Sciencegeeknm

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About Sciencegeeknm

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    Quark
  • Birthday 07/14/1972

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  • Location
    Leicester, England
  • Favorite Area of Science
    Chemistry

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  1. Sciencegeeknm

    Eyring Equation

    Hi, I have another question. From the attachment you can see that at 298 k the delta G is 27.2 and at 353 it is 26.4 There is a difference of 0.8 Kcal. The lecturer says that 0.8 then gives you 0.016 and he has 16 Kcal as the delta S value. Why does 0.8 then give you 0.016? I don’t understand.
  2. Sciencegeeknm

    Eyring Equation

    Thanks! Now I understand.
  3. Sciencegeeknm

    Eyring Equation

    Thanks. Yes using brackets does work better. Once you find the rate it can be used to find the half life of the reactant. I was just wondering on the screenshot attached where the value of 0.693 comes from? Something to do with log2? If I press log and enter 5 a value near to 0.693 is displayed ?
  4. Sciencegeeknm

    Eyring Equation

    Thank you for sending the photo showing your workings. It really helped and now I can reach the right answer using your values and the ones quoted in the lecture. My mistake when doing the RHS of the equation was to divide and multiply the values all in one go. You first have to multiply the gas constant by the temperature and then divide top by bottom. 2 separate steps. You then get 45.94. Then ex 45.94 to get 1.12 ^-20 Then 6.203^12 x 1.12^-20= 0.0000000694736 After some rounding up 7.0^-8 Thanks again for your help. I will now have a better understanding when approaching other equations that come up in the lectures.
  5. Sciencegeeknm

    Eyring Equation

    I know that k is the reaction rate to be solved and kb is the Boltzman constant. The screenshot attached is from the actual lecture. I really have no idea how to use the values given to reach the answer Can you tell me how you reached the correct answer? Sorry to be a pain Thanks
  6. Sciencegeeknm

    Eyring Equation

    Do you have to use the whole equation to reach the reaction rate of 7.0 x10-8? What do You mean when you say the Boltzman constant belongs on the right hand side of the equation? Thanks.
  7. Sciencegeeknm

    Eyring Equation

    Yes it does. I am entering 1.38 ^-23 x 298 = 4.1124 ^-21 divided by 6.63 ^-34 = 6.2027^12 then pressed ex button and entered -27.2 divided by 1.987^-3 x 298 = 2.30x10-7 ?? The answer should be in per seconds. The joules cancel out so is it a kelvin to s-1 conversion? Thanks for your help so far.
  8. Sciencegeeknm

    Eyring Equation

    Thank you. I'm not sure of what steps to use on a calculator. Can you provide more detail? This is not a homework question just a lecture I watched on YouTube. Thanks.
  9. Sciencegeeknm

    Eyring Equation

    I am entering 1.38 x (EE) -23 x 298= 4.1124e-21 divide by 6.63 x 10-34= 6,202,714,932,126.697 Then -27.2 divide by 0.001987 x 298k ?? Then do you multiply one side of the equation with the other? My answers are way off.
  10. Sciencegeeknm

    Eyring Equation

    Hi hypervalant_iodine, i am entering 1.38 (EE button) -23 x 298 and then divide by planks constant. But I get a really large number. Can you send a breakdown of each calculation step you are doing to arrive at the answer? Many thanks.
  11. Sciencegeeknm

    Eyring Equation

    In a lecture I recently watched there was an example of using the Eyring equation. I have attached a picture showing the values plugged into the equation. The activation energy is 27.2 and the temperature used is 298 K. The answer given in the lecture is 7.0 x 10-8 for the reaction rate. I have tried to work this out on a calculator but can't get the answer given? Maybe I am multiplying the wrong values together. I found an online calculator tool which came to the same answer(2nd screenshot attached) Any help would be appreciated.
  12. Sciencegeeknm

    MO(CO)6

    Hi Everyone, I have been looking into the world of octahedral organometallic complexes Molybdenum hexacarbonyl is a white colour and is a d6 compound. The carbonyl ligands are classsed as strong field so would induce a large octahedral splitting of the d orbitals resulting in low spin In low spin the six electrons would completely fill up the 3 t2g orbitals and there would be no unpaired electrons I am right in thinking that because there are no unpaired electrons to make a transition up into the eg orbitals after a photon of light is added, no colour frequency of light is absorbed and therefore all colours are reflected back as white or colourless? Kind Regards Nick.
  13. Sciencegeeknm

    Metal - ligand oxidation states

    I am a bit confused as to how to assign oxidation states to metal ligand complexes. why does the titanium in Ti(NMe2)4 have an oxidation state of +4 but the platinum in Pt(NH3)4 has a +2 oxidation state? Do you need to know how many valence electrons the metal atom has first? Thanks Nick.
  14. On the attached file I have tried to work out the reaction pathway. The reaction given at the top is from a chemistry lecture I watched but no mechanism was given. In the first mechanism I just reacted one of the N20 molecules with hydroxide. The hydroxide attaching to the Nitrogen and is then deprotenated by the second OH- I now have H20 and N03- but it doesn't explain how the second N20 molecule ends up with a second electron on the Nitrogen. In the second mechanism (which I think is correct) the 2 N20 molecules join together to form the dimer N2O4. OH- attacks the anti bonding orbital of the Positive Nitrogen and the 2 electrons from the N-N bond go onto the other N20 molecule. The OH attached is now deprotenated by the second OH- to leave H20,N03- and of course a lone pair on the Nitrogen. It would be great if someone could verify the second mechanism as correct . Regards Nick
  15. Sciencegeeknm

    Solid state peptide synthesis

    Sorry, I didn't mean how does the carboxylate remove the proton from nitrogen, I was referring instead to the Tert-butyl bicarbonate molecule in the by- products box. I was wondering why the oxygen attached to the tert-butyl group deprotanates the OH and starts the decarboxylation reaction?