Consider the gas-phase reaction
Xe + 2F2 ⇌ XeF4
a) Calculate the equilibrium constant from the observation that at some temperature, the extent of reaction is 50% when 0.20 mol of Xe and 0.40 mol of F2 have been mixed in an empty 1.00-L-bulb.
b)How many moles of F2 would have to be added to the equilibrium mixuture of part (a) in order to increase the conversion of Xe to XeF4 to 80%?
Equilibrium
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- Dhamnekar Winod
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Equilibrium
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- Dhamnekar Winod
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Re: Equilibrium
\((a)K= \frac{[XeF_4]}{[Xe][F_2]^2}= \frac{(0.10)}{(010)(0.20)^2}=25,\)
(b) 80% of 0.20 moles of Xe= 0.80 × 0.20 =0.16 mol XeF4
\(\frac{(0.16)}{(0.040)(0.20 + x -0.12)^2}=25\)
x=0.32 mol F2
(b) 80% of 0.20 moles of Xe= 0.80 × 0.20 =0.16 mol XeF4
\(\frac{(0.16)}{(0.040)(0.20 + x -0.12)^2}=25\)
x=0.32 mol F2
Last edited by Dhamnekar Winod on Wed Mar 30, 2022 3:51 am, edited 1 time in total.
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Re: Equilibrium
0,16/(0,04*(0,2+2x)^2) = 25
Solve for x
Solve for x
- Dhamnekar Winod
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Re: Equilibrium
But author gave the answer 0.32 mol of F2 which is correct as per my calculations.
As per your calculation x=0.1 mol of F2
As per your calculation x=0.1 mol of F2
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Re: Equilibrium
But where does 0,12 comes from in your calculation.
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Re: Equilibrium
From ICE table, I found that we have to deduct 0.12 mol from 0.2 mol of F2 and 0.06 mol from 0.10 mol of Xe. That's why my answer matches with author's answer.
Last edited by Dhamnekar Winod on Wed Mar 30, 2022 6:08 am, edited 1 time in total.
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Re: Equilibrium
????and 0.6 mol from 0.10 mol of Xe.
But answer is correct.
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Re: Equilibrium
That's a typographical mistake. I corrected it in my post.
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]