Question

Determine \(\Delta H\) for the reaction $$ \mathrm{CaCO}_{3} \rightarrow \mathrm{CaO}+\mathrm{CO}_{2} $$ given these data: $$ \begin{array}{c} 2 \mathrm{Ca}+2 \mathrm{C}+3 \mathrm{O}_{2} \rightarrow 2 \mathrm{CaCO}_{3} \Delta H=-2,414 \mathrm{~kJ} \\ \mathrm{C}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} \Delta H=-393.5 \mathrm{~kJ} \end{array} $$ \(2 \mathrm{Ca}+\mathrm{O}_{2} \rightarrow 2 \mathrm{CaO} \Delta H=-1,270 \mathrm{~kJ}\)

Short answer

\(\Delta H = +178.5\, \mathrm{kJ}\).

Step-by-step solution

Identify Target Reaction

The target reaction is \[\mathrm{CaCO}_{3} \rightarrow \mathrm{CaO}+\mathrm{CO}_{2}.\] We want to find \(\Delta H\) for this reaction.

List Given Reactions and Their Enthalpies

1. \(2 \mathrm{Ca} + 2 \mathrm{C} + 3 \mathrm{O}_{2} \rightarrow 2 \mathrm{CaCO}_{3}\), \(\Delta H = -2414\, \mathrm{kJ}\).2. \(\mathrm{C} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}\), \(\Delta H = -393.5\, \mathrm{kJ}\).3. \(2 \mathrm{Ca} + \mathrm{O}_{2} \rightarrow 2 \mathrm{CaO}\), \(\Delta H = -1270\, \mathrm{kJ}\).

Determine Individual Reaction Contributions

We aim to find the enthalpy change for the decomposition of \(\mathrm{CaCO}_{3}\). Write out how the given reactions can be used or modified:

Express the Target Reaction Using Given Reactions

To derive the reaction \(\mathrm{CaCO}_{3} \rightarrow \mathrm{CaO} + \mathrm{CO}_{2}\):1. Use the reverse of the first reaction divided by 2: \[\mathrm{CaCO}_{3} \rightarrow \mathrm{Ca} + \mathrm{C} + \frac{3}{2} \mathrm{O}_{2}.\] \(\Delta H = +1207\, \mathrm{kJ}\) (because we reverse and divide)2. Use the second reaction: \[\mathrm{C} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}.\] \(\Delta H = -393.5\, \mathrm{kJ}\)3. Use half of the third reaction: \[\mathrm{Ca} + \frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{CaO}.\] \(\Delta H = -635\, \mathrm{kJ}\) (because we divide by 2)

Calculate \\(\Delta H\\) for the Target Reaction

Add up the modified enthalpy changes:\[\Delta H = +1207\, \mathrm{kJ} + (-393.5\, \mathrm{kJ}) + (-635\, \mathrm{kJ}) = +178.5\, \mathrm{kJ}.\] Therefore, the enthalpy change \(\Delta H\) for the decomposition of \(\mathrm{CaCO}_{3}\) is \(+178.5\, \mathrm{kJ}\).

Key concepts

Enthalpy Change

Enthalpy change is a vital concept in chemistry that represents the heat absorbed or released during a chemical reaction. It is denoted by \( \Delta H \) and gives insights into whether a process is endothermic or exothermic. An endothermic reaction, where energy is absorbed from the surroundings, has a positive \( \Delta H \), whereas an exothermic one, which releases energy, has a negative \( \Delta H \).
In this exercise, we aim to calculate the enthalpy change for the decomposition reaction \( \mathrm{CaCO}_{3} \rightarrow \mathrm{CaO} + \mathrm{CO}_{2} \). To achieve this, Hess's Law, which allows the calculation of an overall enthalpy change through the summation of steps forming a reaction pathway, is utilized.
This means by rearranging and combining various given reactions and their respective \( \Delta H \) values, the overall enthalpy change for a target reaction can be determined. In this case, the known reaction enthalpies are used to find that \( \Delta H = +178.5 \, \mathrm{kJ} \) for the decomposition process, indicating it is endothermic.

Thermochemical Equations

Thermochemical equations are chemical equations that include enthalpy changes, providing a more detailed picture of the energy transformation in a chemical reaction. These equations not only show the reactants and products but also the energy change associated with the reaction.
For the reactions given in the exercise, the thermochemical equations include:

• \( 2 \mathrm{Ca} + 2 \mathrm{C} + 3 \mathrm{O}_{2} \rightarrow 2 \mathrm{CaCO}_{3} \quad \Delta H = -2414 \, \mathrm{kJ} \)
• \( \mathrm{C} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} \quad \Delta H = -393.5 \, \mathrm{kJ} \)
• \( 2 \mathrm{Ca} + \mathrm{O}_{2} \rightarrow 2 \mathrm{CaO} \quad \Delta H = -1270 \, \mathrm{kJ} \)

Each equation indicates the amount of energy gained or lost, which helps in calculating the target reaction enthalpy. By manipulating these equations, such as reversing or dividing, new equations that suit the target reaction can be crafted and an accurate \( \Delta H \) can be calculated.
This highlights the significance of thermochemical equations in providing essential energy data needed for calculations in chemistry, especially in applying Hess's Law.

Chemical Reaction Analysis

Chemical reaction analysis involves understanding and manipulating chemical equations to gain insights into their processes and outcomes. It often involves breaking down or building up reactions using known equations, a crucial step in applying Hess's Law. This exercise demonstrates a clear method of using given known chemical reactions to deduce the unknown enthalpy change of another reaction.
By reversing, dividing, or employing portions of known reactions, each component contributes to form the target reaction. In our case, the given reactions were modified as follows:

• Reversing and halving the first reaction provided the equation for \( \mathrm{CaCO}_{3} \) decomposition.
• Using the second reaction directly depicted the formation of \( \mathrm{CO}_{2} \) from carbon and oxygen.
• Halving the third reaction allowed the formation of \( \mathrm{CaO} \).

Through these modifications, the cumulative enthalpy changes calculated to \( +178.5 \, \mathrm{kJ} \), successfully giving the \( \Delta H \) for the target reaction.
Fundamentally, chemical reaction analysis in this context showcases the power of understanding and manipulating reactions to uncover hidden details about their energetic pathways.