confusedmind09
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hi please can someone talk me through how to answer this? thank you!

(b) The second step in the manufacture of sulfuric acid is the conversion of SO2 into sulfur trioxide, SO3, using Equilibrium 1.
2SO2(g) + O2(g) 2SO3(g)      ∆H = −197 kJ mol−1 Equilibrium 1 An industrial chemist carries out some research into Equilibrium 1.
• The chemist fills a 10.2 dm3 container with SO2(g) at RTP, and then adds 12.0g of O2(g).
• The chemist adds the vanadium(V) oxide catalyst, and heats the mixture. The mixture is allowed to reach equilibrium at a pressure of 2.50atm and a temperature of 1000K.
• A sample of the equilibrium mixture is analysed, and found to contain 0.350 mol of SO3.

(ii) Determine the value of Kp for Equilibrium 1 at 1000K.
Show all your working.
Give your answer to 3 significant figures.
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turkeydinosaur16
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(Original post by confusedmind09)
hi please can someone talk me through how to answer this? thank you!

(b) The second step in the manufacture of sulfuric acid is the conversion of SO2 into sulfur trioxide, SO3, using Equilibrium 1.
2SO2(g) + O2(g) 2SO3(g)      ∆H = −197 kJ mol−1 Equilibrium 1 An industrial chemist carries out some research into Equilibrium 1.
• The chemist fills a 10.2 dm3 container with SO2(g) at RTP, and then adds 12.0g of O2(g).
• The chemist adds the vanadium(V) oxide catalyst, and heats the mixture. The mixture is allowed to reach equilibrium at a pressure of 2.50atm and a temperature of 1000K.
• A sample of the equilibrium mixture is analysed, and found to contain 0.350 mol of SO3.

(ii) Determine the value of Kp for Equilibrium 1 at 1000K.
Show all your working.
Give your answer to 3 significant figures.
1. find the equilibrium moles of each reactant and product using the fact that 0.350 of SO3 is at equilibrium
2. convert these to partial pressures using molar fractions and the 2.5 atm in the question
3. sub into Kp expression
Hope that helps
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confusedmind09
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(Original post by turkeydinosaur16)
1. find the equilibrium moles of each reactant and product using the fact that 0.350 of SO3 is at equilibrium
2. convert these to partial pressures using molar fractions and the 2.5 atm in the question
3. sub into Kp expression
Hope that helps
thank you for your reply! but this is going to sound so stupid, how to I work out the moles of the reactant from 0.350? would i work out the initioal moles by using the 12.0g of oxygen? thank you
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Pigster
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(Original post by confusedmind09)
thank you for your reply! but this is going to sound so stupid, how to I work out the moles of the reactant from 0.350? would i work out the initioal moles by using the 12.0g of oxygen? thank you
This was a tough question and candidates made a range of mistakes answering it and the marking team didn't find it easy to mark.

You will no doubt have been taught to do an ICE table (or equivalent) and yes, you need to work out how many mol of SO2 you start with (using 24) and the amount of mol of O2 (using m/Mr).
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confusedmind09
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thank you!
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m.elle_eni
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Hi what paper was this from? confusedmind09
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trusfratedblink
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hey if you don't mind me asking, which paper is this question from?
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Pigster
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(Original post by trusfratedblink)
hey if you don't mind me asking, which paper is this question from?
It is from one of the papers that OCR have not yet made public.
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trusfratedblink
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(Original post by Pigster)
It is from one of the papers that OCR have not yet made public.
could you please send it to me?
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Pigster
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(Original post by trusfratedblink)
could you please send it to me?
If OCR are not making it available publicly, then neither will I.
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DrewWizKid24
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(Original post by confusedmind09)
hi please can someone talk me through how to answer this? thank you!

(b) The second step in the manufacture of sulfuric acid is the conversion of SO2 into sulfur trioxide, SO3, using Equilibrium 1.
2SO2(g) + O2(g) 2SO3(g)      ∆H = −197 kJ mol−1 Equilibrium 1 An industrial chemist carries out some research into Equilibrium 1.
• The chemist fills a 10.2 dm3 container with SO2(g) at RTP, and then adds 12.0g of O2(g).
• The chemist adds the vanadium(V) oxide catalyst, and heats the mixture. The mixture is allowed to reach equilibrium at a pressure of 2.50atm and a temperature of 1000K.
• A sample of the equilibrium mixture is analysed, and found to contain 0.350 mol of SO3.

(ii) Determine the value of Kp for Equilibrium 1 at 1000K.
Show all your working.
Give your answer to 3 significant figures.
Hey, was the answer 27.2?
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trusfratedblink
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(Original post by Khalid Elmi)
Hey, was the answer 27.2?
yes it was 27.2
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zeeboyeebo
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(Original post by trusfratedblink)
yes it was 27.2
How did u work this out? as i got 1.4^2 / 0.8 x 0.3^2 and ended up with 10.2 as my answer
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Pigster
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(Original post by zeeboyeebo)
How did u work this out? as i got 1.4^2 / 0.8 x 0.3^2 and ended up with 10.2 as my answer
In that case, you're just rubbish as using your calculator.

Have another go. Perhaps use brackets?
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YetiSpaghetti
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(Original post by Pigster)
In that case, you're just rubbish as using your calculator.

Have another go. Perhaps use brackets?
I have gotten 272.222... and I am unsure of why I am off by x10. For my final calculation I got 0.14^2 / 0.03^2 x 0.08; can you explain where I went wrong please?
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Pigster
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(Original post by YetiSpaghetti)
I have gotten 272.222... and I am unsure of why I am off by x10. For my final calculation I got 0.14^2 / 0.03^2 x 0.08; can you explain where I went wrong please?
All your partial pressures are out by a factor of ten.
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YetiSpaghetti
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So was it because my mole calculations were off at equilibrium: 0.075 of SO2, 0.200 of O2 and 0.350 of SO3 - and how would I correct this problem? Or was it my mole fraction calculations where I divided all these values by 2.5 and got 0.03, 0.08, 0.14? Thanks for the help btw
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