Question

Table 1: Temperature, Pressure and Volume Data

Temp of distilled H2O	Room pressure (atm)	Initial Volume of Air (mL)	Final Volume of Air (mL)	Volume of O2 Collected (mL)
21.1°C	1	10mL	60mL	50mL

this experiment was done by mixing hydrogen peroxide (3%) and yeast in a flask where the trapped gas would travel through flexible tubng and displace air in an inverted graduated cylinder that is full of water.

1. Calculate the number of moles of O2 produced using the ideal gas law. Then, use this value to calculate the number of moles of hydrogen peroxide you began the experiment with.

2.   Calculate the number of moles of hydrogen peroxide you would have if you used 5 mL of a pure hydrogen peroxide solution. HINT: The density of hydrogen peroxide is 1.02 g/mL.

4.   Was the calculated percentage of hydrogen peroxide close to the same as the percentage on the label (3%)? Calculate percent error of your value.


5.   Considering that catalysts are not consumed in a reaction, how do you think increasing the amount of catalyst would affect the reaction rate for the decomposition of hydrogen peroxide?

6. How would the number of moles (n) of O2 change if your atmospheric pressure was doubled and all other variables stayed the same?

Answer 1

2 H2O2 (l) --> 2 H2O (l) +O2 (g)

Temperature, T = 21.1 °C = 294.25 K

Volume of O2 collected = 50 mL = 0.05 L

Pressure = 1 atm

1.

P V = n R T

1 * 0.05 = n * 0.0821 * 294.25

Moles of O2 produced, n = 0.00207 moles

Moles of H2O2 reacted = 2n = 0.00414

2.

Mass of H2O2 used = Density * Volume * Purity

= 1.02 g/mL * 5 mL * 100 / 100= 5.1 g

Molar mass of H2O2 = 34 g/mol

Moles of H2O2 = 5.1 g / 34 g/mol

= 0.15

4.

Mass of H2O2 used = Density * Volume * Purity

= 1.02 g/mL * 5 mL * 3 / 100= 0.153 g

Molar mass of H2O2 = 34 g/mol

Moles of H2O2 = 0.153 g / 34 g/mol

= 0.0045

Yes it is close to calculated moles of 0.00414

% error = (0.0045 – 0.00414)/0.0045 * 100

= 8 %

5.

Catalyst helps only in lowering the activation energy of intermediate. It helps enhancing the reaction rate as compared to reaction without catalyst.

Since the reaction rate (mol/s-gram.cat) is independent of amount of catalyst, the rate of decomposition of H2O2 will have no effect on increase in catalyst.

6.

Since the reaction is taking place in liquid phase, increase in pressure will have insignificant impact on reaction rate and hence, number of moles of O2 produced.

Reaction rate will mainly depend on concentration of H2O2 and temperature.