Question

Using the Gibbs function data, determine the equilibrium constant \(K_{P}\) for the dissociation process \(\mathrm{CO}_{2} \rightleftharpoons\) \(\mathrm{CO}+\frac{1}{2} \mathrm{O}_{2}\) at \((a) 298 \mathrm{K}\) and \((b) 1800 \mathrm{K} .\) Compare your results with the \(K_{P}\) values listed in Table \(\mathrm{A}-28\).

Short answer

(a) At 298 K, the calculated equilibrium constant (Kp) is 9.799 x 10^(-25). (b) At 1800 K, the calculated equilibrium constant (Kp) is 2.785 x 10^(-8). These calculated Kp values should be compared with the values listed in Table A-28 to check for accuracy and consistency.

Step-by-step solution

Write the reaction and find the stoichiometric coefficients

We are given the dissociation process as: CO2 \(\rightleftharpoons\) CO + \(\frac{1}{2}\) O2. The stoichiometric coefficients are -1 for CO2, 1 for CO, and \(\frac{1}{2}\) for O2.

Find the values of Gibbs free energy of formation for each species

Using Table A-28, obtain the Gibbs free energy of formation values for each species involved in the reaction at the given temperature (either 298 K or 1800 K).

Calculate the change in Gibbs free energy for the reaction

Use the following formula for calculating the change in Gibbs free energy: \(\Delta G = \sum{v_i \cdot G_i}\), where \(v_i\) is the stoichiometric coefficient and \(G_i\) is the Gibbs free energy of each species.

Calculate the equilibrium constant

Plug the calculated \(\Delta G\) value into the following formula to find the equilibrium constant (\(K_P\)) at the given temperature: \(K_P = e^{-\frac{\Delta G}{RT}}\), where R is the universal gas constant (8.314 J/mol K) and T is the temperature in K. Now we will solve for each temperature: (a) At \(T = 298 \thinspace K\): Step 2: Table A-28 provides the values for Gibbs free energy of formation at 298 K as follows: \(G_{f,CO2} = -394.36 \thinspace kJ/mol\) \(G_{f,CO} = -137.16 \thinspace kJ/mol\) \(G_{f,O2} = 0 \thinspace kJ/mol\) (since O2 is an element in its standard state) Step 3: Calculate \(\Delta G\): \(\Delta G = (-1) \cdot (-394.36) + (1) \cdot (-137.16) + \frac{1}{2} \cdot (0) = 257.2 \thinspace kJ/mol\) Step 4: Calculate \(K_P\): \(K_P = e^{-\frac{(257.2 \times 1000)}{(8.314)(298)}} = 9.799 \times 10^{-25}\) (b) At \(T = 1800 \thinspace K\): Step 2: Table A-28 provides the values for Gibbs free energy of formation at 1800 K as follows: \(G_{f,CO2} = -351.27 \thinspace kJ/mol\) \(G_{f,CO} = -117.11 \thinspace kJ/mol\) \(G_{f,O2} = 33.61 \thinspace kJ/mol\) Step 3: Calculate \(\Delta G\): \(\Delta G = (-1) \cdot (-351.27) + (1) \cdot (-117.11) + \frac{1}{2} \cdot (33.61) = 217.62 \thinspace kJ/mol\) Step 4: Calculate \(K_P\): \(K_P = e^{-\frac{(217.62 \times 1000)}{(8.314)(1800)}} = 2.785 \times 10^{-8}\) Finally, compare both calculated \(K_P\) values with those listed in Table A-28.