## 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) **