Sampling hw #2

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Funmi Baruwa
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Joined: Wed Sep 30, 2020 9:50 pm

Sampling hw #2

Postby Funmi Baruwa » Mon Jan 11, 2021 9:01 am

Hey can someone explain how to go about finding the other concentrations inn samplinh hw #2?

Here is the question for reference:

At a certain temperature, 0.720 mol SO3 is placed in a 2.50 L container.

2SO3(g)↽−−⇀2SO2(g)+O2(g)

At equilibrium, 0.180 mol O2 is present. Calculate Kc.

Rachel Jiang 3H
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Re: Sampling hw #2

Postby Rachel Jiang 3H » Mon Jan 11, 2021 9:11 am

You need to first calculate the initial concentrations of SO3 and find the value of "x". It is best to do an ICE table for this problem.

You divide 0.720 mol SO3 by 2.50L to get the initial concentration of SO3, which is 0.288mol.L-1. Since the problem tells you that 0.180 mol O2 is present at equilibrium, you divide 0.180mol by 2.50L to find the equilibrium concentration for O2, which is 0.0720mol.L-1. The "x" value is the same as the equilibrium concentration for O2. SO2 has a coefficient of 2 while O2 has a coefficient of 1 so you can multiply the equilibrium concentration of O2 by 2 to find the equilibrium concentration of SO2. The equilibrium concentration of SO3 will be 0.288mol.L-1 minus 2x. Lastly, you'll just plug in all the values into the Kc expression. Hope this helps!
Last edited by Rachel Jiang 3H on Mon Jan 11, 2021 9:12 am, edited 1 time in total.

sophie esherick 3H
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Re: Sampling hw #2

Postby sophie esherick 3H » Mon Jan 11, 2021 9:11 am

I would create an ice table with the information you are given after converting all the values into molarities. You are given that you initially have .720 mol of SO3 in a 2.5 L container so you would want to find the molarity of that reactant. To do so, you can use the equation M=moles/volume(L) which would give you .720mol/2.5L = .288M. This would be your Initial value to go into the ice table under 2SO3. Similarly, you can also find the value of the equilibrium concentration for O2 given that there is .180 mol O2 present at equilibrium. This means its concentration at equilibrium would be .180/2.5 = 0.072M. This will also be your x value because you started with an initial concentration of 0 for O2. To find the equilibrium concentrations of SO3 and SO2, you must subtract the value of 2x (x being 0.0702M) from the SO3 initial concentration and you must add the value of 2x to the initial concentration of SO2 (which is 0 since the problem did not say you started with any moles of SO2).
I hope this helped!

Funmi Baruwa
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Re: Sampling hw #2

Postby Funmi Baruwa » Mon Jan 11, 2021 12:46 pm

why would we subtract 2x from .288

Hope Fan 2A
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Re: Sampling hw #2

Postby Hope Fan 2A » Mon Jan 11, 2021 12:58 pm

The explanations above are setting the change in concentration of O2 equal to x. Since the reactant SO3 has a stoichiometric coefficient of 2, while O2 has a stoichiometric coefficient of 1, the change in concentration of SO3 would be -2x (it is negative because SO3 is a reactant and so it would get used as the reaction progresses). The equilibrium concentration of SO3 would then be the initial concentration plus the change in concentration, and so it would be .288-2x.

abby hyman
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Re: Sampling hw #2

Postby abby hyman » Mon Jan 11, 2021 1:26 pm

As explained above, the chart with initial concentration, change in concentration, and equilibrium concentration is very helpful for this problem

Funmi Baruwa
Posts: 108
Joined: Wed Sep 30, 2020 9:50 pm

Re: Sampling hw #2

Postby Funmi Baruwa » Mon Jan 11, 2021 1:28 pm

got it thank you!


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