Question

1. Calculate the rate of the reaction (mol/Lºs) for all five runs. (See egn 26.) (Use an extra page for sample calculations.) Experiment Temperature Initial [H:02] (mol/L) 0.0283 Time of blue colour (3) 521 266 1 22.3 Thiosulphate H:02 moles Reaction rate moles added reacted (mol/ Ls) 0.00251 0.001255 0.0000301 0.00251 0.001255 0.0000589 0.00251 0.001255 0.00009 22.4 2 3 0.0567 0.0851 22.4 173 4 5

Answer 1

Answer :

The rate equation for the given reaction is :

Now, the [H2O2] = initial concentration of H2O2 - moles of H2O2 reacted

When, the initial concentration of H2O2 is 0.0283 mol/L the rate of the reaction is 0.0000301 mol/L• S

Hence, when the initial concentration of H2O2 is 0.0567 mol/L the rate of the reaction is 0.0000589 mol/L•S

Hence,

Comparing, the two equation we get,

Or, a = 0.937 is nearly equal to 1

Hence, it is a first order reaction.

We can now determine the rate constant of the reaction by putting the value of "a" in the third run value.

i.e. when the initial concentration of H2O2 is 0.0851 mol/L, the rate of the reaction is 0.00009 mol/L •S

Hence,

Or, k' = 0.00009/0.097907

Or, k' = 9.19* 10-4 Sec-1

The equation is

Or,

It is a equation of a straight line in the form of

Y =mX +C

Hence, the plot of log(rate) versus log[H2O2] gives :