Equilibrium constant (1 Viewer)

[Snowflek]

Hey guys I'm extremely confused about this topic.
Sulfure dioxide and oxygen react to form sulfure trioxide
2SO2 + O2 --> 2SO3 all gas
One mole of sulfur dioxide is mixed with 0.7 moles of oxygen gas in a 1L container. There is 0.8 mol of SO3 present in the container at equilibrium.
Then the question asks me to complete the ICE table to show concentrations (mol/L) of the reactants and products and then calculate the value of K for the reaction.
So inital for SO2 is 1 and initial for O2 is 0.7 and initial for SO3 is 0 and change would be 0.8.
Can you explain in full detail as i am struggling alot with this understanding how to calculate equilibrium constant. Thank you very much.

 

[jazz519]

First draw the RICE table


Then write out the equilibrium constant expression: K=[so3]^2/([so2]^2[o2])

and then sub in the values from the table for the equilibrium concentrations and write the answer to the appropriate amount of significant figures (also note there's no units for the k constant)

 

[Snowflek]

First draw the RICE table
View attachment 34002

Then write out the equilibrium constant expression: K=[so3]^2/([so2]^2[o2])

and then sub in the values from the table for the equilibrium concentrations and write the answer to the appropriate amount of significant figures (also note there's no units for the k constant)

Is the reason why SO2 is -0.8 because SO3 is+0.8 in which both have molar ratio of 2 and since oxygens molar ratio is 1 its 0.8 divide 2?

 

[jazz519]

Is the reason why SO2 is -0.8 because SO3 is+0.8 in which both have molar ratio of 2 and since oxygens molar ratio is 1 its 0.8 divide 2?

Yep, the concentration changes are proportional to the molar ratios. That's why I'd recommend you include the ratio bit in your table working too, so that you avoid silly mistakes with that

 

[Snowflek]

Yep, the concentration changes are proportional to the molar ratios. That's why I'd recommend you include the ratio bit in your table working too, so that you avoid silly mistakes with that

Thank you very much for your help!

 

[Snowflek]

Yep, the concentration changes are proportional to the molar ratios. That's why I'd recommend you include the ratio bit in your table working too, so that you avoid silly mistakes with that

If the question asks in a 5L container, do I divide the final conecntration by 5?

 

[jazz519]

If the question asks in a 5L container, do I divide the final conecntration by 5?

Yeah you would