IChO 1994 · Theory problem 4

Platinum chemistry and the catalytic oxidation of carbon monoxide

Download PDFIChO 1994

Platinum(IV) oxide is not found in the nature, but it can be prepared in a laboratory. Solid platinum(IV) oxide is in equilibrium with platinum metal and oxygen gas at 1 atm (= \(1.01325\times10^5\) Pa) and 650 °C.

4.1 This suggests that the conditions on the Earth, when the minerals we know were formed, were:

[1] \(p(\ce{O2})=1\) atm, \(t=650\) °C;
[2] \(p(\ce{O2})<1\) atm, \(t<650\) °C;
[3] \(p(\ce{O2})>1\) atm, \(t<650\) °C;
[4] \(p(\ce{O2})<1\) atm, \(t>650\) °C;
[5] \(p(\ce{O2})>1\) atm, \(t>650\) °C

Mark the most probable alternative [1] – [5] on the answer sheet. Please note that the marking of only one alternative will be accepted.

4.2 What are \(\Delta G\) and \(K_p\) for the formation of platinum(IV) oxide at oxygen pressure of 1 atm and temperature of 650 °C?

The preparation of platinum(IV) oxide involves boiling of a solution which contains hexachloroplatinate(IV) ions with sodium carbonate. In this process \(\ce{PtO2 . nH2O}\) is formed and this is in turn converted to platinum(IV) oxide upon subsequent filtering and heat treatment. In the following we assume \(n=4\).

\(\ce{PtO2 . 4H2O}\) or \(\ce{Pt(OH)4 . 2H2O}\) can be dissolved in acids and strong bases.

4.3 Write the balanced equations for the preparation of platinum(IV) oxide according to the procedure given above.

4.4 Write the balanced equations for the dissolution of \(\ce{PtO2 . 4H2O}\) in both hydrochloric acid and sodium hydroxide.

Platinum is mainly found in the nature as the metal (in mixture or in alloying with other precious metals). Platinum is dissolved in aqua regia under the formation of hexachloroplatinate(IV) ions. Aqua regia is a mixture of concentrated hydrochloric and nitric acids in proportion 3 : 1, and of the nitrosylchloride (\(\ce{NOCl}\)) and the atomic chlorine which are formed upon the mixing. The latter is believed to be the active dissolving component.

The hexachloroplatinate(IV) ions can be precipitated as diammonium hexachloroplatinate(IV) and by thermal decomposition of this compound, finely powdered platinum and gaseous products are formed.

4.5 Write the balanced equations for the formation of aqua regia and its reaction with platinum.

4.6 Write the balanced equation of the thermal decomposition of diammonium hexachloroplatinate(IV) at elevated temperature.

From diammonium hexachloroplatinate(IV) we can prepare \(\ce{Pt(NH3)2Cl2}\) which occurs in cis (\(\Delta_fH^0=-467.4\ \mathrm{kJ\,mol^{-1}}\), \(\Delta_fG^0=-228.7\ \mathrm{kJ\,mol^{-1}}\)) and trans (\(\Delta_fH^0=-480.3\ \mathrm{kJ\,mol^{-1}}\), \(\Delta_fG^0=-222.8\ \mathrm{kJ\,mol^{-1}}\)) form.

4.7 The occurrence of the isomers shows that \(\ce{Pt(NH3)2Cl2}\) hasgeometry:

[ 1 ] linear,
[ 2 ] planar,
[ 3 ] tetrahedral,
[ 4 ] octahedral geometry.

Mark the correct alternative of [ 1 ] – [ 4 ] on the answer sheet.

4.8 Is the cis form or trans form thermodynamically more stable?

Platinum is used as a catalyst in modern automobiles. In the catalyst carbon monoxide (\(\Delta_fH^0=-110.5\ \mathrm{kJ\,mol^{-1}}\), \(\Delta_fG^0=-137.3\ \mathrm{kJ\,mol^{-1}}\)) reacts with oxygen to carbon dioxide (\(\Delta_fH^0=-393.5\ \mathrm{kJ\,mol^{-1}}\), \(\Delta_fG^0=-394.4\ \mathrm{kJ\,mol^{-1}}\)).

4.9 Is the reaction spontaneous at 25 °C?

[ 1 ] yes, or
[ 2 ] no.

Is the reaction:

[ 3 ] endothermic, or
[ 4 ] exothermic?

Calculate \(\Delta S^\circ\) for the reaction.

Establish whether the entropy of the reaction system

[5] increases, or
[6] decreases.

4.10 Establish an expression for the temperature dependence of the equilibrium constant in this case.

The overall catalytic reaction is simple, whereas the reaction mechanism in the homogeneous phase is very complicated with a large number of reaction steps, and the course is difficult to control owing to a distinct chain character. With platinum as catalyst the significant reaction steps are: (i) Adsorption of \(\ce{CO}\) and adsorption/dissociation of \(\ce{O2}\) (\(\Delta H=-259\) kJ per mol \(\ce{CO + O}\)), (ii) their activation (105 kJ per mol \(\ce{CO + O}\)) and (iii) the reaction and the desorption of \(\ce{CO2}\) (\(\Delta H=21\) kJ per mol \(\ce{CO2}\)).

A one-dimensional energy-diagram for the platinum catalyzed oxidation of carbon monoxide to dioxide can be represented as:

Figure 1

4.11 Mark the correct alternative of [ 1 ] – [ 4 ] on the answer sheet.

Official English text of the problem.

Answers

Species names and SMILES are collected in the Solution.

  • 4.1: [4], \(p(\ce{O2})<1\ \mathrm{atm}\), \(t>650\,^\circ\mathrm C\), using the intended qualitative oxide-stability argument.
  • 4.2: \(\boxed{\Delta_rG=0,\quad K_p=1}\), taking \(p^\circ=1\ \mathrm{atm}\).
  • 4.3: \[ \ce{[PtCl6]^2- + 2CO3^2- + 4H2O -> PtO2 . 4H2O(s) + 6Cl- + 2CO2(g)} \] \[ \ce{PtO2 . 4H2O(s) ->[\Delta] PtO2(s) + 4H2O(g)} \]
  • 4.4: \[ \ce{PtO2 . 4H2O(s) + 4H+ + 6Cl- -> [PtCl6]^2- + 6H2O} \] \[ \ce{PtO2 . 4H2O(s) + 2OH- -> [Pt(OH)6]^2- + 2H2O} \]
  • 4.5: In the atomic-chlorine description specified in the problem, \[ \ce{HNO3 + 3HCl -> NOCl + 2Cl + 2H2O} \] \[ \ce{Pt + 4Cl + 2Cl- -> [PtCl6]^2-} \]
  • 4.6: A formal primary decomposition is \[ \ce{(NH4)2[PtCl6] ->[\Delta] Pt + 2NH3 + 2HCl + 2Cl2}. \] Allowing subsequent reaction between ammonia and chlorine gives the nitrogen-forming net balance \[ \ce{3(NH4)2[PtCl6] ->[\Delta] 3Pt + 2N2 + 2NH3 + 18HCl}. \] The unspecified gas products require the qualification discussed below.
  • 4.7: [2], specifically square planar.
  • 4.8: Cis, because its standard Gibbs energy of formation is lower.
  • 4.9: [1] yes; [4] exothermic; [6] entropy decreases. For \(\ce{CO + 1/2O2 -> CO2}\), \[ \boxed{\Delta_rH^\circ=-283.0\ \mathrm{kJ\,mol^{-1}},\quad \Delta_rG^\circ=-257.1\ \mathrm{kJ\,mol^{-1}},\quad \Delta_rS^\circ=-86.9\ \mathrm{J\,mol^{-1}\,K^{-1}}.} \] For \(\ce{2CO + O2 -> 2CO2}\), these values double.
  • 4.10: For the reaction producing one mole of carbon dioxide, assuming temperature-independent reaction enthalpy and entropy, \[ \boxed{K_p(T)\simeq \exp\!\left(\frac{3.404\times10^4\ \mathrm K}{T}-10.45\right).} \] Thus \(K_p\) decreases as temperature increases.
  • 4.11: [2]. Relative energy levels are \(0,-259,-154,-304,-283\ \mathrm{kJ\,mol^{-1}}\), in reaction order.

Solution

Chemical names and SMILES

The following identifiers cover the species used in the answers. Dative bonds are written with ->. The platinum(IV) complexes are octahedral; the square-planar stereochemistry of the two platinum(II) isomers is specified explicitly.

Platinum metal and the oxides are extended solids, so their entries represent an atom or formal formula-unit composition, not a discrete molecular structure. SMILES also do not specify gas spin states or the adsorption sites on the catalyst.

Species and name SMILES
\(\ce{Pt}\), platinum [Pt]
\(\ce{PtO2}\), platinum(IV) oxide, formal composition [Pt+4].[O-2].[O-2]
\(\ce{PtO2 . 4H2O}\), hydrated platinum(IV) oxide, formal composition [Pt+4].[O-2].[O-2].O.O.O.O
\(\ce{[PtCl6]^2-}\), hexachloroplatinate(IV) ion [Cl-]->[Pt+4](<-[Cl-])(<-[Cl-])(<-[Cl-])(<-[Cl-])<-[Cl-]
\(\ce{[Pt(OH)6]^2-}\), hexahydroxoplatinate(IV) ion [OH-]->[Pt+4](<-[OH-])(<-[OH-])(<-[OH-])(<-[OH-])<-[OH-]
\(\ce{(NH4)2[PtCl6]}\), diammonium hexachloroplatinate(IV) [NH4+].[NH4+].[Cl-]->[Pt+4](<-[Cl-])(<-[Cl-])(<-[Cl-])(<-[Cl-])<-[Cl-]
cis-\(\ce{Pt(NH3)2Cl2}\), cis-diamminedichloridoplatinum(II), cisplatin [NH3]->[Pt@SP1+2](<-[NH3])(<-[Cl-])<-[Cl-]
trans-\(\ce{Pt(NH3)2Cl2}\), trans-diamminedichloridoplatinum(II), transplatin [NH3]->[Pt@SP2+2](<-[NH3])(<-[Cl-])<-[Cl-]
\(\ce{Na2CO3}\), sodium carbonate [Na+].[Na+].[O-]C(=O)[O-]
\(\ce{NaOH}\), sodium hydroxide [Na+].[OH-]
\(\ce{CO3^2-}\), carbonate ion [O-]C(=O)[O-]
\(\ce{H+}\), hydrogen ion [H+]
\(\ce{Cl-}\), chloride ion [Cl-]
\(\ce{OH-}\), hydroxide ion [OH-]
\(\ce{NH4+}\), ammonium ion [NH4+]
\(\ce{NH4Cl}\), ammonium chloride [NH4+].[Cl-]
\(\ce{H2O}\), water O
\(\ce{HCl}\), hydrogen chloride; hydrochloric acid when dissolved in water Cl
\(\ce{HNO3}\), nitric acid O[N+](=O)[O-]
\(\ce{NOCl}\), nitrosyl chloride O=NCl
\(\ce{Cl}\), atomic chlorine [Cl]
\(\ce{Cl2}\), molecular chlorine ClCl
\(\ce{NH3}\), ammonia N
\(\ce{N2}\), nitrogen N#N
\(\ce{O2}\), oxygen O=O
\(\ce{O}\), atomic oxygen [O]
\(\ce{CO}\), carbon monoxide [C-]#[O+]
\(\ce{CO2}\), carbon dioxide O=C=O

4.1

Consider the equilibrium \[ \ce{PtO2(s) <=> Pt(s) + O2(g)}. \]

Lowering the oxygen partial pressure favors decomposition, which produces oxygen. Using the intended assumption that oxide formation is exothermic, increasing the temperature also favors decomposition.

Thus the combination that most clearly favors native platinum over its oxide is \[ \boxed{p(\ce{O2})<1\ \mathrm{atm},\qquad t>650\,^\circ\mathrm C}, \] namely [4].

4.2

For formation, \[ \ce{Pt(s) + O2(g) <=> PtO2(s)}. \]

The reaction Gibbs energy is zero at equilibrium: \[ \boxed{\Delta_rG=0}. \]

Pure solid activities equal unity. Defining a dimensionless equilibrium constant with \(p^\circ=1\ \mathrm{atm}\), \[ K_p= \frac{a(\ce{PtO2})} {a(\ce{Pt})[p(\ce{O2})/p^\circ]} =\frac{p^\circ}{p(\ce{O2})}. \]

Since the equilibrium oxygen pressure is \(1\ \mathrm{atm}\), \[ \boxed{K_p=1}. \] With this standard-pressure convention, \(\Delta_rG^\circ=-RT\ln K_p=0\) as well.

If an unnormalized, dimensional pressure quotient is used instead, its value is \(1\ \mathrm{atm^{-1}}\).

4.3

Carbonate acts as a base, allowing precipitation of the hydrous oxide while carbon dioxide is released. The net ionic equation is \[ \boxed{\ce{[PtCl6]^2- + 2CO3^2- + 4H2O -> PtO2 . 4H2O(s) + 6Cl- + 2CO2(g)}}. \]

Both sides have charge \(-6\), and all atoms balance. Sodium ions are spectators. Platinum remains in oxidation state \(+IV\), so this is not a redox reaction.

After filtration, heating removes the water: \[ \boxed{\ce{PtO2 . 4H2O(s) ->[\Delta] PtO2(s) + 4H2O(g)}}. \]

4.4

The hydrated oxide is amphoteric.

In hydrochloric acid, protons convert the oxide/hydroxide groups into water, and chloride coordinates the platinum: \[ \boxed{\ce{PtO2 . 4H2O(s) + 4H+ + 6Cl- -> [PtCl6]^2- + 6H2O}}. \]

The total charge is \(-2\) on either side.

In sodium hydroxide, the soluble product is the hexahydroxoplatinate(IV) ion: \[ \boxed{\ce{PtO2 . 4H2O(s) + 2OH- -> [Pt(OH)6]^2- + 2H2O}}. \]

Again, both sides have charge \(-2\). Platinum remains \(+IV\) in both reactions.

4.5

Following the problem’s explicit description in terms of atomic chlorine, mixing the acids gives \[ \boxed{\ce{HNO3 + 3HCl -> NOCl + 2Cl + 2H2O}}. \]

Nitrogen is reduced from \(+V\) in nitric acid to \(+III\) in nitrosyl chloride. Two chloride ions are correspondingly oxidized to two neutral chlorine atoms.

Platinum then reacts according to \[ \boxed{\ce{Pt + 4Cl + 2Cl- -> [PtCl6]^2-}}. \]

Platinum loses four electrons, becoming platinum(IV); four chlorine atoms each gain one electron. Two further chloride ions complete the six-coordinate, octahedral complex.

These equations use the atomic-chlorine model stated in the question; they do not establish a detailed dissolution mechanism.

4.6

A balanced formal decomposition into platinum, ammonia, hydrogen chloride and chlorine is \[ \boxed{\ce{(NH4)2[PtCl6](s) ->[\Delta] Pt(s) + 2NH3(g) + 2HCl(g) + 2Cl2(g)}}. \]

Here platinum is reduced from \(+IV\) to zero, and four chloride ligands are oxidized to two chlorine molecules. The remaining atoms give ammonia and hydrogen chloride.

However, ammonia and chlorine can react further. A nitrogen-forming balance is \[ \ce{4NH3 + 6Cl2 -> 2N2 + 12HCl}. \]

Adding this to three times the formal decomposition above gives \[ \boxed{\ce{3(NH4)2[PtCl6](s) ->[\Delta] 3Pt(s) + 2N2(g) + 2NH3(g) + 18HCl(g)}}. \]

For this net reaction, four nitrogen atoms are oxidized from \(-III\) to zero, supplying the twelve electrons required to reduce three platinum(IV) centers.

Grouping two ammonia and two hydrogen chloride molecules as ammonium chloride gives the corresponding formula-unit balance \[ \ce{3(NH4)2[PtCl6] ->[\Delta] 3Pt + 2N2 + 2NH4Cl + 16HCl}. \]

At elevated temperature, the volatile ammonium chloride can be represented through its dissociation into ammonia and hydrogen chloride. These equations distinguish primary decomposition from secondary gas chemistry; the statement does not specify the final gas composition.

4.7

There are four ligands around platinum. In a tetrahedral arrangement, all pairs of ligand positions are equivalent, so a complex containing two ligands of each kind cannot have distinct cis and trans forms.

A square-planar arrangement permits:

  • Cisplatin: the two chloride ligands are adjacent, with a \(\ce{Cl-Pt-Cl}\) angle of approximately \(90^\circ\); the ammonia ligands are also adjacent.
  • Transplatin: the chloride ligands are opposite, with a \(\ce{Cl-Pt-Cl}\) angle of \(180^\circ\); the ammonia ligands are also opposite.

All four donor atoms and the platinum atom lie in one plane. Therefore the answer is [2], planar, specifically square planar. The stereochemical SMILES are given in the identifier table.

4.8

Thermodynamic stability at fixed temperature and pressure is determined by Gibbs energy. For cis-to-trans conversion, \[ \begin{aligned} \Delta_rG^\circ &=\Delta_fG^\circ(\text{trans})-\Delta_fG^\circ(\text{cis})\\ &=-222.8-(-228.7)\\ &=+5.9\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

Thus the conversion to trans is unfavorable under the standard conditions of the supplied data: \[ \boxed{\text{The cis form is thermodynamically more stable.}} \]

Although the trans form has the lower enthalpy of formation, its entropy contribution makes its Gibbs energy higher.

4.9

Use the reaction normalized to one mole of carbon monoxide: \[ \ce{CO(g) + 1/2O2(g) -> CO2(g)}. \]

The standard formation enthalpy and Gibbs energy of elemental oxygen in its standard state are zero.

Hence \[ \begin{aligned} \Delta_rH^\circ &=-393.5-(-110.5)\\ &=\boxed{-283.0\ \mathrm{kJ\,mol^{-1}}}, \\[4pt] \Delta_rG^\circ &=-394.4-(-137.3)\\ &=\boxed{-257.1\ \mathrm{kJ\,mol^{-1}}}. \end{aligned} \]

Since \(\Delta_rG^\circ<0\), the forward reaction is spontaneous under standard conditions at 25 °C: [1]. Since \(\Delta_rH^\circ<0\), it is exothermic: [4].

Using \[ \Delta_rG^\circ=\Delta_rH^\circ-T\Delta_rS^\circ \] at \(T=298.15\ \mathrm K\), \[ \begin{aligned} \Delta_rS^\circ &=\frac{-283.0-(-257.1)}{298.15} \ \mathrm{kJ\,mol^{-1}\,K^{-1}}\\ &=\boxed{-86.9\ \mathrm{J\,mol^{-1}\,K^{-1}}}. \end{aligned} \]

The system’s entropy decreases: [6], consistent with the reduction from \(1.5\) moles of gas to \(1\) mole of gas.

This does not conflict with spontaneity: the released heat increases the surroundings’ entropy by more than the system’s entropy decreases.

For the integer-coefficient equation \[ \ce{2CO(g) + O2(g) -> 2CO2(g)}, \] the corresponding values are \[ \Delta_rH^\circ=-566.0\ \mathrm{kJ\,mol^{-1}},\quad \Delta_rG^\circ=-514.2\ \mathrm{kJ\,mol^{-1}},\quad \Delta_rS^\circ=-173.7\ \mathrm{J\,mol^{-1}\,K^{-1}}, \] where “per mole” now refers to one mole of reaction as written.

4.10

For the one-carbon-monoxide reaction, \[ K_p= \frac{p(\ce{CO2})/p^\circ} {[p(\ce{CO})/p^\circ][p(\ce{O2})/p^\circ]^{1/2}}. \]

The exact van ’t Hoff relation is \[ \boxed{\frac{d\ln K_p}{dT}=\frac{\Delta_rH^\circ(T)}{RT^2}}. \]

If \(\Delta_rH^\circ\) is treated as constant, integration gives \[ \boxed{ \ln\frac{K_p(T)}{K_p(T_0)} =-\frac{\Delta_rH^\circ}{R} \left(\frac1T-\frac1{T_0}\right) }. \]

Equivalently, taking both reaction enthalpy and entropy as temperature independent, \[ \ln K_p =-\frac{\Delta_rG^\circ}{RT} =-\frac{\Delta_rH^\circ}{RT}+\frac{\Delta_rS^\circ}{R}. \]

Using \(R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\), \[ \ln K_p(T) =\frac{283000}{8.314\,T}-\frac{86.869}{8.314}, \] so \[ \boxed{ K_p(T)\simeq \exp\!\left(\frac{3.404\times10^4\ \mathrm K}{T}-10.45\right) }. \]

At \(298.15\ \mathrm K\), this gives approximately \[ K_p=1.1\times10^{45}. \]

Because the reaction is exothermic, \(d\ln K_p/dT<0\): raising the temperature decreases the equilibrium constant.

If the reaction is instead written as \(\ce{2CO + O2 -> 2CO2}\), its equilibrium constant is the square of the one above: \[ K_{p,2}(T)\simeq \exp\!\left(\frac{6.808\times10^4\ \mathrm K}{T}-20.90\right). \]

4.11

Take the gaseous reactants and free platinum surface as the zero of energy, per mole of carbon monoxide oxidized.

  1. Adsorption and oxygen dissociation lower the energy by \(259\ \mathrm{kJ\,mol^{-1}}\): \[ E_{\text{adsorbed reactants}}=-259\ \mathrm{kJ\,mol^{-1}}. \]

  2. The activation barrier is \(105\ \mathrm{kJ\,mol^{-1}}\) above those adsorbed reactants: \[ E_{\text{transition state}}=-259+105=-154\ \mathrm{kJ\,mol^{-1}}. \]

  3. The final gaseous product must lie at the overall reaction enthalpy calculated in 4.9: \[ E_{\text{gaseous product}}=-283\ \mathrm{kJ\,mol^{-1}}. \]

  4. Desorption requires \(21\ \mathrm{kJ\,mol^{-1}}\), so the adsorbed carbon dioxide lies below the gaseous product: \[ E_{\text{adsorbed product}}=-283-21=-304\ \mathrm{kJ\,mol^{-1}}. \]

The required profile therefore starts at zero, falls to \(-259\), rises to \(-154\), falls to \(-304\), and finally rises to \(-283\ \mathrm{kJ\,mol^{-1}}\).

This is diagram [2]. The catalyst provides a reaction pathway with surface intermediates but does not change the overall reaction enthalpy or equilibrium constant.

The solution was checked against the official solution.