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    (Original post by reneetaylor)
    Otrivine, what is U?
    It is Uranium
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    (Original post by otrivine)
    It is Uranium
    Man. OK, feel pretty silly now.
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    (Original post by reneetaylor)
    Man. OK, feel pretty silly now.
    No, dont be silly its fine
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    (Original post by otrivine)
    No, dont be silly its fine
    Haha, just had a no brainer!

    This page was really helpful by the way!
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    (Original post by reneetaylor)
    Haha, just had a no brainer!

    This page was really helpful by the way!
    You know I saw the mark scheme for our paper in Jan 2013, the answers looked so easy! I felt so upset
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    (Original post by otrivine)
    what you do is 77.6/100 x 17.6=13.something , this is the equilibrium amount/mole of the acid.

    then you use this value for substarting the initial amount of CH3OH and CO. then get concentration and sub the values into the Kc expression.
    thank you very much..i got it now.
    can you help me with one last 1?

    a) MnO4-+ U3+ → Mn2+ + UO2+ in aqueous acid (H+ present)conditions
    b) Cr2O72-+ C2O42- → CO2 + Cr3+ inaqueous acid (H+ present) solutions
    Balance the redox equation
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    (Original post by kayd)
    thank you very much..i got it now.
    can you help me qith one last 1?

    a) MnO4-+ U3+ → Mn2+ + UO2+ in aqueous acid (H+ present)conditions
    b) Cr2O72-+ C2O42- → CO2 + Cr3+ inaqueous acid (H+ present) solutions
    Balance the redox equation
    I can do it tomorrow for you! this is very long!
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    (Original post by otrivine)
    I can do it tomorrow for you! this is very long!

    Already done the half equations for each...just not sure on how to put them together.....Thank you
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    (Original post by kayd)
    thank you very much..i got it now.
    can you help me with one last 1?

    a) MnO4-+ U3+ → Mn2+ + UO2+ in aqueous acid (H+ present)conditions
    b) Cr2O72-+ C2O42- → CO2 + Cr3+ inaqueous acid (H+ present) solutions
    Balance the redox equation

    Already done the half equations for each...just not sure on how to put them together
    Well if you have already done the half-equations all you need to do is make the electrons equal by judicious multiplication of each equation and then add the equations together ...
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    (Original post by charco)
    Well if you have already done the half-equations all you need to do is make the electrons equal by judicious multiplication of each equation and then add the equations together ...
    Thank you
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    (Original post by otrivine)
    A 25.0 cm3 sample of a solution containing both Fe2+ and Fe3+ ions was acidified and titrated with 0.022 mol dm-3 manganate (VII) solution (containing the MnO4- ion) requiring 15.0 cm3. A second sample of the same solution was treated with excess Zn to reduce all the Fe3+ to Fe2+ and then, after filtering, was titrated against the same permanganate solution requiring 19.0 cm3.

    In the first titration only the Fe2+ reacts with acidified MnO4-. Use this information, and a balanced net overall redox reaction, to calculate the concentration of Fe2+ in the solution.


    my answer is 0.066 mol dm-3. I got thi by using the equation
    8H+ + Mno4- + 5fe2+ ----> Mn2+ + 5Fe3+ + 4h2o

    are my correct?
    Please canyou explain how you get 0.066 moles?
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    (Original post by otrivine)
    the first one uses 5dm3 the second one does not or else they would have put it down?
    how did you get this figure 2.91. i got 1.52. but cant seem to get that figure. not sure of working out. can someone help please.
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    (Original post by 123superpower)
    Please canyou explain how you get 0.066 moles?

    can you explain this for me and also Question five Balancing redox equation, am so stuck and our assignment is due today.
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    (Original post by ZakRob)
    A 25.0 cm3 sample of a solution containing both Fe2+ and Fe3+ ions was acidified and titrated with 0.022 mol dm-3 manganate (VII) solution (containing the MnO4- ion) requiring 15.0 cm3. A second sample of the same solution was treated with excess Zn to reduce all the Fe3+ to Fe2+ and then, after filtering, was titrated against the same permanganate solution requiring 19.0 cm3.

    b) In the first titration only the Fe2+ reacts with acidified MnO4-. Use this information, and a balanced net overall redox reaction, to calculate the concentration of Fe2+ in the solution.

    0.066 moldm-3

    After all the Fe3+ has been converted to Fe2+ it is then titrated again with acidified MnO4-. Calculate the concentration of Fe2+ at this point and use this value, and your answer to b), to calculate the concentration of Fe3+ present in the solution before the zinc was added.

    0.0176moldm-3

    Do you want an explanation?

    Yes i want an explanation of how you got theses answers.
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    (Original post by Buduu)
    Can you help me question 3 and show me step by step please, I am struggling. thank you.


    3.Ethanoic acid can be manufactured by the followingreaction, which is carried out between150 °C and 200 °C.
    CH3OH(g)+ CO(g) CH3COOH(g)

    a) Amixture 17.6 moles of methanol and 19.6 moles of carbon monoxide is allowed toreach equilibrium at 175°C in a container with volume 5 dm3. It wasfound that 12.2 moles of ethanoic acid had been formed.
    Write down anexpression for Kc for this reaction


    i) Calculate the concentrations of methanol andcarbon monoxide present at equilibrium. Show your working.
    (4 marks)
    ii) Hence calculate Kc (to 3significant figures) including its units
    (2 marks)

    b) How,if any, would addition of extra carbon monoxide affect the following? Justifyyour answers.

    i) The value of the equilibrium constant forformation of ethanoic acid
    (2 marks)
    ii) The equilibrium yield of ethanoic acid.

    c) Anothersample containing 17.6 moles of methanol and 19.6 moles of carbon monoxide wasallowed to reach equilibrium, but at a lower temperature. This time it wasfound that 77.6% of methanol had reacted.
    (i)Calculate the valueof Kc at the lower temperature. Give your answer to 3 significantfigures
    (ii) Use your answers to aii) and bi) to decide whether the reaction was exothermicor endothermic. Explain your answer



    Hellp i dont get this qustion
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    (Original post by ZakRob)
    It is exothermic. I say this now because I worked out the second Kc value using the volume of 5dm3, which gave a Kc value of 2.91 this compared to the Kc value of 1.52 at a higher temp shows that the exothermic side is favored at lower temperature, and this occurred with the increase of product

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    I'm having difficulties with the Question 3 ii Could you explain to me how to calculate it Cause they've given all the moles so I tried to divide it by the 5dm3 to get the concentration for each products and reactants and then plot it in the Kc expression to get the Kc at equilibrium But I didn't get the value 1.52 so I must be doing something wrong
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    (Original post by Sheba1)
    I'm having difficulties with the Question 3 ii Could you explain to me how to calculate it Cause they've given all the moles so I tried to divide it by the 5dm3 to get the concentration for each products and reactants and then plot it in the Kc expression to get the Kc at equilibrium But I didn't get the value 1.52 so I must be doing something wrong
    are you just using the moles given in the question to calculate concentration? or are you calculating moles at equilibrium first?
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    (Original post by ZakRob)
    A 25.0 cm3 sample of a solution containing both Fe2+ and Fe3+ ions was acidified and titrated with 0.022 mol dm-3 manganate (VII) solution (containing the MnO4- ion) requiring 15.0 cm3. A second sample of the same solution was treated with excess Zn to reduce all the Fe3+ to Fe2+ and then, after filtering, was titrated against the same permanganate solution requiring 19.0 cm3.

    b) In the first titration only the Fe2+ reacts with acidified MnO4-. Use this information, and a balanced net overall redox reaction, to calculate the concentration of Fe2+ in the solution.

    0.066 moldm-3

    After all the Fe3+ has been converted to Fe2+ it is then titrated again with acidified MnO4-. Calculate the concentration of Fe2+ at this point and use this value, and your answer to b), to calculate the concentration of Fe3+ present in the solution before the zinc was added.

    0.0176moldm-3

    Do you want an explanation?
    Hiya I'm doing the same assignment Will you be able to explain how you calculated this as I'm stuck
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    (Original post by bahonsi)
    Hellp i dont get this qustion
    (Original post by Sheba1)
    I'm having difficulties with the Question 3 ii Could you explain to me how to calculate it Cause they've given all the moles so I tried to divide it by the 5dm3 to get the concentration for each products and reactants and then plot it in the Kc expression to get the Kc at equilibrium But I didn't get the value 1.52 so I must be doing something wrong


    So the question was:

    3.Ethanoic acid can be manufactured by the followingreaction, which is carried out between150 °C and 200 °C.
    CH3OH(g)+ CO(g) CH3COOH(g)

    a) Amixture 17.6 moles of methanol and 19.6 moles of carbon monoxide is allowed toreach equilibrium at 175°C in a container with volume 5 dm3. It wasfound that 12.2 moles of ethanoic acid had been formed.
    Write down anexpression for Kc for this reaction


    i) Calculate the concentrations of methanol andcarbon monoxide present at equilibrium. Show your working.
    (4 marks)
    ii) Hence calculate Kc (to 3significant figures) including its units


    Name:  DSC_0263.jpg
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    Firstly draw a table like the one above, with a column for each component in the equation. and three rows for, mol at start, General equation, mol at Equilibrium.

    Name:  DSC_0264.jpg
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    Secondly write down the information in a table like the one in the picture. so now we notice we have one piece of info for equilibrium and two for the start.

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    Now write the general equation. so in our equation its a 1 to 1 to 1 ratio. so for methanol to go from starting mol to equilibrium mol, some of it is used up. and because its 1 to 1, 1x is used up. so the amount of mol at equilibrium is n - x. and x = the amount of mol for a product.
    Name:  DSC_0266.jpg
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    So now you can use the equations do a few sums. and we now have amount of mol at equilibrium.

    Name:  DSC_0267.jpg
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    now to calculate concentration, divide the number of mol by volume.

    Name:  DSC_0268.jpg
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    and finally to calculate Kc, we can use the calculated concentrations in a Kc equation.
 
 
 
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