Calculate equilibrium constant help please! (1 Viewer)

[catherinenguyen255]

Hi all!

Could someone please help me with this question:

1.903 mol sulfur dioxide and 1.856 mol oxygen is placed in a 10L reaction vessel. At equilibrium 0.129 mol sulfur dioxide was left in the container. Calculate the equilibrium constant.

Could you please point me in the right direction.

Thankyou

 

[faisalabdul16]

Hi all!

Could someone please help me with this question:

1.903 mol sulfur dioxide and 1.856 mol oxygen is placed in a 10L reaction vessel. At equilibrium 0.129 mol sulfur dioxide was left in the container. Calculate the equilibrium constant.

Could you please point me in the right direction.

Thankyou

Not sure if you have been taught this but ill explain anyways

What you do is write your equation with states and everything, for this one there all gases.
Then you split each substance into columns and you make 3 rows, which go downwards and you label it ICE.
I = Initial concentration
C = change in concentration
E = concentration at equilibrium.

For this question, note that they don't give you concentrations, they give you mol. So you do the same thing however when calculating equilibrium constant at the end, be sure to divide the mols by the 10L (volume) to make it concentration.

Initially you have 1.903 mol of sulfur dioxide and 1.856 of oxygen and 0 of sulfur trioxide.
When equilibrium is reached you have only 0.129 mol of sulfur dioxide remaining. Therefore the change of the mol of sulfur dioxide is the difference between initial and final. Once you have established the figure for the change in one of the substances, you can use the mol ratios to calculate the mol changes of other substances. For example mol ratio between sulfur dioxide and oxgen is 2:1, therefore you divide the mol change of sulfur dioxide by 2 to get oxygen. And you go on to do same for sulfur trioxide. Then equlibrium mol is just the difference between initial and change.

Then just divide the mol by the volume to get concentration and then just do it Hope you understood

 

[MaccaFacta]

Volume of container = 10 L
Sulfur dioxide = SO2
Oxygen = O2
Sulfur trioxide = SO3

2SO2(g) + O2(g) <> 2SO3(g)

Initial moles of SO2 = 1.903, number of moles of SO2 at equilibrium = 0.129, therefore (1.903 - 0.129) moles of SO2 has been converted into SO3.

1.903 - 0.129 = 1.774 moles of SO2 has been converted to SO3, therefore there must be 1.774 moles of SO3 in the container at equilibrium.

If 1.774 moles of SO2 reacts, then 0.5 x 1.774 moles of O2 (0.887 moles) must have reacted with it to form 1.774 moles of SO3. This means that the number of moles of O2 at equilibrium = 1.856 - 0.887 = 0.969 moles of O2 in the equilibrium mixture.

So this means that in the 10L container, at equilibrium, there are 0.129 moles of SO2, 0.969 moles of O2 and 1.774 moles of SO3.

The concentrations of these species are therefore
[SO2] = 0.129 moles / 10 L = 0.0129 mol/L,
[O2] = 0.969 moles / 10 L = 0.0969 mol/L,
[SO3] = 1.774 moles / 10 L = 0.1774 mol/L

I am going to use "^2" to mean "squared".

K = [SO3]^2 / ([SO2]^2[O2])
= 1.774^2 / (0.0129 x 0.0969)
= 2518

If you're wondering why the K value doesn't have the units of L / mol, it is because HSC equilibrium constant formulas use an approximation. Equilibrium constants don't actually use concentrations, instead they use a similar dimensionless quantity called an "activity". The difference between a concentration and an activity is too small to matter for HSC students, but those doing both Physics and Chemistry might find the lack of a unit troubling.

Hope this helps

 

[uart]

K = [SO3]^2 / ([SO2]^2[O2])
= 1.774^2 / (0.0129 x 0.0969)
= 2518

You got everything spot on there Macca but just messed up the final numerical calc.
I got Kc=1952

BTW. I've got an ulterior motive for butting in here. Can anyone tell if knowledge of equilibrium constants is currently part of the core (non elective) syllabus, or is it only in the Industrial Chem option?

I'm certain I saw an old past paper that had a "which is the correct expression for the equilib const for the above reaction" type question in the multiple choice section. Looking at the current BOS dot point syllabus however I can't see it anywhere except for in this option.

 

[MaccaFacta]

Hey uart. Wish I had a dollar for every time i've stuffed up the arithmetic over the years. But I've checked it three more times and I keep getting 2518. ???

I can help with your question. Before the syllabus changed in 2000 equilibrium constant calculations were in the core, but these days they're only in the Industrial Chemistry option. But Le Chatelier's Principle is still in the core. So all HSC Chemistry students need to know how an equilibrium will shift if you change the concentration of something or, more commonly, the temperature. But only the Industrial Chemistry elective people need to be able to do the Maths.

 

[uart]

Hey uart. Wish I had a dollar for every time i've stuffed up the arithmetic over the years. But I've checked it three more times and I keep getting 2518. ???

It's just the line where the numbers got subbed in for the concentrations. It should read 0.1774^2 / (0.0129^2 x 0.0969) = 1952.

I can help with your question. Before the syllabus changed in 2000 equilibrium constant calculations were in the core, but these days they're only in the Industrial Chemistry option.

Great, thanks for that info. I was trying to help my nephew prepare for his hsc and could see he had no idea what an equilibrium constant was. I wasn't sure whether he needed to or not.

 

[faisalabdul16]

Wow my help want to waste -.-