Problem 138 - General Chemistry: Principles and Modern Applications

Question

The standard molar heats of combustion of C(graphite) and \(\mathrm{CO}(\mathrm{g})\) are -393.5 and \(-283 \mathrm{kJ} / \mathrm{mol}\) respectively. Use those data and that for the following reaction $$\mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{COCl}_{2}(\mathrm{g}) \quad \Delta H^{\circ}=-108 \mathrm{kJ}$$ to calculate the standard molar enthalpy of formation of \(\mathrm{COCl}_{2}(\mathrm{g})\).

Step-by-Step Solution

Answer

The standard molar enthalpy of formation of \(COCl_{2}(g)\) is \(175 kJ/mol\).

Step 1: Identify Given Data

Begin by identifying and writing down the given information. The standard molar heats of combustion of C(graphite) and \(CO(g)\) are \(-393.5 kJ/mol\) and \(-283 kJ/mol\) respectively. The reaction \(CO(g) + Cl_{2}(g) \rightarrow COCl_{2}(g)\) has a heat of reaction \(\Delta H^{\circ} = -108 kJ\).

Step 2: Write The Heat Formation Reactions

The standard molar enthalpy of formation of a compound is the heat absorbed or evolved when 1 mole of the compound is formed from its elements in their standard states. Write heat formation reactions for the known substances: \(C(graphite) + 0.5O_{2}(g) \rightarrow CO(g)\) with \(\Delta H_{f_1} = -283 kJ/mol\), and \(C(graphite) + O_{2}(g) \rightarrow CO_{2}(g)\) with \(\Delta H_{f_2} = -393.5 kJ/mol\).

Step 3: Use Hess's Law To Write The Desired Reaction

Hess's Law states that the heat of any reaction depends only on the initial and final states and is independent of the path or nature of the reaction. The thermochemical equation desired is \(C(graphite) + O_{2}(g) + Cl_{2}(g) \rightarrow COCl_{2}(g)\), from which we'll find \(\Delta H_{f_3}\). By liberally applying Hess's Law, we can subtract the heat enthalpy of formation for \(CO(g)\) from the heat of reaction supplied to isolate the heat enthalpy for \(COCl_{2}(g)\). \(\Delta H_{f_3} = \Delta H_{R} - \Delta H_{f_1}\).

Step 4: Calculate The Enthalpy of Formation

Substitute the known values into the equation from Step 3: \(\Delta H_{f_3} = -108 kJ - (-283 kJ/mol) = 175 kJ/mol\).

Key Concepts

Hess's Law

Hess's Law is a fundamental principle in chemistry that allows us to simplify the calculation of reaction enthalpies. It states that the total enthalpy change for a chemical reaction is the same, regardless of the route by which the chemical change takes place, provided the initial and final conditions are identical. This means that if you know the enthalpy changes for a series of reactions that sum to the desired reaction, you can add them up to find the overall change.

In our example, we're tasked with determining the enthalpy of formation for phosgene \(\text{COCl}_{2}(\text{g})\). We do this by rearranging equations from known enthalpy changes, such as the combustion of carbon and carbon monoxide. By understanding the principle of Hess's Law, you can break down and rebuild reactions, ensuring even complex enthalpy calculations are manageable.

To apply Hess's Law:

• Identify the individual steps that, when combined, create the reaction of interest.
• Use known enthalpy changes for these steps, obtained from experiments or literature.
• Add or subtract these values according to their roles in forming the target reaction.

As a powerful tool in thermochemistry, Hess’s Law confirms that energy conservation universal in chemical reactions, regardless of how those reactions are physically carried out.

Standard Molar Heat of Combustion

The standard molar heat of combustion is the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions (298 K, 1 atm). It provides vital information about the stability and energy content of different substances. For carbon, the value is typically very negative, indicating it releases a substantial amount of energy when burned.

When a material combusts, it often forms carbon dioxide and water—and sometimes other products. These reactions are exothermic, meaning they release heat. For example, the combustion reactions given in the exercise:

• Combustion of graphite: \(\mathrm{C(graphite) + O_{2}(g) \rightarrow CO_{2}(g)\) with \(\Delta H^{\circ}=-393.5\) \(\text{kJ/mol}\)
• Combustion of carbon monoxide: \(\mathrm{CO(g) + 0.5O_{2}(g) \rightarrow CO_{2}(g)\) with \(\Delta H^{\circ}=-283\) \(\text{kJ/mol}\)

By understanding these values, you can deduce other thermochemical properties through related reactions and the application of Hess's Law.

Thermochemical Equations

Thermochemical equations are an extension of standard chemical reaction equations, incorporating energy changes, usually as enthalpy (∆H). They provide insights into the energy profiles of reactions, revealing whether a reaction is exothermic (releases heat) or endothermic (absorbs heat). These equations help visualize and calculate energy dynamics in chemical reactions.

In thermochemical equations:

• Enthalpy changes are given alongside the chemical equations.
• The coefficients express the stoichiometry of reactants and products, adhering to the law of conservation of mass.
• Equations can be manipulated by reversing, multiplying, or adding them, impacting the ∆H values accordingly, essential when using Hess's Law.

In the problem provided, you see the reaction \(\text{CO(g) + Cl}_{2}(\text{g}) \rightarrow \text{COCl}_{2}(\text{g})\) with \(\Delta H^{\circ} = -108\) \(\text{kJ/mol}\). This shows energy is released when \(\text{COCl}_{2}\) forms. By applying thermochemical equations, all pieces including this known enthalpy change contribute to solving for unknowns, like the standard molar enthalpy of formation for \(\text{COCl}_{2}(\text{g})\).