Collision rate termination reaction hydrogen oxygen explosion/combustion

In summary: P/RT) * (4πR^2 * (2kbT/πm)^1/2).Now, we can substitute this value for ν into the definition of the rate constant, k1, to get:k1 = ν * k_d,where k_d is another rate constant that takes into account the effectiveness of the collisions in removing the H radicals from the mixture. Finally, we can rearrange this equation to get:k1 = (12/R^2) * ((kbT)/ (πd^2 * P * √2)) * √(8kbT/πm),which is the desired equation. It is important to note that the last term, √
  • #1
Tigrisje
2
0
Hello,

I'm making a question on the explosion of hydrogen/oxygen. All the things you need to know are the following. The formation of H-radicals in the H2-O2 mixture are responsible for ignition of the explosion. This means the H-radicals need to be formed faster than they are destructed to let the mixture explode. One of the 'termination' reaction in which H-radicals are removed is through collisions with the vessel wall.

H + wall --> removal with rate constant k_1

The question now is: "Textbooks state that the rate constant k1 can be written as:
k_1 = k_d / P with P the total pressure and k_d another rate constant. Assuming that the vessel is spherical with radius R, and d is the diameter of a hydrogen atom, and m its mass. Derive now the following equation:

k1 = (12/ R^2) * ((kb*T ) / (Pi*d^2 * P * Srqt(2) ))* Sqrt( 8*kb*T / Pi*m)


How do I do this? It has something to do with determining the collion rate to the wall and the last Sqrt-term will be from the Boltzmann velocity, but I just can't see where all the terms come from.
 
Physics news on Phys.org
  • #2


Hello,

Thank you for your question. The derivation of the equation for k1 involves several steps, so I will try to explain each one in detail for better understanding.

First, we need to understand the concept of rate constant and how it relates to the collision rate between molecules. The rate constant, k1, for the reaction H + wall --> removal is defined as the proportionality constant between the rate of the reaction and the concentration of H radicals. In other words, it represents the probability of a collision between a H radical and the vessel wall resulting in its removal. This probability is influenced by factors such as the total pressure, P, of the mixture and the size of the vessel.

Next, we need to consider the size of the hydrogen atom and the vessel in which the reaction takes place. Since the vessel is assumed to be spherical with a radius R, the surface area of the vessel is 4πR^2. Now, we need to determine the number of collisions that occur between the H radicals and the vessel wall in a given time. This is known as the collision rate, denoted by Z, and is given by the equation:

Z = (N/V) * (4πR^2 * v),

where N/V is the number of H radicals per unit volume and v is the average velocity of the H radicals. The first term, N/V, can be represented as the concentration of H radicals, [H], which is equal to P/RT according to the ideal gas law. The average velocity, v, can be calculated using the Maxwell-Boltzmann distribution, which gives the probability of finding a molecule with a particular velocity at a given temperature. The equation for v is given by:

v = (2kbT/πm)^1/2,

where kb is the Boltzmann constant, T is the temperature, and m is the mass of the H atom.

Now, we can substitute these values into the equation for Z to get:

Z = (P/RT) * (4πR^2 * (2kbT/πm)^1/2).

Next, we need to determine the collision frequency, denoted by ν, which is the number of collisions per unit time. This can be calculated by multiplying the collision rate, Z, by the cross-sectional area of the hydrogen atom, πd^2/4. This gives us:

ν = (πd^2/4)
 

Related to Collision rate termination reaction hydrogen oxygen explosion/combustion

1. What is a collision rate termination reaction in the context of a hydrogen oxygen explosion/combustion?

A collision rate termination reaction refers to the final stage of a chemical reaction between hydrogen and oxygen, where the reaction rate slows down due to the decreasing number of collisions between reactant molecules. This is also known as the "burnout" stage, where the reactants are mostly consumed and the reaction begins to reach equilibrium.

2. What factors affect the collision rate termination reaction in a hydrogen oxygen explosion/combustion?

The collision rate termination reaction is affected by several factors, including temperature, pressure, and the concentration of reactants. Higher temperatures and pressures can increase the collision rate, while higher concentrations of reactants can lead to more frequent collisions and a faster reaction rate.

3. How does the collision rate termination reaction contribute to the overall explosion/combustion process?

The collision rate termination reaction is a key part of the overall explosion/combustion process, as it helps to determine the rate at which the reactants are consumed and the amount of energy released. This reaction also plays a role in the formation of products and the overall efficiency of the reaction.

4. Can the collision rate termination reaction be controlled or manipulated?

Yes, the collision rate termination reaction can be controlled and manipulated through various means, such as adjusting the temperature, pressure, and concentration of reactants. Additionally, catalysts can also be used to alter the reaction rate and control the termination stage.

5. What are some potential hazards associated with the collision rate termination reaction in a hydrogen oxygen explosion/combustion?

The collision rate termination reaction can be hazardous due to the release of large amounts of energy, which can result in an explosion. Additionally, the rapid consumption of reactants can lead to the production of toxic byproducts, such as carbon monoxide, which can be harmful to humans and the environment.

Similar threads

  • Biology and Chemistry Homework Help
Replies
2
Views
9K
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
Replies
2
Views
10K
  • General Math
4
Replies
125
Views
17K
Replies
4
Views
30K
  • Materials and Chemical Engineering
Replies
2
Views
6K
Back
Top