Question

Calculate \(\Delta H\) for $$\mathrm{Ca}(s)+\frac{1}{2} \mathrm{O}_{2}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{CaCO}_{3}(s)$$given the following reactions: $$\begin{aligned}\mathrm{Ca}(s)+\frac{1}{2} \mathrm{O}_{2}(g) & \longrightarrow \mathrm{CaO}(s) & & \Delta H=-635.1 \mathrm{~kJ} \\\\\mathrm{CaCO}_{3}(s) & \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) & & \Delta H=178.3 \mathrm{~kJ}\end{aligned}$$

Short answer

\(\Delta H = -813.4 \, \text{kJ} \)

Step-by-step solution

- Write the target equation

Identify the target equation for which we need to calculate the enthalpy change\[ \mathrm{Ca}(s) + \frac{1}{2} \mathrm{O}_2(g) + \mathrm{CO}_2(g) \rightarrow \mathrm{CaCO}_3(s) \]

- Write given equations with their enthalpy changes

List the provided reactions and their enthalpy changes:1) \[ \mathrm{Ca}(s) + \frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CaO}(s) \quad \Delta H = -635.1 \, \text{kJ} \]2) \[ \mathrm{CaCO}_3(s) \rightarrow \mathrm{CaO}(s) + \mathrm{CO}_2(g) \quad \Delta H = 178.3 \, \text{kJ} \]

- Reverse the second equation

Reverse the second reaction so that the products become reactants and vice versa:\[ \mathrm{CaO}(s) + \mathrm{CO}_2(g) \rightarrow \mathrm{CaCO}_3(s) \quad \Delta H = -178.3 \, \text{kJ} \]Remember to change the sign of \(\Delta H\).

- Add the equations together

Sum the modified equations to get the target equation:\[ \mathrm{Ca}(s) + \frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CaO}(s) \quad \Delta H = -635.1 \, \text{kJ} \]\[ \mathrm{CaO}(s) + \mathrm{CO}_2(g) \rightarrow \mathrm{CaCO}_3(s) \quad \Delta H = -178.3 \, \text{kJ} \]

- Simplify the complete equation

Combine the left and right sides of the equations:\[ \mathrm{Ca}(s) + \frac{1}{2} \mathrm{O}_2(g) + \mathrm{CO}_2(g) \rightarrow \mathrm{CaCO}_3(s) \]Combine their enthalpy changes to get the final enthalpy change, \(\Delta H\):\[ \Delta H = -635.1 \, \text{kJ} + (-178.3 \, \text{kJ}) = -813.4 \, \text{kJ} \]

Key concepts

Hess's Law

Hess's Law is a fundamental principle in thermochemistry. It states that the total enthalpy change for a chemical reaction is the same, regardless of the pathway or number of steps taken to complete the reaction. This means that if a reaction can be expressed as a sequence of steps, the sum of the enthalpy changes for these steps will be equal to the enthalpy change for the overall reaction. In other words, if you have multiple reactions that can be added together to give a final reaction, the sum of the enthalpy changes of these reactions will equal the enthalpy change of the final reaction.

When solving problems using Hess's Law, you often have to manipulate the given equations—reversing them, and sometimes multiplying by coefficients—to align them with the target reaction. Each manipulation affects the enthalpy change: reversing a reaction changes the sign of \(\text{\Delta H}\), and multiplying by a factor changes the enthalpy change by that factor. Crucially, Hess's Law allows us to calculate enthalpy changes for reactions where direct measurement is challenging.

Thermochemistry

Thermochemistry is the branch of chemistry that deals with the energy changes occurring during chemical reactions. Specifically, it focuses on the enthalpy changes (\(\text{\Delta H}\)) that accompany reactions. Enthalpy is a measure of the total energy of a thermodynamic system, including both internal energy and the energy required to make room for it by displacing its environment. The enthalpy change of a reaction indicates whether a reaction is exothermic (releases heat, \( \text{\Delta H} \) is negative) or endothermic (absorbs heat, \(\text{\Delta H}\) is positive).

In practical terms, understanding thermochemistry allows chemists to predict the heat changes accompanying chemical processes. This is crucial in fields like industrial chemistry, where controlling the temperature of reactions can be vital for safety and efficiency. The problems involving enthalpy changes often require interpreting or combining given data using principles like Hess's Law to find the overall enthalpy change for a desired process.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products through the breaking and formation of chemical bonds. In any reaction, the law of conservation of energy must be obeyed. This means that the total energy of the reactants must equal the total energy of the products plus the energy change of the reaction.

In the context of the exercise, the reactions provided are:

• \(\text{Ca(s) + }\frac{1}{2} \text{O}_{2} \text{(g) }\rightarrow \text{CaO(s)}, \(\text{\Delta H = -635.1 kJ}\)\)
• \(\text{CaCO}_{3}\text{(s) }\rightarrow \text{CaO(s) + CO}_{2}\text{(g)}, \(\text{\Delta H = 178.3 kJ}\)\)

To find the enthalpy change (\text{\Delta H}) for the target reaction \(\text{Ca(s) + }\frac{1}{2} \text{O}_{2}\text{(g) + CO}_{2}\text{(g) }\rightarrow \text{CaCO}_{3}\text{(s)}\), we reverse the second provided reaction, noting that this changes the sign of \(\text{\Delta H}\). Thus, the \(\text{\Delta H}\) for the reverse reaction becomes -178.3 kJ. Summing the enthalpy changes of these adjusted reactions gives the overall enthalpy change for the target reaction:

\(\text{\Delta H}_{\text{total}} = -635.1 \text{ kJ} + (-178.3 \text{ kJ}) = -813.4 \text{ kJ}.\)

Understanding these changes helps in predicting the energy requirements or releases for processes in chemical synthesis and industrial applications.