In: Chemistry
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
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