Equilibrium

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Dhamnekar Winod
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Equilibrium

Post by Dhamnekar Winod »

The reaction of sulfur dioxide gas with oxygen gas to form sulfur trioxide gas is an important step in the production of sulfuric acid.

a) Write the expression for the equilibrium constant.

b)If a 10.0 L vessel contains 15 mol of SO3, 2.0 mol of O2, and 3.0 mol of SO2 at equilibrium at 1373 K, what is the numerical value for the equilibrium constant?

c) How many moles of sulfur dioxide must be forced into reaction vessel to increase the concentration of SO3 to 1.6 mol/L?

I am working on this question.

Any chemistry help will be accepted.
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
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Dhamnekar Winod
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Re: Equilibrium

Post by Dhamnekar Winod »

(a)\(K= \frac{[SO_3]^2}{[SO_2]^2[O_2]}\)

(b)\(\frac{(\frac{15 mol}{10 L})}{(\frac{3 mol}{10 L})^2(\frac{2 mol}{10 L})}=125 \)

(c) I am working on this question.
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Re: Equilibrium

Post by ChenBeier »

a) square at SO3 missing
b) correct

c) (1,6 mol/l)^2 / (0,3+x)^2* (0,2-0,5x) =125
Solve for x
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Dhamnekar Winod
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Re: Equilibrium

Post by Dhamnekar Winod »

My answer to (c) is different than your answer.

My attempt to answer (c) =\(K=\frac{[SO_3]^2}{[SO_2]^2[O_2]}= \frac{ (1.6)^2}{(0.3 + 0.10x)^2(0.2-0.05)} \Rightarrow x=0.695041722823 mol SO_2/L\)
∴ (0.69 mol SO2/L)(10 L)= 6.9 mol of SO2

Is this answer correct? We have to add SO2.
Last edited by Dhamnekar Winod on Wed Mar 30, 2022 4:55 am, edited 2 times in total.
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Re: Equilibrium

Post by ChenBeier »

Who say that I deduct SO2 .
(0,3 +x) is an Addition
Also in your equation the square is missing in (0,3 + 0,1×).
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Re: Equilibrium

Post by Dhamnekar Winod »

I made appropriate changes in my posts as per your suggestions.

Now in answer to (c), There is an increase of 0.1 mol on SO3. As per chemical reaction 2SO2 + O2 →2 SO3, 50% of 0.10 mol increase in SO3 will be deducted from 0.20 mol of O2 and x is the proportion of an increase of 0.10 mol in SO3.
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Re: Equilibrium

Post by ChenBeier »

I am not sure you can use 0.1 for the SO2. If no Addition then 0.1 SO3 requires 0.1 SO2 of the existing amount. But if SO2 should be added then the 0.1 is taken of the Reservoir and the additional amount x = SO2 exists + SO2 New. But it could also nothing is taken and the loss is refilled so the amount doesnt change. Or its only an Addition. Between these is the trues.

I think it is similar to the XeF4 example.
We increase SO3 at 0,1 mol , what means the SO2 would be deducted by 0,1 mol, but we should add SO2 . (0,3-0,1 + x)^2

My result 0,169 mol
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Re: Equilibrium

Post by Dhamnekar Winod »

Correct answer, thanks.
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