Decrease Pressure (Increase Volume) - Le Chatilier's Principle -Change in Equilibrium (1 Viewer)

[frog0101]

Hi,
In terms of collision theory, why does a decrease in pressure (increase in volume) favour the side with less molecules.
Eg. for an increase in pressure:
The system will respond by reducing the pressure, which is done by reducing the total amount of particles in the system, thus shifting the equilibrium position towards the side with the least particles. The side with more particles will experience a greater increase in concentration, due to a volume decrease, hence more of those particles would react.

I understand that increasing the volume favors the side with less molecules - is this because (proportional to the side with more molecules) there is a smaller change in concentration for the side with less, and so the ratio of the rate of reaction for both sides increases to favor side with less molecuels??

 

[bjw22]

Re: Decrease Pressure (Increase Volume) - Le Chatilier's Principle -Change in Equilib

This isn't my number one subject, so personally I simplify things down a bit. For an exam, I reckon it would be suffice to say that by Le Chatelier's Principle, a decrease in pressure results in a shift to the side with more gaseous moles in order to increases the volume produced, hence as in a fixed space resulting in increased pressure - minimise disturbance.

If you want to go deeper by all means go ahead, but I haven't seen a question asking for too much detail, simply asking what happens when?

 

[jazz519]

Re: Decrease Pressure (Increase Volume) - Le Chatilier's Principle -Change in Equilib

Be very careful in your wording it doesn’t relate to the collision theory. Collision theory just says you need certain steric factors (orientations and shape) and energy requirements (activation energy barrier has to be overcome) for a reaction (this is a chemistry field called kinetics). What you are talking about is strictly to do with LCP which is an application of the conservation of mass and conservation of energy laws (I.e. thermodynamics). These two are very different so make sure you don’t use the collision theory wording (you can only use that for describing rate of reaction due to temperature changes and etc which don’t have any bearing on where your equilibrium is or which way it will shift in response to what you do).

Now to answer the question let’s consider the haber process: 3h2 + n2 -><- 2nh3

Deceasing pressure disturbs the equilibrium. So in response lcp will try to do the opposite shift to the side with most gas moles. The reason it shifts to side most gas moles is what does that mean, it means you are making more gas and more gas equals more pressure.

Same thing applied with increasing pressure, so lcp tries to decrease pressure. How can you decrease pressure remove gas from the system. And the best way to do it is to convert those 4 moles of gas on the left to 2 moles of gas. Meaning you get a net decrease in gas moles by 2 moles. Therefore, less gas, less pressure

 

[mariakery1]

Re: Decrease Pressure (Increase Volume) - Le Chatilier's Principle -Change in Equilib

The new syllabus requires you to: explain the overall observations about equilibrium in terms of the collision theory.

Hi,
In terms of collision theory, why does a decrease in pressure (increase in volume) favour the side with less molecules.
Eg. for an increase in pressure:
The system will respond by reducing the pressure, which is done by reducing the total amount of particles in the system, thus shifting the equilibrium position towards the side with the least particles. The side with more particles will experience a greater increase in concentration, due to a volume decrease, hence more of those particles would react.

I understand that increasing the volume favors the side with less molecules - is this because (proportional to the side with more molecules) there is a smaller change in concentration for the side with less, and so the ratio of the rate of reaction for both sides increases to favor side with less molecuels??

An increase in pressure (decrease in volume) favours the side with LESS moles. To explain this using collision theory:

An increase in pressure (decrease in volume) will result in the molecules to have less space to move around and hence the concentration of ALL gaseous substances (both reactants and products) increases. The rate of both the forward and reverse reactions increase due to more successful collisions between its substances. HOWEVER, the system will eventually favour the side with less moles in order to counteract the increase in pressure, since the side with more moles will have more collisions than the side with less moles. Hence, an increase in pressure (decrease in volume) will result in equilibrium to shift towards the side with less moles.

A decrease in pressure (increase in volume) favours the side with MORE moles. To explain this using collision theory:

A decrease in pressure (increase in volume) will result in the molecules to have more space to move around and hence the concentration of ALL gaseous substances (both reactants and products) decreases. The rate of both the forward and reverse reactions decrease due to less successful collisions between its substances. HOWEVER, the system will eventually favour the side with more moles in order to counteract the decrease in pressure (you can’t really explain this part through collision theory or you end up contradicting yourself). Thereby, partially counteracting change in pressure by shifting equilibrium to the side with more moles.

 

[DHRUVtheJAGUAR]

Re: Decrease Pressure (Increase Volume) - Le Chatilier's Principle -Change in Equilib

The new syllabus requires you to: explain the overall observations about equilibrium in terms of the collision theory.



An increase in pressure (decrease in volume) favours the side with LESS moles. To explain this using collision theory:

An increase in pressure (decrease in volume) will result in the molecules to have less space to move around and hence the concentration of ALL gaseous substances (both reactants and products) increases. The rate of both the forward and reverse reactions increase due to more successful collisions between its substances. HOWEVER, the system will eventually favour the side with less moles in order to counteract the increase in pressure, since the side with more moles will have more collisions than the side with less moles. Hence, an increase in pressure (decrease in volume) will result in equilibrium to shift towards the side with less moles.

A decrease in pressure (increase in volume) favours the side with MORE moles. To explain this using collision theory:

A decrease in pressure (increase in volume) will result in the molecules to have more space to move around and hence the concentration of ALL gaseous substances (both reactants and products) decreases. The rate of both the forward and reverse reactions decrease due to less successful collisions between its substances. HOWEVER, the system will eventually favour the side with more moles in order to counteract the decrease in pressure (you can’t really explain this part through collision theory or you end up contradicting yourself). Thereby, partially counteracting change in pressure by shifting equilibrium to the side with more moles.

how could you explain the effect of a change in temperature through the collision theory