## Ozone Library

Half-life of Ozone

The half-life of ozone refers to the amount of time required for half of a given quantity of ozone to decay into oxygen. Ozone is an unstable gas and readily decomposes into oxygen. Therefore, the half-life of ozone is an essential factor to consider when determining its effectiveness in various applications.

## Typical **Gaseous Ozone (Ozone gas)** Half-life time as a Function of Temperature

Half-life time | Temp |
---|---|

∼ 3 months | -50 ºC |

∼ 18 days | -35 ºC |

∼ 8 days | -25 ºC |

∼ 3 days | 20 ºC |

∼ 1.5 hours | 120 ºC |

∼ 1.5 seconds | 250 ºC |

## Typical **Dissolved in Water Ozone** (PH=7) Half-life time as a Function of Temperature

Half-life time | Temp |
---|---|

∼ 30 minutes | 15 ºC |

∼ 20 minutes | 20 ºC |

∼ 15 minutes | 25 ºC |

∼ 12 minutes | 30 ºC |

∼ 8 minutes | 35 ºC |

### The goal is to introduce enough ozone into the water to compensate for the ozone half-life, ensuring a consistent residual ozone concentration.

**For example**, suppose we’re using ozone in an aquaculture system and want to maintain a concentration of 0.4 mg/L of ozone in a 30,000-liter (30 m^3) shrimp pond.

**1. Calculate the initial amount of ozone required:**

Required ozone quantity (NC) = Desired concentration x Pond volume

NC = 0.4 mg/L x 30,000 L = 12,000 mg

Convert from milligrams to grams by dividing by 1,000:

Required ozone quantity (NC) = 12,000 mg / 1,000 = 12 g

So, we would need 12 grams of ozone to maintain the desired concentration in the shrimp pond.

**2. Take into account that ozone degrades over time.**

Let’s assume the half-life of ozone in the pond water is 0.25 hours (∼ 15 minutes at 25ºC).

To determine the amount of ozone to inject per hour for compensation, use the following formula:

Ozone quantity (NC) / Half-life (HL) = Ozone quantity (C)

Where:

– Ozone quantity (NC): The amount of ozone required to maintain the desired concentration without compensating for the half-life.

– Half-life (HL): The half-life of ozone based on water temperature and other conditions.

– Ozone quantity (C): The amount of ozone that needs to be injected to maintain the residual concentration under current water conditions.

Assuming that water conditions remain constant and the half-life of ozone is 0.25 hours (∼ 15 minutes at 25ºC), the formula would be:

**12 g / 0.25 h = 48 g**

We would need to inject 48 grams of ozone per hour (four times more) to compensate for the degradation and maintain a concentration of 0.4 mg/L in the shrimp pond.

**Please note that it’s essential to recognize that not all the ozone generated can be fully dissolved in the water. **To achieve the 48 grams of ozone injection per hour, you may need to invest in a high-concentration ozone generator, such as the Atlas 60, capable of dissolving more than 80% of the produced ozone.

## CHANGES IN OZONE OVER TIME

**1. General equation**

The equation that describes the temporal evolution of the ozone concentration has the following form:

dn/dt= −kn² +q(1)

Where t is the time, k is the reaction rate constant, which is independent of concentration but probably, depends on the temperature, and n is the concentration (number of the molecules per unit volume):

n= (M /μV)* N(2)_{A }

μ is the molar weight of ozone, M is the mass of ozone inside the volume, V, in a given moment, NA is the Avogadro number

In Eq.(1), is the number of molecules delivered per unit of time into the volume and calculated per unit volume:

q= (m/μV) N(3)_{A}

Where m is the mass of ozone delivered into volume, V, per unit of time.

## 2. Closed volume

Consider the case of q = 0. For such a case, integration of Eq. (1) yields

n= (

n/1+_{o}n) (4)_{o }k t

where is the initial concentration. Using (2):

n(5)_{o}= (M_{o}/ μV)* N_{A}

Where M0 is the initial mass of ozone. Eq.(4) can be rewritten as

n= n_{o}/ (1 + t / τ) (6)

where we introduced the time scale parameter τ: when t =τ, the concentration of molecules (and the mass of ozone, for a given volume) decreases by a factor of two. See Eq.(6). Comparing Eqs. (4) and (6) one obtains:

τ= 1 /

nk (7)_{o }

Since k is independent of the concentration, we arrive at an important conclusion: the parameter τ depends on the initial concentration and, for a given volume and pressure, on the initial mass, M0.

τ= (μ / (N(8)_{A }M_{o }k) = k1 / M_{o}

**3. Stationary regime**

Inspecting Eq.(1), one can conclude that the concentration does not change when

q = kn²(9)

When Eq.(9) is satisfied, the rate of concentration changes, *dn*/*dt*, becomes zero, see Eq.(1). Realizing that the concentration is maintained to be constant in time, i.e., and using Eqs. (7) we obtain from (9) that

q = n(10)_{o }/ τ

Using Eqs. (3) and (5) one can rewrite (10) in terms of mass

m = M(11)_{o }/ τ

Thus, to maintain a given mass of ozone, M0, within a fixed volume at a given temperature, one should deliver ozone with the rate (mass per unit time) provided by Eq.(11). The time scale parameter τ on the right-hand side of (11) is determined from an independent experiment.