Question

Calculate \(\Delta G_{\mathrm{rxn}}^{\circ}\) for the reaction: $$\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$Use the following reactions and given \(\Delta G_{\mathrm{rxn}}^{\circ}\) values: \(\mathrm{Ca}(s)+\mathrm{CO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CaCO}_{3}(s) \Delta G_{\mathrm{rxn}}^{\circ}=-734.4 \mathrm{~kJ}\) \(2 \mathrm{Ca}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CaO}(s) \quad \Delta G_{\mathrm{rxn}}^{\circ}=-1206.6 \mathrm{~kJ}\)

Short answer

\(\Delta G_{\mathrm{rxn}}^\circ_{overall} = 734.4 \mathrm{~kJ} - 603.3 \mathrm{~kJ} = 131.1 \mathrm{~kJ}\)

Step-by-step solution

- Write Down Given Reactions and Their \(\Delta G_{\mathrm{rxn}}^\circ\) Values

We are given two reactions with their corresponding \(\Delta G_{\mathrm{rxn}}^\circ\) values:\[\mathrm{Ca}(s)+\mathrm{CO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{CaCO}_{3}(s), \quad \Delta G_{\mathrm{rxn}}^\circ = -734.4 \mathrm{~kJ} \]\[2 \mathrm{Ca}(s) + \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CaO}(s), \quad \Delta G_{\mathrm{rxn}}^\circ = -1206.6 \mathrm{~kJ} \]We are tasked to find the \(\Delta G_{\mathrm{rxn}}^\circ\) for the reaction:\[\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s) + \mathrm{CO}_{2}(g) \]

- Reverse the First Reaction

The first reaction needs to be reversed to match the desired reaction format:\[\mathrm{CaCO}_{3}(s) \rightarrow \mathrm{Ca}(s) + \mathrm{CO}_{2}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \]Since we reversed the reaction, we also need to change the sign of \(\Delta G_{\mathrm{rxn}}^\circ\):\[\Delta G_{\mathrm{rxn}}^\circ_{reversed} = 734.4 \mathrm{~kJ} \]

- Modify the Second Reaction

The second reaction needs to be adjusted so it matches the coefficients of the desired reaction. To accomplish this, we divide the reaction coefficients by 2: \[\mathrm{Ca}(s) + \frac{1}{2}\mathrm{O}_{2}(g) \rightarrow \mathrm{CaO}(s) \]The \(\Delta G_{\mathrm{rxn}}^\circ\) for the modified reaction is also halved:\[\Delta G_{\mathrm{rxn}}^\circ_{modified} = \frac{-1206.6}{2} \mathrm{~kJ} = -603.3 \mathrm{~kJ} \]

- Add the Reactions Together

We add the reversed first reaction and the halved second reaction to get the desired reaction: \[\mathrm{CaCO}_{3}(s) \rightarrow \mathrm{Ca}(s) + \mathrm{CO}_{2}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \]\[\mathrm{Ca}(s) + \frac{1}{2}\mathrm{O}_{2}(g) \rightarrow \mathrm{CaO}(s) \]Summing these reactions gives us the desired overall reaction:\[\mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(s) + \mathrm{CO}_{2}(g) \]The \(\Delta G_{\mathrm{rxn}}^\circ\) for the overall reaction is the sum of the \(\Delta G_{\mathrm{rxn}}^\circ\) values from the modified reactions: \[\Delta G_{\mathrm{rxn}}^\circ_{overall} = 734.4 \mathrm{~kJ} + (-603.3 \mathrm{~kJ}) \]

Key concepts

Thermodynamics

Thermodynamics is a branch of physics that deals with the relationships between heat, work, temperature, and energy. In the context of chemistry, it particularly focuses on how these relationships affect chemical reactions and the physical changes of materials. A fundamental concept in thermodynamics is the law of conservation of energy, which states that within a closed system, energy can change form but cannot be created or destroyed.

When studying chemical reactions, thermodynamics helps us to understand and predict whether a reaction will occur spontaneously. This is assessed by looking at the Gibbs free energy, symbolized as \( \Delta G \), a measure of the potential usefulness of the energy within a chemical system. To calculate the change in Gibbs free energy for a reaction, we often use the formula \( \Delta G = \Delta H - T\Delta S \), where \( \Delta H \) is the change in enthalpy, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the change in entropy.

If the value of \( \Delta G \) is negative, the reaction is spontaneous under constant temperature and pressure, which means it can proceed without energy input from its surroundings. Conversely, if \( \Delta G \) is positive, the reaction is non-spontaneous and requires additional energy to proceed. When \( \Delta G \) equals zero, the system is in equilibrium and no net change occurs.

Chemical Reactions

Chemical reactions involve the transformation of one set of chemical substances to another. During these processes, the bonds between atoms are broken and new bonds are formed, resulting in the production of one or more different substances. In the context of the exercise, the decomposition of calcium carbonate \( \mathrm{CaCO}_{3} \) into calcium oxide \( \mathrm{CaO} \) and carbon dioxide \( \mathrm{CO}_{2} \) is an example of a chemical reaction.

In order to accurately determine whether a reaction will proceed, chemists perform Gibbs free energy calculations. In the given step-by-step solution, the key technique used to calculate the Gibbs free energy of the target reaction is called Hess's law, which states that the total enthalpy change of a reaction is the same regardless of the number of steps the reaction takes. By manipulating and combining given reactions — reversing and halving, in this case — we can determine the Gibbs free energy change for the desired reaction without directly measuring it experimentally.

Enthalpy and Entropy

Enthalpy, symbolized by \( H \), is a measure of the total energy of a thermodynamic system. It includes both the internal energy of the system and the product of its pressure and volume. During a chemical reaction, the enthalpy change \( \Delta H \) shows the amount of heat absorbed or released under constant pressure. If the process releases heat, it is exothermic (\( \Delta H < 0 \) ), and if it absorbs heat, it is endothermic (\( \Delta H > 0 \) ).

Entropy, on the other hand, symbolized by \( S \), represents the degree of randomness or disorder within a system. The second law of thermodynamics states that the total entropy of an isolated system always increases over time. Entropy change \( \Delta S \) indicates whether the disorder of the system has increased (\(\Delta S > 0\)) or decreased (\(\Delta S < 0\)). In a chemical reaction, an increase in the number of gas molecules typically leads to an increase in entropy.

The relationship between enthalpy and entropy is crucial in understanding Gibbs free energy. The balance between these two properties allows scientists to predict the spontaneity of a reaction. While enthalpy relates to the heat involved in the reaction, entropy adds another dimension that accounts for the molecular disorder, resulting in a more comprehensive representation of the reaction's feasibility through the \( \Delta G \) calculation.