Get Answers to all your Questions

header-bg qa

Decomposition of H2O2 follows a first order reaction.  In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be :


  • Option 1)

     6.93×10−2 mol min−1


  • Option 2)

     6.93×10−4 mol min−1


  • Option 3)

    2.66 L min−1 at STP


  • Option 4)

    1.34×10−2 mol min−1



Answers (1)

As we learnt in

Instantaneous Rate -

The rate of a reaction calculated at a particular instant of time is called Instantaneous Rate


- wherein

r_{av}=\frac{-\Delta R}{\Delta t}=\frac{+\Delta P}{\Delta t}

\lim_{\Delta\rightarrow 0}r_{av} = r_{inst}

Formula = \frac{-d[R]}{dt}= \frac{-d[P]}{dt}


 H_{2}O_{2}\rightarrow H_{2}O+\frac{1}{2}O_{2}

-\frac{\Delta \left [ H_{2} O_{2}\right ]}{\Delta t}=\frac{2\Delta O_{2}}{\Delta t}

Rate constant k=\frac{2.303}{t}\:log\frac{R_{0}}{R_t}

R_{o} and R_{t} are initial and final concentration



\frac{-\Delta \left [ H_{2}O_{2} \right ]}{\Delta t}=0.0277\times 0.05=0.00138

\frac{d\left [ O_{2} \right ]}{dt}=\frac{\Delta \left [ H_{2} O_{2}\right ]}{2\Delta t}

=\frac{0.00138}{2}=6.93\times 10^{-4}

Option 1)

 6.93×10−2 mol min−1


Incorrect option

Option 2)

 6.93×10−4 mol min−1


Correct option

Option 3)

2.66 L min−1 at STP


Incorrect option

Option 4)

1.34×10−2 mol min−1


Incorrect option

Posted by

Sabhrant Ambastha

View full answer

Crack JEE Main with "AI Coach"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support