Le Chatelier - changing concentrations and gases (1 Viewer)

[yummy-cookies]

I was wondering if the result of changing the concentration of a gas in an equilbrium reaction involving gases would be affected by the mole ratio of the gases. Is an increase in concentration like an increase in pressure - the reaction favouring the side with less moles?

For example in N<SUB>2</SUB>O<SUB>4(g) </SUB> <--> 2NO<SUB>2(g)</SUB> by increasing the concentration of N<SUB>2</SUB>O<SUB>4 </SUB> will the equilibrium shift to the right?

 

[kwabon]

I was wondering if the result of changing the concentration of a gas in an equilbrium reaction involving gases would be affected by the mole ratio of the gases. Is an increase in concentration like an increase in pressure - the reaction favouring the side with less moles?

For example in N₂O_4(g) <--> 2NO_2(g) by increasing the concentration of N₂O₄ will the equilibrium shift to the right?

yes will shift to the right.

and no for your initial question regarding whether increasing the concentration is related to the increasing the pressure.

just think of it this way, increasing the conc. of the left side (reactants) will shift the equilibrium to the right. and vice versa.
decreasing the conc. of the left side (reactants) will shift the equilibrium to the left.
both obviously due to le chateliers principle. and this is also regarding all equilibrium equations.

 

[gurmies]

Pressure is however, related to volume =] Inversely in fact.

 

[hermand]

I was wondering if the result of changing the concentration of a gas in an equilbrium reaction involving gases would be affected by the mole ratio of the gases. Is an increase in concentration like an increase in pressure - the reaction favouring the side with less moles?

For example in N<SUB>2</SUB>O<SUB>4(g) </SUB> <--> 2NO<SUB>2(g)</SUB> by increasing the concentration of N<SUB>2</SUB>O<SUB>4 </SUB> will the equilibrium shift to the right?

i know what you're trying to say, but it's only like an increase in pressure if that gas is the only one present.

an overall increase in pressure of an only gaseous reaction will make no difference to the equilibrium.

 

[study-freak]

i know what you're trying to say, but it's only like an increase in pressure if that gas is the only one present.

an overall increase in pressure of an only gaseous reaction will make no difference to the equilibrium.

It depends on the mole ratio of gases in the reaction (unless I misunderstood it)

 

[untouchablecuz]

e.g. habor process

N_2(g)+3H_2(g)->2NH_3(g)

1mol+3mol->2mol

4mol->2mol

if pressure increases (i.e. if the gases are compressed), then the equilibrium will move to the right because this favours the production of a smaller amount of particles (and vice versa)

 

[study-freak]

e.g. habor process

N₂+3H₂->2NH₃

1mol+3mol->2mol

4mol->2mol

if pressure increases (i.e. if the gases are compressed), then the equilibrium will move to the right because this favours the production of a smaller amount of gaseous particles (and vice versa)

Just added one word but anyway +1
In the HSC, don't forget to say "gaseous"
I'm pretty sure that it will affect your mark.

 

[untouchablecuz]

ah ok, thanks

 

[hermand]

It depends on the mole ratio of gases in the reaction (unless I misunderstood it)

blehh. you're right.

far out i really should not have neglected chem.