A question on equilibrium constant (1 Viewer)

[Iheartpaulfrank]

okay here it goes,

2SO<SUB>2</SUB><SUB>(g) + O<SUB>2</SUB></SUB><SUB>(g) </SUB>→ 2SO<SUB>3 (g) </SUB>
<SUB>2.00 mol of SO<SUB>2 </SUB></SUB>and 2.00 mol
<SUB> </SUB>and 2.00 mol of O<SUB>2 were placed in a 1.00 L flask and allowed to react at 600C.</SUB>
<SUB>When equilibrium was established 0.150 mol of SO<SUB>2 remained.</SUB></SUB>
<SUB><SUB></SUB></SUB>
<SUB><SUB><?xml:namespace prefix = o ns = "urn:schemas-microsoft-comfficeffice" /><o>(i) Calculate the equlibrium constant for this reaction at 600C. (3 marks)</o></SUB>
</SUB>

 

[pLuvia]

[SO₂]=2mol/L
[O₂]=2mol/L
[SO₃]=0.15mol/L
K=[SO₂]²*[O₂]/[SO₃]²

Calculate for the constant.
I don't think the [SO₃] is right, someone please confirm this with me

 

[Iheartpaulfrank]

that doesn't look right .... does anyone else want to give it a go?

 

[aus_peter2005]

what's the initial concentration of SO3 ???...is it 2 mol/L????

 

[zeek]

By using the initial equation...

2SO₂ + O₂ --> 2SO₃

aswell as the initial figures...

n(SO₂)_initial= 2 mol
n(O₂)_initial= 2 mol
n(SO₂)_final= 0.150 mol
Volume of container = 1.00 L

Step 1
-------------------


If only 0.150 moles of SO₂ remained at equilibrium when 2 moles were initially used, then that must mean 1.850 moles must have reacted (by subtraction).
For O₂, if you observe the equation, you see that there are 2 moles of SO₂ for every one mole of O₂ reacting .: the number of O₂ moles reacting must be half that of the SO₂ .: it is 0.925 moles. Again, by using subtraction, we know that 1.075 moles of O₂ must remain at equilibrium.
Finally, for SO₃, we know that from the equation that 2 moles of SO₂ are required to create 2 moles of SO₃ .: 1.850 moles of SO₂ must have reacted to produce 1.850 moles of SO₃ at equilibrium.

Step 2
---------------

Now that we know the number of moles for each substance during the reaction, we need to calculate the concentration of each substance to find the equilibrium constant.
The formula for the concentration of a substance is given by:
c= n/V
We know that the Volume of the container used in this reaction is 1 L , and we know the respective moles for each substance from step 1, therefore...
c_SO2=0.15 moles/L
c_SO3=1.85 moles/L
c_O2=1.075 moles/L

Step 3
---------------

Finally, we use the concentrations for each substance to find the equilibrium constant. The formula can be easily remembered as the ratio of the products to the reactants, therefore...

K_e= [SO₃]²/([O₂] x [SO₂]²)
=3.4225/(1.075 x 0.0225)
=141.4987....


EDIT:
Some notes that might help explain...
1) You always find the final number of moles of each substance and then the final concentration as equilibrium only occurs at the end
2) Always use a balanced chemical equation
3) K_e >> 1 meaning that it is much greater than 1 and because of this, the reaction is more likely to occur to completion

 

[Iheartpaulfrank]

thank you r3v3ng3 i got the same answer, it's just i got written down that the correct answer was 680.4 or something like that. thank you so much for your effort!


oh and aus peter the initial conc'n of SO3 is 0.