changing equilibrium and pressure (1 Viewer)

[annabananna88]

hey i was hoping someone could help me...
just wondering which way the equilibrium shifts when pressure is increased (i just really dont get it).... if possible in relationship to

2SO2 + O2 <-> 2SO3

(sorry no idea how to do em little numbers on this)

Thanx heaps

 

[pLuvia]

2SO₂ + O₂ <--> 2SO₃

If pressure is increased then the solubility will increase and therefore more SO₂ will dissolve so the equilibrium moves to the right.

When the total pressure of a system at equilibrium is increased, the equilibrium will shift to the side that has less moles of gas.

 

[Dreamerish*~]

Pressure increases equilibrium moves to the right
Pressure decreases equilibrium moves to the left

Those can't be rules though. Since it's an equilibrium, the equation goes both ways. What if I write 2SO₃

O₂ + 2SO₂? Then your statement would no longer be true.
It all depends on how many moles of gas are involved in the reaction. So it would have helped if you put in the states.

So assuming it's a solubility reaction (I have to say I haven't seen SO₃ on its own. Sulfite is an anion - SO₃^2-), we can write the equation as:

2SO_2(g) + O_2(g)

2SO_3(aq)

When the total pressure of a system at equilibrium is increased, the equilibrium will shift to the side that has less moles of gas.

If we look at the above equation, there are 2 moles of sulfur dioxide gas and one mole of oxygen gas on the reactants side, making 3 altogether. On the products side there are no gases. Hence when the total pressure is increased, the equilibrium will shift to the right.

Mind if I ask where the equation is from?

 

[pLuvia]

Ok thanks for clearing that up

 

[Dreamerish*~]

Ok thanks for clearing that up

No worries.

 

[annabananna88]

Thanx heaps thast really helped.... i was doin y7 orrientation stuff last year when we started equilibrium and kinda didnt get it at the time

Thanks Again
Anna

 

[.ben]

Just one question:

if you add more solid to a chemical reaction (reactants side) does that affect the concentration? i've gotten conflicting reports that it does and it doesn't???

thanks

 

[Dreamerish*~]

Just one question:

if you add more solid to a chemical reaction (reactants side) does that affect the concentration? i've gotten conflicting reports that it does and it doesn't???

thanks

It depends. Does the solid react? That is, does it take part in the reaction?

If it doesn't, like dropping a wood block into a fizzy drink, then it shouldn't affect the equilibrium.

 

[.ben]

yes, yes it does.

 

[Dreamerish*~]

yes, yes it does.

If it does, then yes it will affect the equilibrium, because by adding it you're increasing the concentration of a product.

But this is only when you have a solubility reaction. If it's something else, like the precipitation reaction between silver nitrate and sodium chloride, chucking a blob of solid silver into the solution won't affect the equilibrium, because the reactant is silver nitrate, not silver.

 

[.ben]

ok i get it now thanks

 

[OzV]

This reaction is from the manufacturing process of sulphuric acid. It must occur with the addition of a catalyst, at reasonably high temperatures (450 deg C) at a pressure of 1-2 atm. The SO₃ is not the sulfite radical but is sulfur trioxide. The overall reaction including states is:

2SO_2(g) + O_2(g) <--> 2SO_3(g)

The role of the V₂O₅ catalyst is to reduce the time needed to achieve equilibrium by reducing the activation energy, increasing the temperature is needed to increase the efficiency of the catalyst but it also drives the reaction to the left because it is exothermic to the right (it needs to be driven to the right as the SO₃ is then passed on to the next stage to react with oleum), and increasing the pressure pushes the reaction to the right. Because all the reactants and products are gases at these temps and pressures we are looking at gas-phase equilibria and working out which way the reaction will go when the reaction system is at 1-2atm will involve calculating partial pressures and reaction quotients etc...

This is a very hard problem for HSC chem.... Can we explain it in terms of a different reaction? CO₂ + H₂O <--> H₂SO₄ maybe?

 

[Dragie]

uhh I got a question -
If it isn't specified which side is increasing/decreasing in temp/concn/pressure, how do you know to which side the new equilibrium will shift? Does it have something to do with how if (say for pressure) pressure increases, a gas's solubility in water increases therefore there is a shift to the right? I'm just a little confuddled with this at the moment.

 

[insert-username]

If it isn't specified which side is increasing/decreasing in temp/concn/pressure, how do you know to which side the new equilibrium will shift? Does it have something to do with how if (say for pressure) pressure increases, a gas's solubility in water increases therefore there is a shift to the right? I'm just a little confuddled with this at the moment

If the pressure increases, equilibrium will shift to the shift with less gas molecules or less moles of gas. This is because equal volumes of gas at equal temperature and pressure inhabit the same volume. So less gas = less volume = less pressure, to minimise the disturbance (as per Le Chatelier's principle).

2NO₂ (g) N₂O₄ (g)

When the pressure is increased here, equilibrium shifts to the right, where there are less gas molecules.

If the whole system is increasing in temperature, the equilibrium will go in the direction of the endothermic reaction to use up the extra heat. If the temperature is decreasing, equilibrium shifts in the direction of the exothermic reaction, to produce more heat. These two effects are again predicted through Le Chatelier's principle.


I_F

 

[Dragie]

ohhhh ok - thanks for that!

 

[babywwq]

2SO2(g) + O2(g) <--> 2SO3(g)
easy to understand

2mole+1mole=3mole>2mole SO3

thus increase the pressure more mole to less mole