I have been having trouble solving the following problem. The question calls for an estimation of the energy (in kj/mol) required to dissociate ClO in its excited state when it is excited from the v = 0 of the ground state.

My first approach in solving this problem was selecting two transitions (v" = 2 to v' = 7 and v" = 3 to v' = 8). I feel as though my error is in this part of the problem as the question asks for a transition from v = 0 but as far as I can see in the diagram there is no peak for the v = 0 transition.

Following this, I converted the wavelengths for each transition into wavenumbers and then subtracted the wavenumber corresponding to each transition:

$$\nu_{7'0"} = \frac{10^7}{287\ \mathrm{nm}} \approx 34843.205\ \mathrm{cm}^{-1}$$

Then, subtracted $308\ \mathrm{nm}$ in wavenumbers from the above value to give $2375.68\ \mathrm{cm}^{-1}$ for the transition.

Then using the above numbers I used the following formula to find $\omega_e$ and $\omega_ex_e$.

$$G\left(v'\right) = \left(v' + \frac{1}{2}\right)\omega_e - \left(v' + \frac{1}{2}\right)^2 \omega_e x_e$$

Then with $\omega_e$ and $\omega_ex_e$ I used the following formula to find the dissociation energy, $D_e$,

$$D_e = {\omega^2 \over 4\omega_ex_e}$$

Then subtracting the zero-point energy from $D_e$ to get $D_0$ and converting this value into kj/mol just to get an incorrect answer. If anyone could point out where I'm going wrong that would be greatly appreciated.

Spectrum included in the question


1 Answer 1


Quiz 3 has been annoying. From the graph, we know that the molecule dissociates at a wavelength of ~262.5 nm. We know this as this is when there are no more transitions, as indicated by the completely flat line. Using the equation E = (h * c)/ν and substituting in the relevant values, we get E = 7.57E-19 J. Multiplying by Avogadro's number nets us ~456000 J/mol = 456 kJ/mol.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.