Question

The thermodynamic data at 298 K for the decomposition reaction of limestone at equilibrium is given below: \[ \mathrm{CaCO_3(s)} \rightleftharpoons \mathrm{CaO(s)} + \mathrm{CO_2(g)} \] 

 

Hint:

For decomposition reactions of solids, $K_p$ equals the equilibrium pressure of the gaseous product. Use $\Delta G = \Delta H - T\Delta S$ for temperature corrections.

Answer 1

Correct Answer: 4.2

Step 1: Write the decomposition reaction.

⇆ CaCO₃(s) → CaO(s) + CO₂(g)

Step 2: Calculate ΔG°₂₉₈ at 298 K.

ΔG°₂₉₈ = [(-604.0) + (-394.4)] - (-1128.8) = +130.4 kJ mol^-1

Step 3: Calculate ΔH° and ΔS°.

ΔH° = [(-635.1) + (-393.5)] - (-1206.9) = +178.3 kJ mol^-1

ΔS° = (ΔH° - ΔG°₂₉₈) / 298 = ((178.3 - 130.4)×10³) / 298 = 160.7 J mol^-1K^-1

Step 4: Estimate ΔG°₁₂₀₀.

ΔG°₁₂₀₀ = ΔH° - TΔS° = 178.3×10³ - 1200(160.7) = -14.5×10³ J mol^-1

Step 5: Compute equilibrium constant.

ΔG° = -RT ln K_p → K_p = e^{-ΔG° / RT}

K_p = e^{14500 / (8.314 × 1200)} = e^1.45 = 4.26

Step 6: Conclusion.

Since K_p = P_CO2, partial pressure of CO₂ = 1.01 atm (approx.).