Question

The following data was collected in an experiment. The initial amount of hydrogen peroxide is 0.0035 mol. Find the rate constant

Moles of O2        Time ( in minutes)

0.0000482           0.008

0.0003544           1.798

0.0006828           4.027

0.0010133            7.758

Answer 1

The decomposition of H2O2 is first order and this is confirmed in the subsequent calculatons.

H2O2---> H2O+1/2O2

1 mole of oxygen is formed from 2 moles of H2O2 is decompsed. So moles of O2 will have to be multiplied with 2 to get moels of H2O2 consumed. So moles of H2O2 available = mole of H2O2 initially present- mole of H2O2 consuemed. Percentage conversion XA= 1-moles of H2O2 at any time/moles of H2O2 at zero time. The calculations are tabulated and a plot of -ln(1-XA) vs t is drawn whose slope gives the rate constant.

	Initial moles	0.0035
Moesof O2 formed	Moles of H2O2consumed	Time (minutes)	moles of H2O2 present	XA	1-XA	(-ln(1-XA)	t
0.0000482	0.0000964	0.008	0.0034036	0.027542857	0.972457	0.027929	0.008
0.000354	0.000708	1.798	0.002792	0.202285714	0.797714	0.226005	1.798
0.0006828	0.0013656	4.027	0.0021344	0.390171429	0.609829	0.494577	4.027
0.0010133	0.0020266	7.758	0.0014734	0.579028571	0.420971	0.86519	7.758

t The plot is given below from which rate constant K= 8.717min-1