IChO 2017 · Theory problem 3

Thermodynamics of chemical reactions

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Part A.

Methanol is produced commercially by using a mixture of carbon monoxide and hydrogen over zinc oxide/copper oxide catalyst:

\[ \ce{CO(g) + 2 H2(g) -> CH3OH(g)}. \]

The standard enthalpy of formation (\(\Delta H_f^\circ\)) and the absolute entropy (\(S^\circ\)) for each of the three gases at room temperature (298 K) and at a standard pressure of 1 bar are given as follows.

Gas \(\Delta H_f^\circ\) (kJ mol\(^{-1}\)) \(S^\circ\) (J K\(^{-1}\) mol\(^{-1}\))
\(\ce{CO(g)}\) −111 198
\(\ce{H2(g)}\) 0 131
\(\ce{CH3OH(g)}\) −201 240

3.1 Calculate \(\Delta H^\circ\), \(\Delta S^\circ\), \(\Delta G^\circ\), and \(K_p\) for the reaction at 298 K.

(If you are unable to calculate \(K_p\) at 298 K in problem 3.1, use \(K_p=9\times10^5\) later on.)

3.2 A commercial reactor is operated at a temperature of 600 K. Calculate the value of \(K_p\) at this temperature, assuming that \(\Delta H^\circ\) and \(\Delta S^\circ\) are independent on temperature.

(If you are unable to calculate \(K_p\) at 600 K in problem 3.2, use \(K_p=1.0\times10^{-2}\) later on.)

3.3 Production of methanol in industry is based on flowing of the gas comprising 2.00 moles of \(\ce{H2}\) for each mole of \(\ce{CO}\) into the reactor. The mole fraction of methanol in the exhaust gas from the reactor was found to be 0.18. Assuming that equilibrium is established, what is the total pressure in the reactor at a high temperature of 600 K?

Part B.

3.4 Consider the following closed system at 300 K. The system comprises 2 compartments, separated by a closed valve, which has negligible volume (see next figure). At the same pressure \(P\), compartment A and compartment B contain 0.100 mol argon gas and 0.200 mol nitrogen gas, respectively. The volumes of the two compartments, \(V_A\) and \(V_B\), are selected so that the gases behave as ideal gases.

Figure 1

After opening the valve slowly, the system is allowed to reach equilibrium. It is assumed that the two gases form an ideal gas mixture. Calculate the change in Gibbs free energy at 300 K, \(\Delta G\).

If you are unable to calculate \(K_p\) at 298 K in problem 3.1, use \(K_p=9\times10^5\) later on.

Official English text of the problem.

Answers

  • 3.1: \(\Delta H^\circ=-90\ \mathrm{kJ\,mol^{-1}}\); \(\Delta S^\circ=-220\ \mathrm{J\,K^{-1}\,mol^{-1}}\); \(\Delta G^\circ=-24.4\ \mathrm{kJ\,mol^{-1}}\); \(K_p(298\ \mathrm K)=1.9\times10^4\).
  • 3.2: \(K_p(600\ \mathrm K)=2.2\times10^{-4}\).
  • 3.3: \(P=1.0\times10^2\ \mathrm{bar}\).
  • 3.4: \(\Delta G=-476\ \mathrm J=-0.476\ \mathrm{kJ}\) for the whole system.

Solution

Use \(R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\) and standard pressure \(p^\circ=1\ \mathrm{bar}\). Equilibrium constants defined using \(p_i/p^\circ\) are dimensionless.

3.1 — Thermodynamic quantities and equilibrium constant at 298 K

Carbon monoxide ([C-]#[O+]) reacts with hydrogen ([H][H]) to produce methanol (CO):

\[ \ce{CO(g) + 2 H2(g) -> CH3OH(g)}. \]

The standard reaction enthalpy is the sum of the product formation enthalpies minus that of the reactants, including stoichiometric coefficients:

\[ \begin{aligned} \Delta H^\circ &=\Delta H_f^\circ(\ce{CH3OH(g)}) -\Delta H_f^\circ(\ce{CO(g)}) -2\Delta H_f^\circ(\ce{H2(g)})\\ &=-201-(-111)-2(0)\\ &=\boxed{-90\ \mathrm{kJ\,mol^{-1}}}. \end{aligned} \]

Similarly,

\[ \begin{aligned} \Delta S^\circ &=S^\circ(\ce{CH3OH(g)}) -S^\circ(\ce{CO(g)}) -2S^\circ(\ce{H2(g)})\\ &=240-198-2(131)\\ &=\boxed{-220\ \mathrm{J\,K^{-1}\,mol^{-1}}}. \end{aligned} \]

Thus, at 298 K,

\[ \begin{aligned} \Delta G^\circ &=\Delta H^\circ-T\Delta S^\circ\\ &=-90-(298)(-0.220)\\ &=-24.44\ \mathrm{kJ\,mol^{-1}}\\ &\approx\boxed{-24.4\ \mathrm{kJ\,mol^{-1}}}. \end{aligned} \]

Using \(\Delta G^\circ=-RT\ln K_p\),

\[ K_p =\exp\!\left(\frac{24440}{8.314\times298}\right) =1.9235\times10^4 \approx\boxed{1.9\times10^4}. \]

3.2 — Equilibrium constant at 600 K

With the specified assumption that \(\Delta H^\circ\) and \(\Delta S^\circ\) are temperature independent,

\[ \begin{aligned} \Delta G^\circ(600\ \mathrm K) &=-90-(600)(-0.220)\\ &=+42.0\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

Therefore,

\[ K_p(600\ \mathrm K) =\exp\!\left(-\frac{42000}{8.314\times600}\right) =2.2052\times10^{-4} \approx\boxed{2.2\times10^{-4}}. \]

The decrease in \(K_p\) with increasing temperature is consistent with the reaction being exothermic.

The optional fallback constants printed in the question differ from those calculated from the table. The following calculation uses the calculated value \(K_p(600\ \mathrm K)=2.2052\times10^{-4}\).

3.3 — Total reactor pressure

The feed contains hydrogen and carbon monoxide in the same \(2:1\) ratio in which they are consumed. Consequently, the unreacted gases retain this ratio at equilibrium:

\[ y_{\ce{H2}}=2y_{\ce{CO}}. \]

Since \(y_{\ce{CH3OH}}=0.18\),

\[ y_{\ce{CO}}+y_{\ce{H2}}=1-0.18=0.82. \]

Hence,

\[ y_{\ce{CO}}=\frac{0.82}{3}=0.273333, \qquad y_{\ce{H2}}=\frac{2(0.82)}{3}=0.546667. \]

For ideal gases, \(p_i=y_iP\), so

\[ \begin{aligned} K_p &=\frac{p_{\ce{CH3OH}}/p^\circ} {(p_{\ce{CO}}/p^\circ)(p_{\ce{H2}}/p^\circ)^2}\\ &=\frac{y_{\ce{CH3OH}}} {y_{\ce{CO}}y_{\ce{H2}}^2} \left(\frac{p^\circ}{P}\right)^2. \end{aligned} \]

Solving for \(P\),

\[ \begin{aligned} P &=p^\circ \sqrt{\frac{y_{\ce{CH3OH}}} {K_p\,y_{\ce{CO}}y_{\ce{H2}}^2}}\\ &=(1\ \mathrm{bar}) \sqrt{\frac{0.18} {(2.2052\times10^{-4})(0.273333)(0.546667)^2}}\\ &=99.96\ \mathrm{bar}. \end{aligned} \]

Thus,

\[ \boxed{P=1.0\times10^2\ \mathrm{bar}}. \]

3.4 — Gibbs free energy change on mixing

The gases are argon, \(\ce{Ar}\) ([Ar]), and nitrogen, \(\ce{N2}\) (N#N).

Initially, both compartments have the same temperature and pressure. The ideal-gas law therefore gives

\[ V_A=\frac{n_{\ce{Ar}}RT}{P}, \qquad V_B=\frac{n_{\ce{N2}}RT}{P}. \]

After opening the valve, the total amount and total volume are unchanged. At the same temperature, the final total pressure is

\[ P_f=\frac{(n_{\ce{Ar}}+n_{\ce{N2}})RT}{V_A+V_B}=P. \]

The final mole fractions are

\[ y_{\ce{Ar}}=\frac{0.100}{0.300}=\frac13, \qquad y_{\ce{N2}}=\frac{0.200}{0.300}=\frac23. \]

For an ideal gas,

\[ \mu_i(T,p_i)=\mu_i^\circ(T)+RT\ln\!\left(\frac{p_i}{p^\circ}\right). \]

Each gas initially has pressure \(P\), and its final partial pressure is \(y_iP\). Therefore,

\[ \begin{aligned} \Delta G &=\sum_i n_i\left(\mu_{i,f}-\mu_{i,i}\right)\\ &=RT\sum_i n_i\ln\!\left(\frac{y_iP}{P}\right)\\ &=RT\left(n_{\ce{Ar}}\ln y_{\ce{Ar}} +n_{\ce{N2}}\ln y_{\ce{N2}}\right). \end{aligned} \]

Substitution gives

\[ \begin{aligned} \Delta G &=(8.314)(300) \left[0.100\ln\!\left(\frac13\right) +0.200\ln\!\left(\frac23\right)\right]\ \mathrm J\\ &=-476.28\ \mathrm J. \end{aligned} \]

Thus, for the whole system,

\[ \boxed{\Delta G=-476\ \mathrm J=-0.476\ \mathrm{kJ}}. \]

The negative value reflects spontaneous mixing. For this ideal-gas mixture, \(\Delta H_{\mathrm{mix}}=0\), and the entropy increases by

\[ \Delta S_{\mathrm{mix}}=-\frac{\Delta G}{T} =1.59\ \mathrm{J\,K^{-1}}. \]

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.