Chapter 17, Problem 17.29QP

Consider the decomposition of calcium carbonate:

${\text{CaCO}}_3(s)\stackrel{}{\rightleftharpoons }\text{CaO}(s)+{\text{CO}}_2(g)$

Calculate the pressure in atm of CO₂ in an equilibrium process (a) at 25°C and (b) at 800°C. Assume that ΔH° = 177.8 kJ/mol and ΔS° = 160.5 J/K · mol for the temperature range.

Interpretation Introduction

Interpretation:

The equilibrium pressure ($Kp$) values has to be calculated given the calcium carbonate $CaC{O}_3$ decomposition reaction at ${25}^{°}C$.

Concept Information:

Equilibrium constant: Concentration of the products to the respective molar concentration of reactants it is called equilibrium constant.  If the K value is less than one the reaction will move to the left side and the K values is higher (or) greater than one the reaction will move to the right side of reaction.

Kp:  The equilibrium constant calculated from the partial pressures of a reaction equation. It is used to express the relationship between product pressures and reactant pressures. It is unites number, although it relates the pressures.

$\begin{array}{l}\text{Kp}\text{=}\text{Kc}{\text{(RT)}}^{\text{Δn}}\text{(or})\\ Kc=\frac{\text{Kp}}{{\text{(RT)}}^{\text{Δn}}}(or)Kp={\text{(RT)}}^{\text{Δn}}\\ \text{Δn}\text{=Product - Reactant}\end{array}$

Thermodynamics is the branch of science that relates heat and energy in a system.  The laws of thermodynamics explain the fundamental quantities such as temperature, energy and randomness in a system.  Entropy is the measure of randomness in a system.  For a spontaneous process there is always a positive change in entropy. Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter G.  All spontaneous process is associated with the decrease of free energy in the system.  The equation given below helps us to calculate the change in free energy in a system.

$\text{ΔG}°\text{ = }\Delta {\rm H}°\text{- T}\Delta \text{S}°$

Where,

$\text{ΔG}°$ is the change in free energy of the system

$\Delta {\rm H}°$ is the change in enthalpy of the system

$\text{T}$ is the absolute value of the temperature

$\Delta \text{S}°$ is the change in entropy in the system

Explanation of Solution

The calcium carbonate decomposition equilibrium expression given by,

$\begin{array}{l}{\text{CaCO}}_{\text{3}}(s)\rightleftharpoons \text{CaO(s)}\text{+}{\text{CO}}_{\text{2}}\text{(g)}\\ \text{Kp}{\text{= CO}}_{\text{2}}[\because \text{Consider}\text{gases}\text{molecule}\text{only}]\\ \text{ΔH}°\text{=177}\text{.8}\text{KJ/mol}\\ \text{Kp}\text{=}\text{Kc}{\text{[8}\text{.314atm/K}\text{.atm}\text{×}\text{T]}}^{\text{Δn}}\\ \text{Kp}\text{=}\text{Kc}{\text{(RT)}}^{\text{Δn}}\text{(or})\\ Kc=\frac{\text{Kp}}{{\text{(RT)}}^{\text{Δn}}}(or)Kp={\text{(RT)}}^{\text{Δn}}\\ \text{Here}\text{Δn}\text{=}1\text{-0}\text{=}\text{1 }\text{(Product - Reactant)}\\ \text{and}\text{given}\text{Temprature}{25}^{\text{0}}\text{C}\\ \text{T}\text{=}\text{(25+ 273)}\text{K}\\ =298\text{K}\\ \text{ΔG}=\text{ΔH-}\text{TΔS }\\ {\text{ΔG}}^0={\text{ΔH}}^0\text{-}{\text{TΔS}}^0-----[1]\\ {\text{Calculate the 25}}^{\text{0}}\text{C }\\ \text{The respactive values are substituted in equation (1)}\\ {\text{ΔG}}^0={\text{ΔH}}^0\text{-}{\text{TΔS}}^0=(177.8\times {10}^{3}J/mol)-(298K)(160.5\text{J/K}\text{.mol})\\ {\text{ΔG}}^0=130.0\times {10}^3\text{J/mol}\\ (P_{C{O}_2})=Kp=e^{\frac{{\text{-ΔG}}^{\text{0}}}{\text{RT}}}=e^{\frac{-130.0\times {10}^3\text{J/mol}}{\text{(8}\text{.314J/K}\text{.mol)(298K)}}}\\ (P_{C{O}_2})=Kp=e^{-52.47}\\ \text{Taking}\text{for}\text{antilog}\text{both}\text{sides, we }\text{get}\text{. }\\ \text{Kp= }\underset{\_}{\text{1}\text{.6}\times {\text{10}}^{\text{-23}}\text{atm}}\end{array}$

Interpretation Introduction

Interpretation:

The equilibrium pressure ($Kp$) values has to be calculated given the enthalpy $(\Delta H°)$  and entropy $(\Delta S°)$ calcium carbonate decomposition reaction at ${800}^{°}C$.

Concept Information:

Equilibrium constant: Concentration of the products to the respective molar concentration of reactants it is called equilibrium constant.  If the K value is less than one the reaction will move to the left side and the K values is higher (or) greater than one the reaction will move to the right side of reaction.

Kp:  The equilibrium constant calculated from the partial pressures of a reaction equation. It is used to express the relationship between product pressures and reactant pressures. It is unites number, although it relates the pressures.

$\begin{array}{l}\text{Kp}\text{=}\text{Kc}{\text{(RT)}}^{\text{Δn}}\text{(or})\\ Kc=\frac{\text{Kp}}{{\text{(RT)}}^{\text{Δn}}}(or)Kp={\text{(RT)}}^{\text{Δn}}\\ \text{Δn}\text{=Product - Reactant}\end{array}$

Thermodynamics is the branch of science that relates heat and energy in a system.  The laws of thermodynamics explain the fundamental quantities such as temperature, energy and randomness in a system.  Entropy is the measure of randomness in a system.  For a spontaneous process there is always a positive change in entropy. Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter G.  All spontaneous process is associated with the decrease of free energy in the system.  The equation given below helps us to calculate the change in free energy in a system.

$\text{ΔG}°\text{ = }\Delta {\rm H}°\text{- T}\Delta \text{S}°$

Where,

$\text{ΔG}°$ is the change in free energy of the system

$\Delta {\rm H}°$ is the change in enthalpy of the system

$\text{T}$ is the absolute value of the temperature

$\Delta \text{S}°$ is the change in entropy in the system

Explanation of Solution

First we calculate the enthalpy and entropy values for given decomposition reaction at $800°C$.

$\begin{array}{l}{\text{CaCO}}_{\text{3}}(s)\rightleftharpoons \text{CaO(s)}\text{+}{\text{CO}}_{\text{2}}\text{(g)}\\ \text{Kp}{\text{= CO}}_{\text{2}}\\ \text{ΔH}\text{=177}\text{.8}\text{KJ/mol}\\ \text{Kp}\text{=}\text{Kc}{\text{[8}\text{.314}\text{.atm/K}\text{.atm}\text{×}\text{T]}}^{\text{Δn}}\\ \text{Kp}\text{=}\text{Kc}{\text{(RT)}}^{\text{Δn}}\text{(or})\\ Kc=\frac{\text{Kp}}{{\text{(RT)}}^{\text{Δn}}}(or)Kp={\text{(RT)}}^{\text{Δn}}\\ (P_{C{O}_2})=Kp=e^{\frac{{\text{-ΔG}}^{\text{0}}}{\text{RT}}}\\ \text{Here}\text{Δn}\text{=}1\text{-0}\text{=}\text{1 }\text{(Product - Reactant)}\\ \text{and}\text{given}\text{Temprature}{350}^{\text{0}}\text{C}\\ \text{T}\text{=}\text{(800+ 273)}\text{K}\\ =1073\text{K}\\ \text{ΔG}=\text{ΔH-}\text{TΔS }\\ {\text{ΔG}}^{°}={\text{ΔH}}^{°}\text{-}{\text{TΔS}}^{°}-----[2]\\ \text{The respactive values are substituted for equation (2)}\\ {\text{ΔG}}^{°}={\text{ΔH}}^{°}\text{-}{\text{TΔS}}^{°}=(177.8\times {10}^{3}J/mol)-(1073K)(160.5\text{J/K}\text{.mol})\\ {\text{ΔG}}^0=5.584\times {10}^3\text{J/mol}\\ (P_{C{O}_2})=Kp=e^{\frac{{\text{-ΔG}}^{\text{0}}}{\text{RT}}}=\frac{-5.53\times {10}^3\text{J/mol}}{\text{(8}\text{.314J/K}\cdot \text{mol)(1073K)}}\\ (P_{C{O}_2})=Kp=e^{-0.625}\\ \text{Kp= 0}\text{.535 atm}\end{array}$ 

Calcium carbonate, partial pressure $(Kp)$ value is $\underset{\_}{\text{0}\text{.535 atm}}$