IChO 2002 · Theory problem 8

Lighting Lamps

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Theme IV - Chemistry Related to Light and Energy

Chemistry plays a major role in meeting our needs of light and energy. Our life is unthinkable without artificial light and energy for mobility.

PROBLEM 8

LIGHTING LAMPS

Since 1891 lighting lamps have been manufactured in The Netherlands. The improvement today in comparison to the first lamp is enormous, especially with the introduction of the gas discharge lamps. The life-time has increased by orders of magnitude. The colour is also an important aspect. Rare earth metal compounds like \(\ce{CeBr3}\) are now included to reach a colour temperature of 6000 K in the lamp. These compounds are ionic solids at room temperature, and upon heating they sublime partially to give a vapour of neutral metal halide molecules. To achieve a high vapour pressure, the sublimation enthalpy should be as low as possible.

8.1 Give a thermochemical cycle (Law of Hess) for sublimation of \(\ce{CeBr3}\), via a vapour of mononuclear ions. (\(H_{\mathrm l}=H_{\mathrm{lattice}}\); \(H_{\mathrm e}=H_{\mathrm{electrostatic}}\); \(H_{\mathrm s}=H_{\mathrm{sublimation}}\); \(H\) is not absolute, \(H\) means \(\Delta H\))

Figure 1

The lattice energy of the solid can be calculated using the Born–Landé formula:

\[ H_{\mathrm l} =f\frac{Z_+Z_-Ae^2}{r_++r_-}\left(1-\frac1n\right) \]

The factor \(fe^2\) (necessary in order to calculate the lattice energy in \(\mathrm{kJ\,mol^{-1}}\)) amounts to 139 when the ionic radii are substituted in nm. The Madelung constant \(A\) for the lattice is 2.985. The Born exponent \(n\) is 11. The charges of the ions \(Z_+\) and \(Z_-\) are integer numbers (\(Z_-\) is negative). For the calculation of the energy of gaseous \(\ce{CeBr3}\) (when formed from ions) the same Born-Landé formula can be used without \(A\). The structure of \(\ce{CeBr3}\) in the gas phase is planar triangular. The radius of \(\ce{Ce^3+}\) is 0.115 nm and of \(\ce{Br^-}\) is 0.182 nm.

8.2 Calculate the enthalpy of sublimation of \(\ce{CeBr3}\) (in integers; be aware of the signs!)

Attempts to make a better lamp have been undertaken by adding a stoichiometric amount of \(\ce{CsBr}\) to the \(\ce{CeBr3}\) in the lamp leading at room temperature to solid \(\ce{CsCeBr4}\). When the sublimation temperature decreases the life time of the lamp will increase likewise. The \(\ce{CsCeBr4}\) lattice has a \(\ce{NaCl}\) structure with \(\ce{Cs+}\) as cations and tetrahedral \(\ce{CeBr4^-}\) as complex anions. Sublimation of \(\ce{CsCeBr4}\) leads to a vapour of \(\ce{CsBr}\) and \(\ce{CeBr3}\) molecules.

8.3 Give the reaction equations of the thermochemical cycle (Law of Hess) for this process in which some steps involve \(\ce{CeBr4^-}\) ions, mononuclear ions and/or neutral molecules in the gas phase.

Figure 2

8.4 Calculate the enthalpy of sublimation of \(\ce{CsCeBr4}\) (in integers).

Use the Born–Landé formula for all steps in the process and report the separate energies also (be aware of the signs!). The Madelung constant for \(\ce{NaCl}\) is 1.75. The Cs–Ce distance in the lattice is 0.617 nm. The \(\ce{CeBr4^-}\) anion is a tetrahedron in which the ratio between the edge and the distance between a corner of the tetrahedron and the centre of gravity (body-radius) amounts to \((2\sqrt6)/3=1.633\). The Born exponent of \(\ce{CsBr}\) is 11. The radius of \(\ce{Cs+}\) is 0.181 nm.

8.5 Conclusion in relation to the previous answers: Was adding \(\ce{CsBr}\) a good idea? Mark the correct answer.

  • [ ] Adding \(\ce{CsBr}\) is counterproductive.
  • [ ] Adding \(\ce{CsBr}\) has no influence.
  • [ ] Adding \(\ce{CsBr}\) is advantageous.
  • [ ] From these data no clear answer can be given.

Official English text of the problem.

Answers

  • 8.1 \[ \begin{aligned} \ce{CeBr3(s) &-> Ce^3+(g) + 3Br^-(g)} &&\Delta H=-H_{\mathrm l},\\ \ce{Ce^3+(g) + 3Br^-(g) &-> CeBr3(g)} &&\Delta H=H_{\mathrm e}. \end{aligned} \] Thus \(\ce{CeBr3(s) -> CeBr3(g)}\), with \[ \boxed{H_{\mathrm s}=-H_{\mathrm l}+H_{\mathrm e}}. \]

  • 8.2 \[ H_{\mathrm l}=-3810,\qquad H_{\mathrm e}=-3092,\qquad \boxed{H_{\mathrm s}=+718\ \mathrm{kJ\,mol^{-1}}}. \] All listed energies are in \(\mathrm{kJ\,mol^{-1}}\).

  • 8.3 A consistent ordering of the four steps is \[ \begin{aligned} 1:\quad&\ce{CsCeBr4(s) -> Cs+(g) + CeBr4^-(g)},\\ 2:\quad&\ce{Cs+(g) + CeBr4^-(g) -> Cs+(g) + Ce^3+(g) + 4Br^-(g)},\\ 3:\quad&\ce{Ce^3+(g) + 3Br^-(g) -> CeBr3(g)},\\ 4:\quad&\ce{Cs+(g) + Br^-(g) -> CsBr(g)}. \end{aligned} \] Their sum is \[ \boxed{\ce{CsCeBr4(s) -> CeBr3(g) + CsBr(g)}}. \]

  • 8.4 For the ordering above: \[ \boxed{ H_1=+358,\quad H_2=+3542,\quad H_3=-3092,\quad H_4=-348 \quad(\mathrm{kJ\,mol^{-1}}) } \] \[ \boxed{H_{\mathrm{total}}=+460\ \mathrm{kJ\,mol^{-1}}}. \]

  • 8.5 Mark “Adding \(\ce{CsBr}\) is advantageous.”

Solution

8.1 — Thermochemical cycle for cerium(III) bromide

The species are cerium(III) ions, \(\ce{Ce^3+}\), SMILES [Ce+3]; bromide ions, \(\ce{Br^-}\), SMILES [Br-]; and neutral gaseous cerium(III) bromide, \(\ce{CeBr3}\), connected SMILES [Ce](Br)(Br)Br. The gas molecule is trigonal planar, with Ce at the centre and three Br atoms at the vertices of an equilateral triangle. SMILES does not specify this geometry.

Solid cerium(III) bromide is an extended ionic lattice; its formula-unit composition can be represented by [Ce+3].[Br-].[Br-].[Br-].

The given Born–Landé expression uses signed ionic charges. Consequently, \(H_{\mathrm l}<0\) describes formation of the lattice from separated gaseous ions. Breaking the lattice therefore requires \(-H_{\mathrm l}>0\).

Fill the first reaction box with

\[ \ce{CeBr3(s) -> Ce^3+(g) + 3Br^-(g)}, \qquad \Delta H=-H_{\mathrm l}. \]

Fill the second box with association of the gaseous ions into a neutral molecule:

\[ \ce{Ce^3+(g) + 3Br^-(g) -> CeBr3(g)}, \qquad \Delta H=H_{\mathrm e}. \]

Adding these equations cancels all gaseous ions. The bottom box is therefore

\[ \ce{CeBr3(s) -> CeBr3(g)}, \qquad \boxed{H_{\mathrm s}=-H_{\mathrm l}+H_{\mathrm e}}. \]

8.2 — Sublimation enthalpy of cerium(III) bromide

Define the common factor

\[ C=fe^2\left(1-\frac1n\right) =139\left(1-\frac1{11}\right) =126.3636\ \mathrm{kJ\,nm\,mol^{-1}}. \]

The Ce–Br separation is

\[ r=0.115+0.182=0.297\ \mathrm{nm}. \]

Using the given lattice expression once per mole of \(\ce{CeBr3}\),

\[ H_{\mathrm l} =C\frac{(3)(-1)(2.985)}{0.297} =-3810.06\ \mathrm{kJ\,mol^{-1}}. \]

For the gaseous molecule, sum the energies of every distinct ion pair, using the prescribed expression without a Madelung constant.

There are three attractive Ce–Br pairs. There are also three repulsive Br–Br pairs, whose separation follows from the \(120^\circ\) Br–Ce–Br angle:

\[ d_{\mathrm{Br-Br}} =\sqrt{r^2+r^2-2r^2\cos120^\circ} =\sqrt3\,r =0.514419\ \mathrm{nm}. \]

Thus

\[ \begin{aligned} H_{\mathrm e} &=C\left[ 3\frac{(3)(-1)}{r} +3\frac{(-1)(-1)}{\sqrt3\,r} \right]\\ &=\frac{126.3636}{0.297} \left(-9+\frac3{\sqrt3}\right)\\ &=-3092.27\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

The attractive contribution is \(-3829.20\ \mathrm{kJ\,mol^{-1}}\), while bromide–bromide repulsion contributes \(+736.93\ \mathrm{kJ\,mol^{-1}}\).

Hess’s law gives

\[ \begin{aligned} H_{\mathrm s} &=-(-3810.06)+(-3092.27)\\ &=717.78\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

Therefore, to the requested integer,

\[ \boxed{H_{\mathrm s}\bigl(\ce{CeBr3}\bigr)=+718\ \mathrm{kJ\,mol^{-1}}}. \]

8.3 — Thermochemical cycle for caesium tetrabromidocerate(III)

The additional species are:

Species Name SMILES Geometry
\(\ce{Cs+}\) Caesium ion [Cs+] Monatomic
\(\ce{CeBr4^-}\) Tetrabromidocerate(III) ion [Ce-](Br)(Br)(Br)Br Ce at the centre of a regular tetrahedron of Br atoms
\(\ce{CsBr(g)}\) Caesium bromide molecule [Cs]Br Diatomic

The connected SMILES encode connectivity and overall charge. In particular, the formal charge on [Ce-] does not denote the oxidation state: Ce is \(+3\) in \(\ce{CeBr4^-}\). The tetrahedral geometry must be stated separately.

Solid caesium tetrabromidocerate(III), \(\ce{CsCeBr4}\), has the formula-unit representation [Cs+].[Ce-](Br)(Br)(Br)Br. Its extended rock-salt lattice consists of caesium cations and tetrahedral complex anions.

Fill the four reaction boxes as follows.

  1. Separate the lattice into gaseous constituent ions: \[ \ce{CsCeBr4(s) -> Cs+(g) + CeBr4^-(g)}, \qquad \Delta H=H_1. \]

  2. Dissociate the complex anion into mononuclear ions: \[ \ce{Cs+(g) + CeBr4^-(g) -> Cs+(g) + Ce^3+(g) + 4Br^-(g)}, \qquad \Delta H=H_2. \] The caesium ion is a spectator and contributes zero net enthalpy to this step.

  3. Form one gaseous cerium(III) bromide molecule: \[ \ce{Ce^3+(g) + 3Br^-(g) -> CeBr3(g)}, \qquad \Delta H=H_3. \]

  4. Form one gaseous caesium bromide molecule: \[ \ce{Cs+(g) + Br^-(g) -> CsBr(g)}, \qquad \Delta H=H_4. \]

All intermediate ions cancel on addition:

\[ \ce{CsCeBr4(s) -> CeBr3(g) + CsBr(g)}, \]

\[ \boxed{H_{\mathrm{total}}=H_1+H_2+H_3+H_4}. \]

8.4 — Sublimation enthalpy of caesium tetrabromidocerate(III)

Use \(C=126.3636\ \mathrm{kJ\,nm\,mol^{-1}}\) throughout.

For step 1, the lattice contains ions of charges \(+1\) and \(-1\). Their separation is the given Cs–Ce distance, \(0.617\ \mathrm{nm}\). Reversing lattice formation gives

\[ H_1 =-C\frac{(+1)(-1)(1.75)}{0.617} =+358.41\ \mathrm{kJ\,mol^{-1}}, \]

so

\[ \boxed{H_1=+358\ \mathrm{kJ\,mol^{-1}}}. \]

For step 2, calculate the energy of forming the tetrahedral complex anion from its mononuclear ions, then reverse the sign.

The four Ce–Br distances are \(r=0.297\ \mathrm{nm}\). Each of the six Br–Br pairs lies along an edge of the tetrahedron, so

\[ d_{\mathrm{Br-Br}}=1.633r =1.633(0.297) =0.485001\ \mathrm{nm}. \]

The complex-ion formation energy is

\[ \begin{aligned} E_4 &=C\left[ 4\frac{(3)(-1)}{r} +6\frac{(-1)(-1)}{1.633r} \right]\\ &=\frac{126.3636}{0.297} \left(-12+\frac6{1.633}\right)\\ &=-3542.34\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

Dissociation is the reverse process:

\[ H_2=-E_4=+3542.34\ \mathrm{kJ\,mol^{-1}}, \]

\[ \boxed{H_2=+3542\ \mathrm{kJ\,mol^{-1}}}. \]

For step 3, the formation of gaseous \(\ce{CeBr3}\) was calculated in 8.2:

\[ H_3=-3092.27\ \mathrm{kJ\,mol^{-1}}, \qquad \boxed{H_3=-3092\ \mathrm{kJ\,mol^{-1}}}. \]

For step 4, the caesium–bromide separation is

\[ r_{\mathrm{Cs-Br}}=0.181+0.182=0.363\ \mathrm{nm}. \]

There is one attractive ion pair and no Madelung factor:

\[ H_4 =C\frac{(+1)(-1)}{0.363} =-348.11\ \mathrm{kJ\,mol^{-1}}, \]

\[ \boxed{H_4=-348\ \mathrm{kJ\,mol^{-1}}}. \]

Adding the unrounded values,

\[ \begin{aligned} H_{\mathrm{total}} &=358.41+3542.34-3092.27-348.11\\ &=460.37\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

Therefore,

\[ \boxed{H_{\mathrm s}\bigl(\ce{CsCeBr4}\bigr)=+460\ \mathrm{kJ\,mol^{-1}}}. \]

This value is per mole of solid \(\ce{CsCeBr4}\), producing one mole each of gaseous \(\ce{CeBr3}\) and \(\ce{CsBr}\).

8.5 — Effect of adding caesium bromide

Both sublimation processes produce one mole of gaseous \(\ce{CeBr3}\) per mole of starting cerium-containing solid, but

\[ 460\ \mathrm{kJ\,mol^{-1}} < 718\ \mathrm{kJ\,mol^{-1}}. \]

According to the enthalpy criterion specified in the problem, adding \(\ce{CsBr}\) therefore makes vaporization more favourable and permits operation at a lower temperature, increasing lamp life.

Mark the third option: “Adding \(\ce{CsBr}\) is advantageous.”

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.