Leaving Cert Chemistry Higher free practice questions with answers

Question 1 (50 marks) [Atomic structure, energetics, bonding and a standard solution]

(a) The visible line emission spectra of salts of two elements, X and Y, are shown. The table gives the wavelengths of the three strongest visible lines for each of three elements. (i) Explain how a line in an atomic emission spectrum is produced. Refer to the relationship Em − En = hf in your answer. (ii) Identify element X. Give a reason for your answer. (iii) Identify element Y. Give a reason for your answer. (iv) Calculate the frequency of the light of wavelength 589 nm. (c = 3.00 × 10⁸ m s⁻¹) (v) Calculate the energy of one photon of this light, and hence the energy of one mole of these photons in kJ mol⁻¹. (h = 6.63 × 10⁻³⁴ J s; Avogadro constant = 6.02 × 10²³ mol⁻¹) X Y 400 450 500 550 600 650 700 wavelength (nm) Line emission spectra of salts of X and Y.

(a) Answer: (i) Heat or an electric current excites electrons to higher energy levels; when an electron falls from a higher level (Em) to a lower level (En) it emits a photon of energy hf = Em − En, so each line has one fixed frequency. (ii) X is sodium: its lines are at 568, 589 and 616 nm. (iii) Y is strontium: its lines are at 461, 605 and 650 nm. (iv) f = c/λ = 5.09 × 10¹⁴ Hz. (v) E = hf = 3.38 × 10⁻¹⁹ J per photon; per mole 3.38 × 10⁻¹⁹ × 6.02 × 10²³ = 2.03 × 10⁵ J = 203 kJ mol⁻¹.

• Approach: link each spectral line to one electron transition, then match the line positions to the reference table, then use f = c/λ and E = hf.

• (i) Excited electrons (heat or electricity supplied) occupy higher energy levels. When an electron drops from level m to a lower level n the energy difference is released as one photon, Em − En = hf, where f is the photon's frequency. Only certain energy differences exist, so only certain frequencies, and therefore a few sharp lines, appear. 3 marks.

• (ii) The lines in spectrum X are at 568, 589 and 616 nm, which match the sodium row exactly. (iii) The lines in spectrum Y are at 461, 605 and 650 nm, which match strontium. 1 mark for each element, 1 mark for each reason.

• (iv) λ = 589 nm = 589 × 10⁻⁹ m. f = c/λ = 3.00 × 10⁸ ÷ 5.89 × 10⁻⁷ = 5.09 × 10¹⁴ Hz. 1 mark.

• (v) E = hf = 6.63 × 10⁻³⁴ × 5.09 × 10¹⁴ = 3.38 × 10⁻¹⁹ J per photon (1). For one mole multiply by Avogadro's constant: 3.38 × 10⁻¹⁹ × 6.02 × 10²³ = 2.03 × 10⁵ J mol⁻¹ = 203 kJ mol⁻¹ (1).

(b) Enthalpy changes of combustion are measured in the laboratory, and Hess's law is used to find enthalpy changes that cannot be measured directly. The standard enthalpy changes of combustion are:

carbon, −393.5 kJ mol⁻¹; hydrogen, −285.8 kJ mol⁻¹; ethanol, −1367 kJ mol⁻¹. (i) State Hess's law. (ii) Use the data to calculate the enthalpy change of formation of ethanol, C₂H₅OH(l), from its elements:

2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l) (iii) A student burned ethanol in a spirit burner under a metal can of water and calculated a heat of combustion that was smaller (less exothermic) than the value above. Give two reasons for this difference.