O_{2}]}{dt}\) 

Hence as per the differential rate expression of first order kinetics, \( -\frac{d[H_{2}O_{2}]}{dt}\) = \(k[H_{2}O_{2}]\)

As, we need the amount of the oxygen, So, therefore we by combining both above equation then we get, 

⇒\(\frac{d[O_{2}]}{dt}= -\frac{1}{2}\frac{d[H_{2}O_{2}]}{dt}=\frac{1}{2} k[H_{2}O_{2}]\)

Therefore, When the concentration of the H₂O₂ reach around 0.05M,\(\frac{d[O_{2}]}{dt}\) = \(\frac{1}{2}\times0.0277\times0.05\)

So, after calculation is done we will have \(\frac{d[O_{2}]}{dt}\) = \(6.93\times 10^{-4}molmin^{-1}\) 

Therefore the correct option will be B. \(6.93\times 10^{-4}molmin^{-1}\)