The ‘initial rate proportional to 1/t’ concept is applicable to a chemical kinetics experiment that has the initial concentration of a reactant varied but the measured amount of a product fixed. One such experiment is the decomposition of hydrogen peroxide over a platinum catalyst, while another is the iodine clock. For the decomposition experiment, we have:

Using various initial concentrations of *H _{2}O_{2}*, different times taken to collect a fixed volume of oxygen are recorded. The diagram below illustrates the setup, where the fixed volume of oxygen produced under constant temperature is measured by a fixed change in pressure of the constant-volume system.

Since the reaction is very slow, we assume that the initial concentration of *H _{2}O_{2}* for each experiment remains relatively unchanged for the volume of oxygen generated (the change in pressure of the system can be fixed at an amount that does not exceed three minutes to occur). Using eq10, and assuming gases behave ideally, we can write

To produce a fixed volume of oxygen, the change in amount of oxygen, and hence the change in pressure of the system, is constant regardless of the concentration of *H _{2}O_{2}* used. Thus,

where C is a constant and is equal to .

This answers the question of why initial rate is proportional to 1/*t*.

To develop an equation that that allows us to further analyse the decomposition reaction, we propose the following rate law:

where [*H _{2}O_{2}*]

*is the initial concentration of*

_{0 }*H*.

_{2}O_{2}Taking the natural logarithm for both sides of eq28 and eq29, we have

and

Substitute eq31 in eq30

Therefore, using the time measurements for the fixed change in pressure to occur for different initial concentrations of *H _{2}O_{2}*, a plot of versus gives a gradient of the order of the reaction,

*i*, and a vertical intercept of where

*k*is the rate constant.

###### Question

Eq32 is derived using the rate equation . What if we define the rate equation as ?

###### Answer

To produce a fixed volume of oxygen, the change in amount of *H _{2}O_{2}*, , is a constant value regardless of the initial amounts of

*H*. For example, to produce 1 mole of oxygen, the change in amount of

_{2}O_{2}*H*is always -2 moles irrespective of how many moles of

_{2}O_{2}*H*there are to begin with. Furthermore, the same volume of

_{2}O_{2}*H*is used as we vary [

_{2}O_{2 }*H*]. So, the 2

_{2}O_{2}^{nd}term on RHS of eq33 is a constant, which makes eq33, like eq32, a linear function.

Comparing eq33 with eq32,