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I'm asking this here because it's a maths type question
I seem to get a much warmer response here despite making silly mistakes:

My friend told me he is going to use 365nm light to irradiate Ibotenic acid -> Muscazone
I mentioned to him that this is barely below the visible spectrum and that you need a specific amount of energy to break bonds using photons, so I went on a little adventure:

First I had to calculate the bond energy of each species which was very tedious, I wonder if there is a better way:

Ibotenic acid:
O-H  = 2*463
C--N = 615
C-C  = 3*346
C--C = 602
C-O  = 3*358
N-O  = 201
C-N  = 2*305
C--O = 799
N-H  = 2*391  
{Total bond energy = 6037 kj/mol}

Muscazone:
O-H  = 463
C-O  = 3*358
C-C  = 3*346
C--C = 602
N-O  = 201
C-N  = 4*305
C--O = 799
N-H  = 3*391

Total Bond energy = 6570 kj/mol

$$ E=6037000-6570000=-533000j\cdot mol^{-1} $$

planck equation:

$$ \lambda = \frac{hc}{E} $$

Dimensional analysis:
$$ \lambda = \frac{j\cdot s\cdot m}{j\cdot s\cdot n} \Rightarrow \lambda = \frac{m}{n} $$


Seems that I can't get rid of n or mol so I used avogadros number:

$$ \lambda = \frac{(6.63\cdot 10^{-34}) (2.99\cdot 10^{8})}{(-533000) (6.02\cdot 10^{23})} $$

$$ \lambda = 224nm $$


Is my reasoning, maths and answer correct?

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  • $\begingroup$ The calculation of $\lambda$ is wrong. Check the calculation of $\frac{(6.63\times10^{-34})(2.99\times10^8)}{(-533000)(6.02\times10^{23})}$ $\endgroup$
    – jjagmath
    Mar 10, 2021 at 16:47
  • $\begingroup$ hmmm? help me out @jjagmath $\endgroup$
    – Nickotine
    Mar 10, 2021 at 17:32
  • $\begingroup$ $\frac{(6.63\times10^{-34})(2.99\times10^8)}{(-533000)(6.02\times10^{23})} = -6.17819\times 10^{-55}$ I have no idea how you got 224. $\endgroup$
    – jjagmath
    Mar 10, 2021 at 17:46
  • $\begingroup$ I just done the calculation again and I got -0.0000002238 so 224nm, but I done it on my phone so perhaps it's came out wrong @jjagmath $\endgroup$
    – Nickotine
    Mar 10, 2021 at 17:56
  • $\begingroup$ You are calculating $\frac{(6.63\times10^{-34})(2.99\times10^8)(6.02\times10^{23})}{(-533000)}$ $\endgroup$
    – jjagmath
    Mar 10, 2021 at 18:04

1 Answer 1

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Here are just a couple of notes that might help you understand the problem better:

  • There is no better way to calculate the total bond energy $E_{mole}$
  • Since the final energy is larger, it means that the molecules must absorb energy to transform into the final product. That's the meaning of the $-$ sign.
  • One photon is absorbed by one molecule, so you need to calculate the bond energy per molecule $E_{molecule}=E_{mole}/N_A$
  • Transform this energy to wavelength $$\lambda=\frac{hc}{E_{molecule}}=\frac{hcN_A}{E_{mole}}$$
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  • $\begingroup$ I thought the negative sign meant the reaction gives out energy? Before I do something stupid what is ${N_A}$? avogadros number? Also did I actually calculate anything at all or was it just completely wrong? @Andrei $\endgroup$
    – Nickotine
    Mar 10, 2021 at 17:28
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    $\begingroup$ @Nickotine you are right. Higher bond energy means more stable. So you probably just need to calculate the wavelength to break only certain bonds. But the chemistry of that is out of my expertise. And yes, $N_A$ is Avogadro's number $\endgroup$
    – Andrei
    Mar 10, 2021 at 18:38
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    $\begingroup$ $[h]=Js$, $[c]=m/s$, $[E_{mole}]=J/mole$, $[N_A]=1/mole$. To get Joules, you need to take $J/mole$ and multiply with $mole$, or equivalently divide by $1/mole$, so $$[\lambda]=m=\frac{Js\cdot m/s}{(J/mole)/(1/mole)}=\frac{Js\cdot m/s\cdot 1/mole}{J/mole}$$ $\endgroup$
    – Andrei
    Mar 11, 2021 at 1:17
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    $\begingroup$ ${\lambda = \frac{j\cdot s\cdot m}{j\cdot s\cdot (1/n)}}=\frac{j\cdot s}{j \cdot s}\cdot\frac{m}{\frac{1}{n}}=n\cdot m$ $\endgroup$ Mar 13, 2021 at 21:33
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    $\begingroup$ No you can't. If your age depends on how many $years$ you were alive, then it depends on it. That's it :) $\endgroup$ Mar 13, 2021 at 22:23

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