From the following data: (i) (g) + (g) ⇌ (g) + (g); K_2000K = 4.4; (ii) 2(g) ⇌ 2(g) + (g); K_2000K = — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Analyze the given reactions and their equilibrium constants We are given the following three gas-phase equilibrium reactions and their respective equilibrium constants at a high temperature of \(T = 2000\text{ K}\): (i) \(\ce{H2(g) + CO2(g) <=> H2O(g) + CO(g)} \quad \text{with} \quad K_1 = 4.4\) (ii) \(\ce{2H2O(g) <=> 2H2(g) + O2(g)} \quad \text{with} \quad K_2 = 5.31 \times 10^{-10}\) (iii) \(\ce{2CO(g) + O2(g) <=> 2CO2(g)} \quad \text{with} \quad K_3 = 2.24 \times 10^{22}\) Step 2 - Analyze the chemical nature of reaction (iii) Reaction (iii) is the combustion of carbon monoxide gas to form carbon dioxide: \[\ce{2CO(g) + O2(g) <=> 2CO2(g)}\] By definition, combustion reactions involve the oxidation of a fuel (in this case, \(\ce{CO}\)) with oxygen (\(\ce{O2}\)) to produce highly stable oxidized products (\(\ce{CO2}\)). In \(\ce{CO2}\), carbon forms extremely strong and stable double bonds with oxygen. Because the bonds in the product (\(\ce{CO2}\)) are significantly stronger and lower in potential energy than those in the reactants (\(\ce{CO}\) and \(\ce{O2}\)), energy is released in the process. Therefore, the combustion of carbon monoxide is highly exothermic: \[\Delta H^\circ < 0\] Step 3 - Analyze thermodynamics and the magnitude of the equilibrium constant We can also understand this from a thermodynamic standpoint. The standard Gibbs free energy change (\(\Delta G^\circ\)) of a reaction is related to its equilibrium constant (\(K\)) by: \[\Delta G^\circ = -RT \ln K\] Let us calculate the value of \(\Delta G^\circ\) for reaction (iii) at \(T = 2000\text{ K}\): \[\Delta G^\circ = -2.303 R T \log_{10} K_3\] Substituting the values: \[\Delta G^\circ = -2.303 \times 8.314\text{ J K}^{-1}\text{ mol}^{-1} \times 2000\text{ K} \times \log_{10}(2.24 \times 10^{22})\] \[\Delta G^\circ = -2.303 \times 8.314 \times 2000 \times 22.35\text{ J mol}^{-1}\] \[\Delta G^\circ \approx -8.56 \times 10^{5}\text{ J mol}^{-1} = -856\text{ kJ mol}^{-1}\] The large negative value of \(\Delta G^\circ\) indicates that the reaction is extremely spontaneous and thermodynamically favored. The relation between standard Gibbs free energy, enthalpy, and entropy is: \[\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\] For the reaction \(\ce{2CO(g) + O2(g) <=> 2CO2(g)}\), the number of gaseous moles decreases (\(\Delta n_g = 2 - 3 = -1\)). A decrease in the number of gaseous moles means a decrease in entropy, so: \[\Delta S^\circ < 0\] Since \(\Delta S^\circ\) is negative, the term \(-T\Delta S^\circ\) is positive: \[-T\Delta S^\circ > 0\] At a very high temperature like \(2000\text{ K}\), the positive term \(-T\Delta S^\circ\) becomes quite large. For \(\Delta G^\circ\) to still remain highly negative (\(-856\text{ kJ mol}^{-1}\)), the standard enthalpy change (\(\Delta H^\circ\)) must be extremely negative: \[\Delta H^\circ \ll 0\] This thermodynamic analysis proves mathematically that the reaction is strongly exothermic. Step 4 - Evaluate the options * **(A) Exothermic**: Correct. As demonstrated both chemically (combustion reaction) and thermodynamically (exceptionally large equilibrium constant \(K\) overcoming a negative entropy change at \(2000\text{ K}\)), the reaction releases heat (\(\Delta H^\circ < 0\)). * **(B) Endothermic**: Incorrect. Endothermic reactions absorb heat (\(\Delta H^\circ > 0\)). An endothermic reaction with a negative entropy change would have a positive \(\Delta G^\circ\) at high temperatures and would not be favored. * **(C) Thermal**: Incorrect. "Thermal" simply means relating to heat, but is not a valid classification for the thermodynamic direction of heat flow of a chemical reaction. * **(D) Cannot say**: Incorrect. The combustion of CO and the provided thermodynamic parameters are sufficient to determine that the reaction is exothermic. The correct option is A. \[\boxed{\text{A}}\]