First slide

Electromagnetic radiation and spectra

Question

The longest wavelength of light capable of breaking a single (C-CI) bond in Cl₂ is if
$Cl - Cl (g) ⟶ 2 Cl (g); ΔH = 242 kJ {mol}^{- 1}$

Moderate

A

494 pm
B

494 nm
C

494 $\overset{\circ}{A}$
D

247 nm

Solution

$\begin{matrix} ΔH = 242 kJ {mol}^{- 1} \\ = 242 \times 10^{3} J {mol}^{- 1} = \frac{242 \times 10^{3} J}{N_{0}} J {molecule}^{- 1} \\ = \frac{242 \times 10^{3} J}{6 .02 \times 10^{23}} J per (Cl - Cl) bond \end{matrix}$
$Also ΔH = hv = \frac{hc}{λ}$
$\begin{matrix} ∴ λ = \frac{hc}{ΔH} = \frac{6 .62 \times 10^{- 34} Js \times 3 \times 10^{8} {ms}^{- 1} \times 6 .02 \times 10^{23}}{242 \times 10^{3} J} \\ = 4 .94 \times 10^{- 7} m = 494 \times 10^{- 9} m = 494 nm \end{matrix}$

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