Chemistry 3/4 2013 Thread

[Homer]

Consider the following equilibria.
i   H2(g) + CO2(g) (forth and back arrow) H2O(g) + CO(g);  (change of)H = +42 kJ mol–1
ii   N2O4(g) (forth and back arrow) 2NO2(g);  (change of)H = +58 kJ mol–1
iii   H2(g) + F2(g) (forth and back arrow) 2HF(g);  (change of)H = –536 kJ mol–1

How would you alter:
the volume of each mixture in order to produce a net forward reaction?

[psyxwar]

Consider the following equilibria.
i   H2(g) + CO2(g) (forth and back arrow) H2O(g) + CO(g);  (change of)H = +42 kJ mol–1
ii   N2O4(g) (forth and back arrow) 2NO2(g);  (change of)H = +58 kJ mol–1
iii   H2(g) + F2(g) (forth and back arrow) 2HF(g);  (change of)H = –536 kJ mol–1

How would you alter:
the volume of each mixture in order to produce a net forward reaction?

i. change in volume shouldn't affect equilibrium position
ii. increase volume
iii. change in volume shouldn't affect equilibrium position

Not sure if this is what you're looking for though.

[Homer]

yeah these are the answers but i wanted to know why?

[psyxwar]

yeah these are the answers but i wanted to know why?

Well, have you looked at Le Chatellier's principle? It basically states that the system will act to partially oppose changes in the system, sort of like homeostasis I guess (if you do biology). For example, if I was to increase the temperature of a reaction, the equilibrium position of the reaction would shift in order to oppose this change; that is, to act to decrease the system's temperature. So if I had an endothermic reaction, the equilibrium position would shift to the right. I hope that makes sense.

Anyways, if we apply Le Chatellier's principle to pressure (as pressure is inversely proportional to volume) then:

Increasing volume (and thus decreasing pressure) means the system wants to oppose this by increasing pressure. More molecules = more pressure, so the equilibrium position shifts to the side with the most moles of gas.

N2O4(g) (forth and back arrow) 2NO2(g)

On the left we have one mole of gas and on the right there is 2 moles. This means that the equilibrium will shift to the right in order to partially offset the affect of decreasing pressure.

[Homer]

i see thanks psyxwar

just want to make sure for iii)  change in volume shouldn't affect equilibrium position - is it because of number of moles of either side (1 moles (H2,F2) =2, and 2 moles of HF) are the same?

[psyxwar]

Yup.

[Homer]

The equilibrium constant for the reaction given by the equation:
2HI(g) (forth and back arrow) H2(g) + I2(g)

A mixture was prepared by placing 4.0 mol of HI in a 2.0 L vessel at 330°C. At equilibrium 0.44 mol of H2 and 0.44 mol of I2 were present. Calculate the value of the equilibrium constant at this temperature.

[psyxwar]

Use stoich to figure out how much HI is present, work out the concentrations of everything and then just calculate.

[Homer]

haha sorry but all i could find was; there will be 0.88 mols of HI reacting, and there are 4 mols provided. I dont know know what to do after

[Edward21]

Well, have you looked at Le Chatellier's principle? It basically states that the system will act to partially oppose changes in the system, sort of like homeostasis I guess (if you do biology). For example, if I was to increase the temperature of a reaction, the equilibrium position of the reaction would shift in order to oppose this change; that is, to act to decrease the system's temperature. So if I had an endothermic reaction, the equilibrium position would shift to the right. I hope that makes sense.

Anyways, if we apply Le Chatellier's principle to pressure (as pressure is inversely proportional to volume) then:

Increasing volume (and thus decreasing pressure) means the system wants to oppose this by increasing pressure. More molecules = more pressure, so the equilibrium position shifts to the side with the most moles of gas.

N2O4(g) (forth and back arrow) 2NO2(g)

On the left we have one mole of gas and on the right there is 2 moles. This means that the equilibrium will shift to the right in order to partially offset the affect of decreasing pressure.

Woah grad year 2014, you're way ahead of your time, well done! Next year will be a complete revision course for you! 

[psyxwar]

A mixture was prepared by placing 4.0 mol of HI in a 2.0 L vessel at 330°C. At equilibrium 0.44 mol of H2 and 0.44 mol of I2 were present. Calculate the value of the equilibrium constant at this temperature.

Okay so, 0.88 mol reacted as you said, because for every one mol H2 produced you have 2 mol HI reacting.

4.00 -0.88 = 3.12 mol

So:
3.12 mol HI, therefore c(HI) = 1.56M
 0.44 mol H2 and 0.44 mol I2, therefore concentrations of both H2 and I2 are 0.22M

K = (0.22M*0.22M)/((1.56M)^2)
K = 1.99 x 10^-2
K = 2.0 x 10^-2

[Homer]

hey sorry for the dumb questions but, why do you use 3.12 instead of 0.88. Isn't the 0.88mol involved in the reaction? What does the 3.12mol do and why do we use its concentration for the calculation in the equilibrium constant? Thanks

[psyxwar]

We're interested in the equilibrium concentrations of everything and not how much has reacted. For example, to calculate the K value of an equilibrium mixture when given equilibrium concentrations for a given temperature, you'd just sub those into the equilibrium equation. We're interested in what's left at equilibrium and not what has reacted to get to that point.

[9_7]

Hey guys, when does Le Chateliers principle apply? Help!!

[clıppy]

Hey guys, when does Le Chateliers principle apply? Help!!

I'm definitely probably wrong about this, but let's see if I understood what was taught to me.
So Chatelier's principle is mainly used in equilibrium reactions and says that any change to a reaction at equilibrium is counteracted 'partway'

So lets say we have a reaction at equilibrium. We add more product. This addition of product is counteracted by reacting backwards and creating more reactants, however, this new equilibrium is not at the same ratio we began with.
We could say the original equilibrium ratio was 2:1 (reactants:products) and after adding product to make this 2:5, the new equilibrium was reached at a 3:2 ratio. It reacted backwards making more reactants than products but it didn't get back to it's original ratio, hence counteracted 'partway'

Hopefully that helps in some small way and if I'm wrong (or you/I need better clarification) someone with more knowledge can post and help out.

EDIT: Woot 100th post