Question

The value of \(\Delta_{\mathrm{f}} H^{\circ}\) of \(\mathrm{U}_{3} \mathrm{O}_{8}(\mathrm{~s})\) is \(-853.5 \mathrm{~kJ}\) \(\mathrm{mol}^{-1} . \Delta H^{\circ}\) for the reaction: \(3 \mathrm{UO}_{2}(\mathrm{~s})\) \(+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{U}_{3} \mathrm{O}_{8}(\mathrm{~s})\), is \(-76.00 \mathrm{~kJ} .\) The value of \(\Delta_{\mathrm{f}} H^{\circ}\) of \(\mathrm{UO}_{2}(\mathrm{~s})\) is (a) \(-259.17 \mathrm{~kJ}\) (b) \(-310.17 \mathrm{~kJ}\) (c) \(+259.17 \mathrm{~kJ}\) (d) \(930.51 \mathrm{~kJ}\).

Short answer

\(-259.17 \mathrm{~kJ}\)

Step-by-step solution

Understand the Hess's Law

According to Hess's Law, the total enthalpy change during a chemical reaction is the same whether the reaction occurs in one step or in a series of steps. This means that we can use the enthalpy of formation of products and reactants to calculate the enthalpy change for the overall reaction.

Write the equation for the enthalpy change of the reaction

The enthalpy change for the given reaction can be expressed as the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants. In formula terms, \( \Delta H^\circ_{reaction} = \Delta_{\mathrm{f}} H^\circ_{\mathrm{U}_{3}\mathrm{O}_{8}} - 3 \times \Delta_{\mathrm{f}} H^\circ_{\mathrm{UO}_{2}} - \Delta_{\mathrm{f}} H^\circ_{\mathrm{O}_{2}} \). Note that the enthalpy of formation of an element in its standard state, like \(\text{O}_2(g)\), is zero.

Plug in the known values and solve for the unknown

The \(\Delta H^\circ_{reaction}\) for the reaction is given as \(-76.00\mathrm{~kJ}\), and the \(\Delta_{\mathrm{f}} H^\circ\) of \(\mathrm{U}_{3}\mathrm{O}_{8}(\mathrm{~s})\) is \(-853.5\mathrm{~kJ}\). Since the enthalpy of formation of \(O_2(g)\) is 0 (as it's in its standard state), we have \(-76.00 = -853.5 - 3 \times \Delta_{\mathrm{f}} H^\circ_{\mathrm{UO}_{2}} - 0\).

Solve for \(\Delta_{\mathrm{f}} H^\circ_{\mathrm{UO}_{2}}\)

Rearranging the previous step's equation to solve for \(\Delta_{\mathrm{f}} H^\circ_{\mathrm{UO}_{2}}\), we get \(\Delta_{\mathrm{f}} H^\circ_{\mathrm{UO}_{2}} = \frac{-853.5 - (-76.00)}{3}\). Solving this gives us \(\Delta_{\mathrm{f}} H^\circ_{\mathrm{UO}_{2}} = \frac{-777.5}{3}\), which equals approximately \(-259.17 \mathrm{~kJ}\).

Key concepts

Hess's Law

Hess's Law is a fundamental principle in thermochemistry that states the total enthalpy change during the course of a chemical reaction is the same, regardless of the number of steps in which the reaction occurs. This principle is invaluable because it allows scientists to determine enthalpy changes for reactions where direct measurement is challenging.

Using Hess's Law involves combining known enthalpy changes of individual reactions to find the enthalpy change of an overall reaction. To put it into practice, one must ensure that the combined reactions add up to the overall desired reaction, with substances not present in the final equation cancelling out. It's a bit like a puzzle, aligning reactions to reveal the complete picture of energy change. In the context of our exercise, Hess's Law helps us to determine the enthalpy change for a complex reaction from simpler reactions with known enthalpy values.

Enthalpy of Formation

The enthalpy of formation, denoted as \( \Delta_{\mathrm{f}} H^\circ \), is a specific type of enthalpy change. It represents the heat change that occurs when one mole of a compound is formed from its elements in their standard states. A key point to remember is that the reference form for each element is its most stable form under standard conditions, which for many diatomic gases is the gaseous form. For instance, the standard state of oxygen is \( O_2(g) \).

Additionally, the standard enthalpy of formation for any element in its most stable form is zero. This is because by definition, elements in their standard states have not undergone formation and therefore there is no heat change associated. This simplifies calculations significantly, as seen in the given exercise where the enthalpy of formation for oxygen gas is zero.

Chemical Reaction Enthalpy

Chemical reaction enthalpy, often just called the reaction enthalpy, \( \Delta H^\circ_{reaction} \), is the heat absorbed or released during a chemical reaction at constant pressure. It's a reflection of the energy needed to break bonds in the reactants compared to the energy released when new bonds form in the products. When a reaction emits heat, it is exothermic, and \( \Delta H^\circ_{reaction} \) is negative; if it absorbs heat, it is endothermic, with a positive \( \Delta H^\circ_{reaction} \).

In textbooks and on exams, students might encounter questions that ask for this calculation, often given the enthalpies of formation of reactants and products. In these cases, we apply Hess's Law, where \( \Delta H^\circ_{reaction} = \sum \Delta_{\mathrm{f}} H^\circ_{products} - \sum \Delta_{\mathrm{f}} H^\circ_{reactants} \) as the most efficient route to the answer.

Standard State

The term 'standard state' refers to a set of conditions of a material at a specific temperature, pressure, and state (solid, liquid, or gas) that is most stable or commonly referred to under normal conditions. For example, the standard state for a substance at room temperature (25°C or 298.15 K) and atmospheric pressure (1 bar or 100 kPa) is typically used in thermochemistry.

The significance of the standard state lies in its use for calculating standard enthalpy changes, such as the standard enthalpy of formation. When all substances in a reaction are in their standard states, the resulting enthalpy change is the 'standard enthalpy change' of the reaction. This allows chemists to have common reference points, enabling comparison and calculation of enthalpy changes under consistent conditions.