# HOME - TEACHING

## Catalase and capillaries

$$2\ H_2O_2 \rightarrow O_2 + 2 \ H_2O$$

We know

• the quantity of $$H_2O_2$$
• We want the measure the reaction speed
• We can observe the volume displaces according to the time

### For $$O_{2}$$ :

• $$P \times V = n \times R \times T$$
P = 1atm

• $$V = \pi \times r^{2} \times h$$
With R the radius of the capillary, and h its length.
For example with r = 0.5mm and h = 1cm
$$V = 3,14 \times 0,5.10^{-3} \times 1.10^{-2} = 1,57.10^{-5}\ m^3 = 1,57.10^{-2} L$$

• $$R = 0,08206 \ L·atm·mol^{-1}·K^{-1}$$
We can simplify by choosing T = 298K (27°C)

• $$n = \frac{P \times V}{R \times T}$$

• $$n = \frac{1 \times 1,57.10^{-2}}{0,08206 \times 298} = 0.0064 moles$$ of $$O_2$$ in 1 cm of capillary which is 6.4 mmol

### For $$H_{2}O$$ :

• 1 L = 1 kg
• $$M_{H_{2}O} = 18\ g.mol^{-1}$$

• $$Vm = \frac{1000}{18} = 55.5\ mol.L^{-1}$$

• $$n = V/Vm = 1,57.10^{-2}/55.5 = 0.0002828 moles$$ in 1 cm of capillary which is 0.2828 mmol

### –> We can neglect the volume occupied by $$H_2O$$

So now we have the following stoechiometry : $$2\ H_2O_2 \approx O_2$$

So $$n_{H_2O_2} \approx \frac{n_{O_2}}{2}$$

### Putting everything together

• What we measure : speed of deplacement in cm/s
• What we want to know : the catalytic activity in mole/s

$$catalitic \ activity (mole.s^{-1}) = speed (s^{-1}) * \frac{6.4}{2} \times 10^{-3}$$

### Calculator of catalitic activity

Speed in capillary (cm/s)

mol/s