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I have that $\pu{100 ml}$ of an aqueous solution containing $\pu{1.0 mM}$ $\ce{H2O2}$, $\pu{100 mM}$ $\ce{CH3OH}$ and $\pu{200 mM}$ DMSO are exposed to monochromatic light with a wavelength of $300$ nm. The incident light intensity is $\pu{2mW/cm2}$ and the exposed area is $\pu{4cm2}$. The absorbance of the solution at $300$ nm (which depends only on $\ce{H2O2}$) was measured to be $0.25$. Only the homolysis of $\ce{H2O2}$ to two hydroxyl radicals needs to be taken into consideration. The hydroxyl radicals formed react with $\ce{CH3OH}$ at a rate constant of $\pu{1.0E8M^{-1}s^{-1}}$ and with DMSO at a rate constant of $\pu{2.0E9M^{-1}s^{-1}}$. The product of the reaction with $\ce{CH3OH}$ is formaldehyde ($\ce{CH2O}$). The concentration of formaldehyde as a function of reaction time was measured and is given below. Based on the data below, determine the quantum yield for photoinduced homolysis of $\ce{H2O2}$. $$ \begin{array}{rl} \hline \textrm{time}/\pu{h} & [\ce{CH2O}]/\pu{M} \\ \hline 0 & 0 \\ 1 & \pu{7.7E-7} \\ 2 & \pu{1.5E-6} \\ 3 & \pu{2.3E-6} \\ 4 & \pu{3.1E-6} \\ 5 & \pu{3.8E-6} \\ \hline \end{array} $$

I hope this isn't closed off as "off-topic" since it is quite a long question but I genuinely don't know where to start on this. I feel that there is so much information and I can't quite determine what is useful and what has just been added to confuse us.

I wanted to start by writing all the reactions and then use the steady-state assumption to determine the rate equation for $\ce{CH2O}$ but even there I got stuck. From what I have understood two of the reactions are:

$$\ce{H2O2 -> 2OH-}$$

$$\ce{2OH- + CH3OH -> CH2O + 2 H2O}$$

But I don't know how the third reaction with DMSO will look like and can't seem to find anything online. I don't know if the way I am trying to solve this problem is the correct way since I don't really know if I am supposed to solve for $\ce{CH2O}$ and then plot it to get something from the plot so I would truly appreciate if someone could push me in the right direction or link helpful examples or other useful information that might help me solve this. Thank you!

EDIT

With your help I have tried to being answering the problem but I come across some things that I don't know how to tackle. I began with calculating that the incident light intensity should be $2$ mW/cm$^2$ * $4$cm$^2$ = $8$ mW. By using Lambert-Beers law I calculated the intensity of the outgoing light as:

$$log\frac{I_0}{I} = A$$

$$I = \frac{I_0}{10^A} = \frac{8mW}{10^{0.25}} = 4.49 = 4.5 mW$$

Then the amount of absorbed light is: ΔI = I$_0$ - I = $8 - 4.5 = 3.5 mW$

So from this I can now calculate the denominator of the quantum yield fraction (I call it b for amount of absorbed photons) from the equation:

$$b = \frac{absorbed light}{energy per mol photons} = \frac{ΔIλ}{hcN_A} = \frac{3.5*10^{-3}Js^{-1}* 300 * 10^{-9}m}{6.625*10^{-34}Js^{-1}*3*10^8ms^{-1}*6.02*10^{23}mol}$$ $$= 8.78*10^{-9} smol^{-1}$$

I don't think this is the correct answer since the units don't add up. Should I maybe take the intensity times the number of hours ($5$h) to get the intensity only in Joules?

Then if I write out the rate equations:

$$\frac{d[CH_2O]}{dt}=k_2[OH^*]^2[CH_3OH]$$

$$\frac{d[OH^*]}{dt}=k_1[H_2O_2]-k_2[OH^*]^2[CH_3OH] → [OH^*]^2 = \frac{k_1[H_2O_2]}{k_2[CH_3OH]}$$

which then gives:

$$\frac{d[CH_2O]}{dt}=k_1[H_2O_2]$$

I don't really understand what use I have of this in order to calculate the number of molecules of the reactant which is used..

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  • $\begingroup$ Well, not writing anion instead of radical would be a start... BTW why do you know how to make a table and not about mhchem or mathrm? $\endgroup$
    – Mithoron
    May 14 at 17:30
  • $\begingroup$ This is a messy problem. As to DMSO as long as it does not produce formaldehyde it reduces this yield by a lot $10^8/(10^8+20\cdot10^8)$. Next the product yield is tiny anyway so $[H_2O_2],[CH_3OH],[DMSO]$ are effectively constant. You know from absorbance and Beers law what fraction of energy is absorbed, you should assume each photon breaks a bond. Write down the rate equations. Ignore product from DMSO. You will need to integrate to get total yield as I assume illumination is continuous. $\endgroup$
    – porphyrin
    May 14 at 18:04
  • $\begingroup$ If DMSO interfere I think it should also be counted. The quantum yield would be the practical one on the products of interest. How many of the product formaldehyde / how many photons. $\endgroup$
    – Alchimista
    May 15 at 11:08
  • $\begingroup$ @porphyrin Thank you for your help! I have added some of my new calculations and questions to my original question under edit. I would truly appreciate it if You could take a look. $\endgroup$
    – confused
    May 15 at 12:37
  • $\begingroup$ You are at steady state so $d[OH]/dt =2 k_{abs}[H2O2] - k_2[OH][MeOH]=0$. You get $k_{abs}$ from the number of photons/sec converted into 'moles of photons'/sec (Einsteins?) and you can then use data in table. You can assume pseudo first order as the amount of OH produced is tiny compared to other species. $\endgroup$
    – porphyrin
    May 18 at 7:27

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