Consider the decomposition of barium carbonate: $$ \mathrm{BaCO}_{3}(s) \rightleftharpoons \mathrm{BaO}(s)+\mathrm{CO}_{2}(g) $$ Using data from Appendix \(\mathrm{C}\) , calculate the equilibrium pressure of \(\mathrm{CO}_{2}\) at (a) 298 \(\mathrm{K}\) and \((\mathbf{b}) 1100 \mathrm{K} .\)

$$ \Delta G^{\circ}_{reaction} = 177.7 \: \text{kJ/mol} $$ #Step 2: Calculate Kp at 298 K and 1100 K# Using the formula: $$ K_p = e^{-\frac{\Delta G^\circ_{reaction}}{RT}} $$ we can find the Kp at 298 K and 1100 K: At 298 K: $$ K_p(298 \: \text{K}) = e^{-\frac{177.7 \times 10^3 \: \text{J/mol}}{8.314 \: \text{J/mol} \cdot \text{K} \times 298 \: \text{K}}} $$ $$ K_p(298 \: \text{K}) = 4.42 \times 10^{-12} $$ At 1100 K: $$ K_p(1100 \: \text{K}) = e^{-\frac{177.7 \times 10^3 \: \text{J/mol}}{8.314 \: \text{J/mol} \cdot \text{K} \times 1100 \: \text{K}}} $$ $$ K_p(1100 \: \text{K}) = 4.40 \times 10^{-2} $$ #Step 3: Find the Equilibrium Pressure of CO2# From the balanced equation, we know that the stoichiometric coefficients are all equal to 1. Thus, for every mole of BaCO3 that decomposes, one mole of CO2 is produced. We can then equate the Kp values with the equilibrium partial pressure of CO2 (P_CO2): At 298 K: $$ K_p(298 \: \text{K}) = P_{CO2} $$ $$ 4.42 \times 10^{-12} = P_{CO2}(298 \: \text{K}) $$ At 1100 K: $$ K_p(1100 \: \text{K}) = P_{CO2} $$ $$ 4.40 \times 10^{-2} = P_{CO2}(1100 \: \text{K}) $$ So, the equilibrium pressure of CO2 at 298 K is \(4.42 \times 10^{-12}\: \text{atm}\) and at 1100 K is \(4.40 \times 10^{-2}\: \text{atm}\).