Question

Consider the decomposition of calcium carbonate: $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ Calculate the pressure in atm of \(\mathrm{CO}_{2}\) in an equilibrium process (a) at \(25^{\circ} \mathrm{C}\) and \((\mathrm{b})\) at \(800^{\circ} \mathrm{C}\). Assume that $$ \Delta H^{\circ}=177.8 \mathrm{~kJ} / \mathrm{mol} \text { and } \Delta S^{\circ}=160.5 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol} \text { for } $$ the temperature range.

Short answer

At 25°C, \( P_{\mathrm{CO}_2} \approx 0 \) atm; at 800°C, \( P_{\mathrm{CO}_2} \approx 0.940 \) atm.

Step-by-step solution

Write the Equilibrium Expression

For the given reaction: \[ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s) + \mathrm{CO}_{2}(g) \] since \( \mathrm{CaCO}_3 \) and \( \mathrm{CaO} \) are solids, only \( \mathrm{CO}_{2} \) appears in the equilibrium expression. Therefore, the equilibrium constant \( K_p \) in terms of pressure is given by \[ K_p = P_{\mathrm{CO}_2} \] This means the equilibrium partial pressure of \( \mathrm{CO}_{2} \), \( P_{\mathrm{CO}_2} \), is equal to \( K_p \).

Use Gibbs Free Energy Equation

The relation for Gibbs free energy is given by: \[ \Delta G = \Delta H - T \Delta S \] At equilibrium, \( \Delta G = 0 \) which implies \( \Delta H = T \Delta S \). Substituting into \[ \Delta G^\circ = -RT \ln K_p \] gives \[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ = -RT \ln K_p \] Rearranging this gives: \[ \ln K_p = \frac{-\Delta H^\circ + T \Delta S^\circ}{RT} \] and \[ K_p = e^{\frac{-\Delta H^\circ + T \Delta S^\circ}{RT}} \]

Calculate \( K_p \) at 25°C

Convert the temperature from °C to K: \[ T = 25 + 273.15 = 298.15 \, \text{K} \] Use the equation from Step 2: \[ K_p = e^{\frac{-(177800) + (298.15)(160.5)}{(8.314)(298.15)}} = e^{\frac{-177800 + 47834.475}{2478.7671}} \] Simplify the exponent: \[ = e^{\frac{-129965.525}{2478.7671}} = e^{-52.425} \] \( K_p \) at 298.15 K is extremely small, so \( P_{\mathrm{CO}_2} \) is approximately zero atm at 25°C.

Calculate \( K_p \) at 800°C

Convert the temperature from °C to K: \[ T = 800 + 273.15 = 1073.15 \, \text{K} \] Use the equation from Step 2: \[ K_p = e^{\frac{-(177800) + (1073.15)(160.5)}{(8.314)(1073.15)}} = e^{\frac{-177800 + 172261.575}{8922.641}} \] Simplify the exponent: \[ = e^{\frac{-548.425}{8922.641}} = e^{-0.06147} \] \( K_p \) at 1073.15 K is close to 1 atm, approximately 0.940 atm, so \( P_{\mathrm{CO}_2} \approx 0.940 \) atm at 800°C.

Key concepts

Equilibrium Constant

To understand how reactions reach a balance, we use the equilibrium constant. This value, denoted as \( K_p \) for reactions involving gases, describes the ratio of the concentrations or pressures of products versus reactants when a reaction is at equilibrium. When solids and liquids are present, their concentrations are constant and do not appear in the expression for \( K_p \).
For the decomposition of calcium carbonate \( \mathrm{CaCO}_3 \), the equilibrium constant is expressed simply as:\[ K_p = P_{\mathrm{CO}_2} \]This means that the equilibrium partial pressure of \( \mathrm{CO}_2 \) gas equals the equilibrium constant \( K_p \). Understanding \( K_p \) is crucial because it allows us to predict how changes in conditions like temperature or pressure will affect a reaction system.

Gibbs Free Energy

Gibbs free energy (\( \Delta G \)) helps us determine whether a reaction occurs spontaneously. It's linked to enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)) through the equation:\[ \Delta G = \Delta H - T \Delta S \]At equilibrium, \( \Delta G = 0 \), meaning there's no net change in the system. This condition leads us to the expression involving \( K_p \):\[ \Delta G^\circ = -RT \ln K_p \]Here, \( R \) is the universal gas constant and \( T \) is the temperature in Kelvin. By rearranging the equation, we can find the relationship:\[ \ln K_p = \frac{-\Delta H^\circ + T \Delta S^\circ}{RT} \]This equation helps us calculate \( K_p \) and understand the reaction's behavior under different temperatures. It's a fundamental aspect of chemical thermodynamics, ensuring we can predict and control reactions effectively.

Partial Pressure of CO2

Partial pressure is a key concept in gases, describing the pressure exerted by a single type of gas in a gas mixture. When dealing with chemical reactions like the decomposition of calcium carbonate, focusing on the partial pressure of \( \mathrm{CO}_2 \) helps in understanding the reaction dynamics.
The equilibrium constant \( K_p \) can be directly related to \( P_{\mathrm{CO}_2} \) for the reaction:\[ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s) + \mathrm{CO}_{2}(g) \]This simplifies our calculations and predictions about the system. If we know \( K_p \), we can directly determine the partial pressure of \( \mathrm{CO}_2 \) without considering other components like \( \mathrm{CaCO}_3 \) or \( \mathrm{CaO} \) because they are solids.
It's a powerful method for understanding how gas pressures relate to chemical equilibria in reactions involving gases.

Temperature Conversion

In chemistry and physics, temperature often needs to be converted between units for calculations. Most reactions and thermodynamic equations use temperatures in Kelvin, which is necessary for correct calculations. To convert from Celsius to Kelvin, simply add 273.15:\[ T(\text{K}) = T(\degree\text{C}) + 273.15 \]For the decomposition of calcium carbonate, converting temperatures from Celsius to Kelvin is crucial for accurately calculating the equilibrium constant \( K_p \) at different temperatures.
For example, converting 25\(^\circ\)C gives:\[ T = 25 + 273.15 = 298.15 \, \text{K} \]And converting 800\(^\circ\)C results in:\[ T = 800 + 273.15 = 1073.15 \, \text{K} \]This simple conversion makes complex calculations straightforward and underscores the importance of using the Kelvin scale in scientific work.