Calculating Wavelength Required To Break Bonds Using The Photon Energy Equation: ${λ = \frac{hc}{E}}$

Question

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

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

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

$$ \lambda = 224nm $$

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Is my reasoning, maths and answer correct?

Answer 1

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 $$\lambda=\frac{hc}{E_{molecule}}=\frac{hcN_A}{E_{mole}}$$