Question

For a reaction, \(\mathrm{CaCO}_{3(\mathrm{~s})} \rightarrow \mathrm{CaO}_{(0)}+\mathrm{CO}_{2(g)}\) \(\Delta_{j} H^{\circ}(\mathrm{CaO})=-635.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\Delta_{f} H^{\circ}\left(\mathrm{CO}_{2}\right)=-393.5 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}\) and \(\Delta_{f} H^{\circ}\left(\mathrm{CaCO}_{3}\right)=-1206.9 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}\) Which of the following is a correct statement? (a) A large amount of heat is evolved during the decomposition of \(\mathrm{CaCO}_{3}\). (b) Decomposition of \(\mathrm{CaCO}_{3}\) is an endothermic process and heat is provided for decomposition. (c) The amount of heat evolved cannot be calculated from the data provided. (d) \(\Delta_{r} H^{\circ}=\Sigma \Delta_{f} H^{\circ}\) (reactants) \(-\Sigma \Delta_{f} H^{\circ}\) (products)

Short answer

The correct statement is (b): Decomposition of \(\mathrm{CaCO}_{3}\) is an endothermic process and heat is provided for decomposition, because the standard reaction enthalpy \(\Delta_{r} H^\circ\) is positive (178.3 kJ/mol).

Step-by-step solution

Write the reaction and associate the given enthalpies

For the reaction \(\mathrm{CaCO}_{3(\mathrm{s})} \rightarrow \mathrm{CaO}_{(\mathrm{s})}+\mathrm{CO}_{2(\mathrm{g})}\), we have the following enthalpies given: \(\Delta_{j} H^{\circ}(\mathrm{CaO}) = -635.1 \ \mathrm{kJ} \cdot \mathrm{mol}^{-1}\), \(\Delta_{f} H^{\circ}(\mathrm{CO}_{2}) = -393.5 \ \mathrm{kJ} \cdot \mathrm{mol}^{-1}\), and \(\Delta_{f} H^{\circ}(\mathrm{CaCO}_{3}) = -1206.9 \ \mathrm{kJ} \cdot \mathrm{mol}^{-1}\).

Define the standard reaction enthalpy

The standard reaction enthalpy \(\Delta_r H^\circ\) can be calculated using Hess's law, which relates the enthalpy changes of individual steps in a reaction to the overall enthalpy change. The formula is \(\Delta_{r} H^\circ = \Sigma \Delta_{f} H^\circ \text{(products)} - \Sigma \Delta_{f} H^\circ \text{(reactants)}\).

Calculate the standard reaction enthalpy

Use the values given for the formation enthalpies to calculate \(\Delta_{r} H^\circ\): \(\Delta_{r} H^\circ = (-635.1 \ \mathrm{kJ}) + (-393.5 \ \mathrm{kJ}) - (-1206.9 \ \mathrm{kJ}) = 178.3 \ \mathrm{kJ}\cdot \mathrm{mol}^{-1}\). The positive value indicates that heat is absorbed, and therefore the decomposition of \(\mathrm{CaCO}_{3}\) is an endothermic process.

Analyze the result and choose the correct statement

Since the standard reaction enthalpy is positive, the correct statement regarding the decomposition of \(\mathrm{CaCO}_{3}\) is (b): Decomposition of \(\mathrm{CaCO}_{3}\) is an endothermic process and heat is provided for the decomposition.

Key concepts

Hess's Law

Hess's law is a principle that asserts the total enthalpy change for a chemical reaction is the same, regardless of the number of steps or its pathway, provided that the initial and final conditions are the same. This principle allows us to calculate the enthalpy change for a reaction by using known enthalpies of a series of intermediate reactions that sum up to the overall reaction.

For example, if a reaction can be expressed as the sum of two or more other reactions, the enthalpy change of the main reaction is the sum of the enthalpy changes of those reactions. This law is particularly useful in cases where the direct measurement of heat released or absorbed in a reaction is challenging. By rearranging and combining known reactions (which have measured enthalpies), we can indirectly determine the enthalpy of complex reactions.

Enthalpy Change

Enthalpy change, denoted as \(\Delta H\), is the amount of heat absorbed or released by a system under constant pressure during a chemical reaction. It is a central concept in thermodynamics and is used extensively to understand energy transfers in chemical processes.

When \(\Delta H\) is negative, the process is exothermic, meaning it releases heat to the surroundings. Conversely, a positive \(\Delta H\) indicates an endothermic process, where the system absorbs heat from its surrounding environment. Enthalpy change can be determined experimentally, or calculated using such methods as Hess's law, bond enthalpies, or standard formation enthalpies.

Endothermic Process

An endothermic process is a type of chemical reaction that absorbs heat from its surroundings, resulting in an increase in enthalpy (\(\Delta H > 0\)). This absorption of heat often causes the surrounding environment to become cooler. Common examples of endothermic processes include the melting of ice, the evaporation of water, and chemical reactions such as photosynthesis.

In an endothermic process, the energy required for the reaction to occur is typically provided in the form of thermal energy. It is important to note that endothermic reactions require a continuous supply of energy to proceed, and they often occur less spontaneously than exothermic reactions, which release energy.

Reaction Enthalpy Calculation

To calculate the enthalpy change of a reaction, also known as reaction enthalpy, one can apply the knowledge of Hess's law and standard formation enthalpies of the reactants and products involved in the reaction. The standard formation enthalpy relates to the enthalpy change that occurs when one mole of a compound is formed from its elements in their most stable form at standard conditions (usually 1 bar pressure and 25°C or 298 K).

The formula to calculate the standard reaction enthalpy is \(\Delta_r H^\circ = \Sigma \Delta_f H^\circ \text{(products)} - \Sigma \Delta_f H^\circ \text{(reactants)}\). By knowing the standard enthalpies of formation, you can sum up the enthalpies of the products and subtract the sum of the enthalpies of the reactants to find the overall reaction enthalpy. Positive or negative results indicate whether the reaction is endothermic or exothermic, respectively.