Home / AP Chemistry: 3.13 Beer-Lambert Law – Exam Style questions with Answer- FRQ

AP Chemistry: 3.13 Beer-Lambert Law – Exam Style questions with Answer- FRQ

Question

Answer the following questions regarding light and its interactions with molecules, atoms, and ions.

(a) The longest wavelength of light with enough energy to break the Cl-Cl bond in \(Cl_2(g)\) is 495 nm.

(i) Calculate the frequency, in \(s^{-1}\), of the light.

(ii) Calculate the energy, in J, of a photon of the light.

(iii) Calculate the minimum energy, in kJ \(mol^{-1}\), of the Cl-Cl bond. 

(b) A certain line in the spectrum of atomic hydrogen is associated with the electronic transition in the H from the sixth energy level (n = 6) to the second energy level (n = 2). atom

(i) Indicate whether the H atom emits energy or whether it absorbs energy during the transition. Justify your answer.

(ii) Calculate the wavelength, in nm, of the radiation associated with the spectral line.

(iii) Account for the observation that the amount of energy associated with the same electronic transition (n = 6 to rz = 2) in the \(He^+\) ion is greater than that associated with the corresponding transition in the H atom.

▶️Answer/Explanation

Answer:

(a) (i) \(v=\frac{c}{\lambda}=\frac{3.00 \times 10^{17}nm/sec}{495nm}(or,=\frac{3.00 \times 10^8m/sec}{495 \times 10^{-9}m})=6.06 \times 10^{14} sec^{-1}\)
(ii) \(E=hv=(6.626 \times 10^{-34}J sec)(6.06 \times 10^{14} sec^{-1}) = 4.02 \times 10^{-19}J\)
(iii) \((4.02 \times 10^{-19}J)(6.022 \times 10^{-23}mol^{-1})(0.00100kJ/J)=242 kJ/mol\)

(b) (i) Energy is emitted.
The n = 6 state is at a higher energy than the n=2 state. Going from a high energy state to low energy state means that energy must be emitted.
(ii) \(E_2=\frac{-2.178 \times 10^{-18} J}{2^2}=-5.45 \times 10^{-19}J, E_6 = \frac{-2.178 \times 10^{-18}J}{6^2}=-6.05 \times 10^{-20}J\)
\(\Delta E = E_6 – E_2 = -6.05 \times 10^{-20}J- (-5.45 \times 10^{-19}J) = 4.84 \times 10^{-19}J)= 4.84 \times 10^{-19}J\)
OR,
\(\Delta E = 2.178 \times 10^{-18}(\frac{1}{2^2}-\frac{1}{6^2})J = 2.178 \times 10^{-18}(0.2222)J = 4.84 \times 10^{-19}J\)
\(E=\frac{hc}{\lambda} \Rightarrow \lambda = \frac{hc}{E}\)
OR,
\(v=\frac{E}{h}=\frac{4.84 \times 10^{-19}J}{6.626 \times 10^{-34}J sec}=7.30 \times 10^{14} sec^{-1}\)
\(\lambda = \frac{(6.626 \times 10^{-34}J sec)(3.00 \times 10^{17} nm sec^{-1}}{4.84 \times 10^{-19}J}=411 nm\)
OR,
\(\lambda = \frac{(6.626 \times 10^{-34}J sec)(3.00 \times 10^{17}nm sec^{-1})}{4.84 \times 10^{-19}J}=411 nm\)
(iii) The positive charge holding the electron is greater for \(He^+\), which has a 2+ nucleus, than for H with its 1+ nucleus. The stringer attraction means that it requires more energy for the electron to move to higher energy levels. Therefore, transitions form high energy state to lower states will be more energetic for \(He^+\) than for H.

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