Chemistry Labs

Problem 1

The dissociation of molecular chlorine is an endothermic process, ΔH=243.6\Delta H = 243.6 kJ mol−1^{-1}. The dissociation can also be attained by the effect of light. 1.1 At what wavelength can the dissociating effect of light be expected? 1.2 Can this effect also be obtained with light whose wavelength is smaller or larger than the calculated critical wavelength? 1.3 What is the energy of the photon with the critical wavelength? When light that can effect the chlorine dissociation is incident on a mixture of gaseous chlorine and hydrogen, hydrogen chloride is formed. The mixture is irradiated with a mercury UV lamp (λ=253.6\lambda = 253.6 nm, power input 10 W). An amount of 2 % of the energy supplied is absorbed by the gas mixture in a 10 litre vessel. Within 2.5 seconds of irradiation, 65 millimoles of HCl are formed. 1.4 How large is the quantum yield (the number of product molecules per absorbed photon)? 1.5 How can the value obtained be qualitatively explained? Describe the reaction mechanism.
Step 1 of 5: Critical wavelength
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Intuition

One photon must carry at least the energy needed to break one mole of Cl–Cl bonds divided by Avogadro's number.

λc=NAhcΔH=6.02×1023×6.6×10−34×3×1082.436×105=4.91×10−7 m=491 nm\lambda_c = \dfrac{N_A h c}{\Delta H} = \dfrac{6.02\times 10^{23}\times 6.6\times 10^{-34}\times 3\times 10^{8}}{2.436\times 10^{5}} = 4.91\times 10^{-7}\ \text{m} = 491\ \text{nm}
Analysis

Set the photon energy equal to the bond energy per molecule: hν=ΔH/NAh\nu = \Delta H/N_A, so λc=NAhc/ΔH≈491\lambda_c = N_A h c/\Delta H \approx 491 nm. Any photon with λ<491\lambda < 491 nm is even more energetic and can also dissociate ClX2\ce{Cl2}; longer-wavelength photons are too weak.

Common pitfall. Divide ΔH\Delta H (per mole) by NAN_A before comparing with hνh\nu — comparing a molar energy directly to a photon energy gives a wavelength a factor 102310^{23} too small.