Question

Consider the reaction $$ \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) $$ Use the appropriate tables to calculate (a) \(\Delta G^{\circ}\) at \(552^{\circ} \mathrm{C}\) (b) \(K\) at \(552^{\circ} \mathrm{C}\)

Short answer

Question: Calculate the Gibbs free energy change (ΔG°) and the equilibrium constant (K) for the given reaction at 552°C. Answer: The Gibbs free energy change (ΔG°) for the given reaction at 552°C is 28.4 kJ/mol, and the equilibrium constant (K) is approximately 0.366.

Step-by-step solution

Convert the temperature to Kelvin

To convert the given temperature from Celsius to Kelvin, simply add 273.15 to it. This is the temperature we will be using for our calculations: $$ T=552^\circ C+273.15=825.15 K $$

Find the standard Gibbs free energies of formation

Look up the standard Gibbs free energies of formation for each substance involved in the given reaction. There are tables available in textbooks, published articles or on the internet. The values are usually given in kJ/mol. For the given reaction, we'll find the following values: $$ \Delta G_{f}^{\circ}(\mathrm{CO}(g))=-137.2 \;\mathrm{kJ/mol} $$ $$ \Delta G_{f}^{\circ}(\mathrm{H}_{2} \mathrm{O}(g))=-228.6 \;\mathrm{kJ/mol} $$ $$ \Delta G_{f}^{\circ}(\mathrm{CO}_{2}(g))=-394.4 \;\mathrm{kJ/mol} $$ $$ \Delta G_{f}^{\circ}(\mathrm{H}_{2}(g))=0 \;\mathrm{kJ/mol} $$

Calculate the Gibbs free energy change (ΔG°)

Use the following equation to find ΔG° for the reaction: $$ \Delta G^{\circ}=\sum \Delta G_{f}^{\circ}(\textrm{products})-\sum\Delta G_{f}^{\circ}(\textrm{reactants}) $$ Plugging in the values obtained in step 2: $$ \Delta G^{\circ}=[\Delta G_{f}^{\circ}(\mathrm{CO}_{2}(g))+\Delta G_{f}^{\circ}(\mathrm{H}_{2}(g))]-[\Delta G_{f}^{\circ}(\mathrm{CO}(g))+\Delta G_{f}^{\circ}(\mathrm{H}_{2} \mathrm{O}(g))] $$ $$ \Delta G^{\circ}=[(-394.4)+(0)]-[-137.2-228.6]=28.4 \;\mathrm{kJ/mol} $$ So, the Gibbs free energy change is 28.4 kJ/mol. (Answer for part a)

Calculate the equilibrium constant (K)

Now use the relationship between ΔG° and K to find the equilibrium constant for the reaction: $$ \Delta G^{\circ}=-RT\ln(K) $$ Rearrange the equation to solve for K: $$ K=e^{-\frac{\Delta G^{\circ}}{RT}} $$ Plug in the values for ΔG°, R, and T: $$ K=e^{-\frac{28400\;\mathrm{J/mol}}{8.314\;\mathrm{J/(mol\cdot K)} \cdot 825.15\;\mathrm{K}}} $$ $$ K=e^{-1.008}=0.366 $$ Therefore, the equilibrium constant (K) at 552°C is approximately 0.366. (Answer for part b)