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

k=\frac{2.303}{50}\:log\frac{0.50}{0.125}=\frac{2.303}{50}log4

k=0.0277

\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

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