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March 29, 2024, 08:57:48 am

Author Topic: Industrial chem, equilibrium constants  (Read 899 times)  Share 

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mylinh-nguyen

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Industrial chem, equilibrium constants
« on: July 26, 2017, 11:37:31 pm »
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Hi can someone explain why only temperature has an effect on equilibrium constants but not pressure, conc etc..

kemi

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Re: Industrial chem, equilibrium constants
« Reply #1 on: July 27, 2017, 02:06:09 am »
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Consider:

2A + B <=> C

K = [C]/[A^2][B.]

If C were to increase suddenly, more of A & B would be formed, proportional to how much C was added. So if C increases, and then decreases as it forms more A and B, it will eventually RESTORE the original proportion between conc. of products and reactants. Yes, conc. will change, as more of one substance has been added, but the ratio between concentrations will be re-established.

Now, consider the same reaction and assume all substances are gaseous.

Pressure increases when volume decreases.
Pressure decreases when volume increases.

If I suddenly decreased the volume of the reaction vessel, wouldn't the conc. of all gases increase proportionally? Remember - C = n/V, so if I halve V, C will double.

But wait - the equil. shifts to the right, because there are less moles, to reduce pressure. Yes, to establish the original value of K, as it has decreased in this case with the doubling of concentrations (the square on the bottom means conc will be 4x original conc of A). So as C increases, the ratio increases to re-establish K.

Now temperature DOES change K because it drives the equilibrium to favour a certain reaction, as the forward and reverse reactions have different activation energies. As heat is continually liberated in a system with an exothermic reaction, it will reverse to absorb the heat, and continue to do so, hence increasing concentrations of reactants, decreasing concentrations of products and ultimately adjusting their proportion, which is K.

2A + B <=> C + Heat

If I add heat, K will decrease because there will be a constantly higher conc. of reactants than products, and reactants sit in the denominator of the K expression.

If I take away heat, K will increase as products are favoured and heat liberated (i.e. fwd rxn favoured). The product sits in the numerator. Also remember that as the product increases, reactants decrease and vice versa.

Well that's my basic understanding. Any experts who want to correct or polish that up are most welcome!

This isn't in the syllabus but I think I spotted a sneaky CSSA q on this...



« Last Edit: July 31, 2017, 02:03:53 pm by kemi »
HSC 2017

- X1Eng - X1Math - Chem - Bio (3rd in NSW) -

99.50 :D

mylinh-nguyen

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Re: Industrial chem, equilibrium constants
« Reply #2 on: July 28, 2017, 11:39:51 pm »
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Consider:

2A + B <=> C


K = [C]/[A^2] (the B in the denominator appears invisible on my screen?)

If C were to increase suddenly, more of A & B would be formed, proportional to how much C was added. So if C increases, and then decreases as it forms more A and B, it will eventually RESTORE the original proportion between conc. of products and reactants. Yes, conc. will change, as more of one substance has been added, but the ratio between concentrations will be re-established.

Now, consider the same reaction and assume all substances are gaseous.

Pressure increases when volume decreases.
Pressure decreases when volume increases.

If I suddenly decreased the volume of the reaction vessel, wouldn't the conc. of all gases increase proportionally? Remember - C = n/V, so if I halve V, C will double.

But wait - the equil. shifts to the right, because there are less moles, to reduce pressure. Yes, to establish the original value of K, as it has decreased in this case with the doubling of concentrations (the square on the bottom means conc will be 4x original conc of A). So as C increases, the ratio increases to re-establish K.

Now temperature DOES change K because it drives the equilibrium to favour a certain reaction, as the forward and reverse reactions have different activation energies. As heat is continually liberated in a system with an exothermic reaction, it will reverse to absorb the heat, and continue to do so, hence increasing concentrations of reactants, decreasing concentrations of products and ultimately adjusting their proportion, which is K.

2A + B <=> C + Heat

If I add heat, K will decrease because there will be a constantly higher conc. of reactants than products, and reactants sit in the denominator of the K expression.

If I take away heat, K will increase as products are favoured and heat liberated (i.e. fwd rxn favoured). The product sits in the numerator. Also remember that as the product increases, reactants decrease and vice versa.

Well that's my basic understanding. Any experts who want to correct or polish that up are most welcome!

This isn't in the syllabus but I think I spotted a sneaky CSSA q on this...




I sorta get it thanksss!!

kemi

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Re: Industrial chem, equilibrium constants
« Reply #3 on: July 31, 2017, 02:04:37 pm »
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Sorry, it's a lot to take but no problem! ;D
HSC 2017

- X1Eng - X1Math - Chem - Bio (3rd in NSW) -

99.50 :D