IChO 2017 · Theory problem 1

Production of propene using heterogeneous catalysts

Download PDFIChO 2017

Propene or propylene is one of the most valuable chemicals for the petrochemical industry in Thailand and around the world. One good example of the commercial use of propene is for the production of polypropylene (PP).

Part A.

Propene can be synthesized via a direct dehydrogenation of propane in the presence of a heterogeneous catalyst. However, such a reaction is not economically feasible due to the nature of the reaction itself. Provide a concise explanation to each of the questions below. Additional information: \(H_{\mathrm{bond}}(\ce{C=C})=1.77H_{\mathrm{bond}}(\ce{C-C})\), \(H_{\mathrm{bond}}(\ce{H-H})=1.05H_{\mathrm{bond}}(\ce{C-H})\), and \(H_{\mathrm{bond}}(\ce{C-H})=1.19H_{\mathrm{bond}}(\ce{C-C})\), where \(H_{\mathrm{bond}}\) refers to average bond enthalpy of the indicated chemical bond.

1.1 What is the enthalpy change of the direct dehydrogenation of propane? Show your calculation and express your answer in terms of \(H_{\mathrm{bond}}(\ce{C-C})\).

1.2 It is difficult to increase the amount of propene by increasing pressure at constant temperature. Which law or principle can best explain this phenomenon? Select your answer by marking “✓” in one of the open circles.

  • ☐ Boyle’s law
  • ☐ Charles’ law
  • ☐ Dalton’s law
  • ☐ Raoult’s law
  • ☐ Le Chatelier’s principle

1.3 Initially, the system is in equilibrium. Consistent with question 1.1, what is/are correct set(s) of signs for the following thermodynamic variables of the system for the direct dehydrogenation of propane? Select your answer(s) by marking „✓” in any of the open square(s).

\(\Delta H\) \(\Delta S\) \(\Delta G\) \(T^{*}\)
☐ − + + lower
☐ − + − higher
☐ − − + lower
☐ − − − higher
☐ + + + lower
☐ + + − higher
☐ + − + lower
☐ + − − higher

☐ None of the above is correct

\(^*\) Relative to the initial temperature at the same partial pressure.

Part B.

A better reaction to produce large quantity of propene is the oxidative dehydrogenation (ODH) using solid catalysts, such as vanadium oxides, under molecular oxygen gas. Although this type of reaction is still under intense research development, its promise toward the production of propene at an industrial scale eclipses that of the direct dehydrogenation.

1.4 The overall rate of propane consumption in the reaction is

\[ r_{\ce{C3H6}} = \frac{1}{ \dfrac{p^{0}}{k_{\mathrm{red}}p_{\ce{C3H6}}} + \dfrac{p^{0}}{k_{\mathrm{ox}}p_{\ce{O2}}} } \]

where \(k_{\mathrm{red}}\) and \(k_{\mathrm{ox}}\) are the rate constants for the reduction of metal oxide catalyst by propane and for the oxidation of the catalyst by molecular oxygen, respectively, and \(p^{o}\) is the standard pressure of 1 bar. Some experiments found that the rate of oxidation of the catalyst is 100,000 times faster than that of the propane oxidation. The experimental

\[ r_{\ce{C3H8}} = k_{\mathrm{obs}}\frac{p_{\ce{C3H6}}}{p^{0}} \]

where \(k_{\mathrm{obs}}\) is the observed rate constant (\(0.062\ \mathrm{mol\,s^{-1}}\)). If the reactor containing the catalyst is continuously passed through with propane and oxygen at a total pressure of 1 bar, determine the value of \(k_{\mathrm{red}}\) and \(k_{\mathrm{ox}}\) when the partial pressure of propane is 0.10 bar. Assume that the partial pressure of propene is negligible.

Part C.

The metal oxide catalyst contains oxygen atoms on its surface that serve as active sites for the ODH. Denoting red* as a reduced site and \(\ce{O(s)}\) as an oxygen atom on the surface of the catalyst, one of the proposed mechanisms for the ODH in the presence of the catalyst can be written as follows:

Figure 1

Given

\[ \beta= \frac{\text{number of reduced sites}} {\text{total number of active sites}} \]

The rate laws for the above 3 steps are:

\[ r_1=k_1p_{\ce{C2H8}}(1-\beta) \]

\[ r_2=k_2p_{\ce{C3H6}}(1-\beta) \]

and

\[ r_3=k_3p_{\ce{O2}}\beta \]

1.5 Assuming that the amount of oxygen atoms on the surface stays constant at any time of reaction, calculate \(\beta\) as a function of

\[ k_1,\ k_2,\ k_3,\ p_{\ce{C2H8}},\ p_{\ce{C3H6}},\ \text{and }p_{\ce{O2}}. \]

Official English text of the problem.

Answers

  • 1.1: \(\boxed{\Delta H\approx+0.36H_{\mathrm{bond}}(\ce{C-C})}\). The reaction is endothermic.
  • 1.2: Check Le Chatelier’s principle.
  • 1.3: Check the fifth and sixth rows: \((+,+,+,\text{lower})\) and \((+,+,-,\text{higher})\), in the order \((\Delta H,\Delta S,\Delta G,T^*)\).
  • 1.4: \(\boxed{k_{\mathrm{red}}\approx0.062\ \mathrm{mol\,s^{-1}}}\) and \(\boxed{k_{\mathrm{ox}}\approx6.9\times10^{2}\ \mathrm{mol\,s^{-1}}}\).
  • 1.5: Using the corrected propane subscript, \[ \boxed{ \beta= \frac{k_1p_{\ce{C3H8}}+9k_2p_{\ce{C3H6}}} {k_1p_{\ce{C3H8}}+9k_2p_{\ce{C3H6}}+2k_3p_{\ce{O2}}} }. \]

Solution

The molecular species used below are propane CCC, propene C=CC, hydrogen [H][H], molecular oxygen O=O, water O, and carbon dioxide O=C=O. The adsorbed oxygen atoms and reduced catalyst sites belong to an unspecified solid surface, so their surface bonding has no uniquely defined molecular SMILES.

1.1 — Enthalpy of direct dehydrogenation

The balanced reaction is

\[ \ce{C3H8(g) -> C3H6(g) + H2(g)}. \]

Propane contains two carbon–carbon single bonds and eight carbon–hydrogen bonds. The products contain one carbon–carbon single bond, one carbon–carbon double bond, six carbon–hydrogen bonds, and one hydrogen–hydrogen bond.

Using average bond enthalpies,

\[ \begin{aligned} \Delta H &\approx \left[2H_{\mathrm{bond}}(\ce{C-C}) +8H_{\mathrm{bond}}(\ce{C-H})\right]\\ &\quad- \left[H_{\mathrm{bond}}(\ce{C-C}) +H_{\mathrm{bond}}(\ce{C=C}) +6H_{\mathrm{bond}}(\ce{C-H}) +H_{\mathrm{bond}}(\ce{H-H})\right]\\ &= H_{\mathrm{bond}}(\ce{C-C}) +2H_{\mathrm{bond}}(\ce{C-H}) -H_{\mathrm{bond}}(\ce{C=C}) -H_{\mathrm{bond}}(\ce{H-H}). \end{aligned} \]

Let \(D=H_{\mathrm{bond}}(\ce{C-C})\). The supplied ratios give

\[ \begin{aligned} \Delta H &\approx[1+2(1.19)-1.77-1.05(1.19)]D\\ &=0.3605D\\ &\approx\boxed{+0.36H_{\mathrm{bond}}(\ce{C-C})}. \end{aligned} \]

The result has the same molar-energy units as the bond enthalpy. Its positive sign identifies an endothermic reaction.

1.2 — Effect of pressure

The forward reaction increases the number of gas molecules from one to two:

\[ \ce{C3H8(g) <=> C3H6(g) + H2(g)}. \]

According to Le Chatelier’s principle, increasing pressure by compression at constant temperature favors the side with fewer gas molecules. The equilibrium therefore shifts toward propane, reducing the equilibrium amount of propene.

Check Le Chatelier’s principle.

1.3 — Thermodynamic signs after changing temperature

From 1.1,

\[ \Delta H>0. \]

At the initial equilibrium temperature \(T_0\),

\[ \Delta G(T_0)=\Delta H-T_0\Delta S=0. \]

Consequently,

\[ \Delta S(T_0)=\frac{\Delta H}{T_0}>0, \]

consistent with the increase in the number of gaseous molecules.

At fixed partial pressures,

\[ \left(\frac{\partial\Delta G}{\partial T}\right)_{\{p_i\}} =-\Delta S<0. \]

Thus, changing the temperature from the initial equilibrium gives:

Temperature change \(\Delta H\) \(\Delta S\) \(\Delta G\)
Lower temperature + + +
Higher temperature + + −

Check the fifth and sixth rows. These signs describe the temperature perturbation at the specified partial pressures. Once the composition reaches its new equilibrium, the reaction Gibbs energy is again zero.

1.4 — Reduction and oxidation rate constants

The balanced overall oxidative dehydrogenation is

\[ \ce{2 C3H8(g) + O2(g) -> 2 C3H6(g) + 2 H2O(g)}. \]

The kinetic expressions must use the propane partial pressure, \(p_{\ce{C3H8}}\), wherever they describe propane consumption.

For the stated propane–oxygen mixture with negligible products,

\[ p_{\ce{O2}}=1.00-0.10=0.90\ \mathrm{bar}. \]

The experimental consumption rate is

\[ r=k_{\mathrm{obs}}\frac{p_{\ce{C3H8}}}{p^0} =(0.062\ \mathrm{mol\,s^{-1}}) \frac{0.10\ \mathrm{bar}}{1\ \mathrm{bar}} =6.2\times10^{-3}\ \mathrm{mol\,s^{-1}}. \]

Define the two pressure-weighted kinetic terms

\[ a=k_{\mathrm{red}}\frac{p_{\ce{C3H8}}}{p^0}, \qquad b=k_{\mathrm{ox}}\frac{p_{\ce{O2}}}{p^0}. \]

Then the given rate expression becomes

\[ r=\frac{1}{1/a+1/b}=\frac{ab}{a+b}. \]

Interpreting the stated speed comparison as \(b=10^5a\),

\[ r=\frac{a}{1+10^{-5}} =\frac{k_{\mathrm{red}}}{1+10^{-5}} \frac{p_{\ce{C3H8}}}{p^0}. \]

Comparison with the experimental expression gives

\[ k_{\mathrm{red}} =k_{\mathrm{obs}}(1+10^{-5}) \approx\boxed{0.062\ \mathrm{mol\,s^{-1}}}. \]

The oxidation constant follows from the same rate comparison:

\[ \frac{k_{\mathrm{ox}}p_{\ce{O2}}} {k_{\mathrm{red}}p_{\ce{C3H8}}}=10^5, \]

so

\[ \begin{aligned} k_{\mathrm{ox}} &=10^5k_{\mathrm{red}} \frac{p_{\ce{C3H8}}}{p_{\ce{O2}}}\\ &=10^5(0.06200062) \frac{0.10}{0.90}\ \mathrm{mol\,s^{-1}}\\ &\approx\boxed{6.9\times10^2\ \mathrm{mol\,s^{-1}}}. \end{aligned} \]

Both constants have units of \(\mathrm{mol\,s^{-1}}\), because the pressure factors in these expressions are dimensionless.

1.5 — Steady-state fraction of reduced sites

The three supplied reactions are

\[ \ce{C3H8(g) + O(s) ->[k_1] C3H6(g) + H2O(g) + red^{*}} \qquad (1) \]

\[ \ce{C3H6(g) + 9 O(s) ->[k_2] 3 CO2(g) + 3 H2O(g) + 9 red^{*}} \qquad (2) \]

\[ \ce{O2(g) + 2 red^{*} ->[k_3] 2 O(s)} \qquad (3) \]

Here \(\ce{O(s)}\) denotes an oxygen atom on the catalyst surface, as defined in the question.

Each occurrence of step 1 consumes one surface oxygen atom; step 2 consumes nine; step 3 supplies two. Constancy of the surface oxygen population therefore requires

\[ 2r_3-r_1-9r_2=0. \]

Using the supplied rate laws, with the propane subscript corrected,

\[ 2k_3p_{\ce{O2}}\beta = k_1p_{\ce{C3H8}}(1-\beta) +9k_2p_{\ce{C3H6}}(1-\beta). \]

Factor the right-hand side:

\[ 2k_3p_{\ce{O2}}\beta = \left(k_1p_{\ce{C3H8}}+9k_2p_{\ce{C3H6}}\right)(1-\beta). \]

Solving for \(\beta\),

\[ \boxed{ \beta= \frac{k_1p_{\ce{C3H8}}+9k_2p_{\ce{C3H6}}} {k_1p_{\ce{C3H8}}+9k_2p_{\ce{C3H6}}+2k_3p_{\ce{O2}}} }. \]

The factors 9 and 2 arise from the oxygen-atom balance. The pressure and coverage exponents remain those explicitly supplied in the question.

The solution was checked against the official solution.