Kinetics of Decomposition of Hydrogen Peroxide | Chemical Kinetics | Physical Chem

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Hydrogen peroxide with formula H2O2 is the simplest peroxide, a compound with an oxygen-oxygen single bond.
Hydrogen peroxide is unstable and slowly decomposes in the presence of light. Because of its instability, hydrogen peroxide is typically stored in dark bottles.

H_2O_2 \overset { Pt }{ \longrightarrow } H_2O + \frac12O_2

Hydrogen peroxide decomposes to produce water and half of the oxygen is liberated out. In this case, the concentration of H2O2 is found periodically by titrating it with KMnO4 solution.
At a fixed interval of time say every 10 min 5 ml of the reaction mixture is taken out form the RBF and it is titrated against KMnO4 solution from the burette.

So at the start as the concentration of H_{ 2 }{ O }_{ 2 } is high, It needs more amount of KMn{ O }_{ 4 } to get equivalence point. Or we can say the volume of KMn{ O }_{ 4 } is proportional to initial concentration of H_{ 2 }{ O }_{ 2 }. left( { Initial concentration of H_{ 2 }{ O }_{ 2 } (a) } right) propto left( { Volume of KMn{ O }_{ 4 } used at the start time { V }_{ 0 } } right) i.e. a propto { V }_{ 0 } ...(1) As the reaction proceed the concentration of H_{ 2 }{ O }_{ 2 } goes on decreasing and thus the amount of KMn{ O }_{ 4 } needed will also go on decreasing, Or the volume of KMn{ O }_{ 4 } left( { V }_{ t } right) needed at time t will be proportional to the concentration of H_{ 2 }{ O }_{ 2 } remaining i.e. (a - x) as x is the amount reacted. left( { Concentration of H_{ 2 }{ O }_{ 2 } remaining left( a - x right) } right) propto left( { Volume of KMn{ O }_{ 4 } left( { V }_{ t } right) used at time t } right) i.e. left( a - x right) propto { V }_{ t } ...(2) Substituting these values in the first order equation. k = frac { 2.303 }{ t } log { frac { a }{ a-x } } we get, boxed { k = frac { 2.303 }{ t } log { frac { { V }_{ 0 } }{ { V }_{ t } } } } The decomposition of hydrogen peroxide, tested by applying this equation, is found to be of the first order.

Numerical: From the following data, show that the decomposition of hydrogen peroxide is a reaction of the first order: Time (minutes): 0 10 20 Vol. of KMn{ O }_{ 4 } (ml): 46.1 29.8 19.3 Solution: If the reaction is of the first order, it must obey the equation, k = frac { 2.303 }{ t } log { frac { a }{ a-x } } or boxed { k = frac { 2.303 }{ t } log { frac { { V }_{ 0 } }{ { V }_{ t } } } } The value of k at each time can be calculated as follows: 1) at time t = 10 min volume of KMn{ O }_{ 4 } used { V }_{ t } = 29.8ml k = frac { 2.303 }{ 10 } log { frac { 46.1 }{ 29.8 } } = 0.2303times log { left( 1.547 right) } = 0.0436{ min }^{ -1 } 2) at time t = 20 min volume of KMn{ O }_{ 4 } used { V }_{ t } = 19.3ml k = frac { 2.303 }{ 20 } log { frac { 46.1 }{ 19.3 } } = 0.115times log { left( 2.389 right) } = 0.0435{ min }^{ -1 } Since the value of k is nearly constant, given reaction is of first order. The average value of the rate constant = 0.044{ min }^{ -1 }