Question

The standard enthalpy change \(\Delta H^{\circ}\) for the thermal decomposition of silver nitrate according to the following equation is \(+78.67 \mathrm{~kJ}\) : \(\mathrm{AgNO}_{3}(s) \longrightarrow \mathrm{AgNO}_{2}(s)+\frac{1}{2} \mathrm{O}_{2}(g)\) The standard enthalpy of formation of \(\mathrm{AgNO}_{3}(s)\) is \(-123.02 \mathrm{~kJ} / \mathrm{mol} .\) Calculate the standard enthalpy of formation of \(\mathrm{AgNO}_{2}(s)\).

Short answer

The standard enthalpy of formation of \( \mathrm{AgNO}_{2}(s) \) is \( 201.69 \, \text{kJ/mol} \).

Step-by-step solution

Understand the Given Information

We know the standard enthalpy change for the reaction: \( \Delta H^{\circ} = +78.67 \, \text{kJ/mol} \). We also know the standard enthalpy of formation of \( \mathrm{AgNO}_{3}(s) \) is \( -123.02 \, \text{kJ/mol} \). We need to find the standard enthalpy of formation of \( \mathrm{AgNO}_{2}(s) \).

Apply Hess's Law

Hess's Law states that the total enthalpy change during a reaction is the same, no matter how the reaction occurs. Use it to find the unknown enthalpy of formation.

Write the Equation for the Reaction

Write the balanced reaction along with the formation enthalpies: \[ \Delta H^{\circ}_{\text{reaction}} = \Delta H^{\circ}_{\text{f, products}} - \Delta H^{\circ}_{\text{f, reactants}} \]Substitute known values:\[ 78.67 \, \text{kJ/mol} = \Delta H^{\circ}_{\text{f, } \mathrm{AgNO}_{2}(s)} + \frac{1}{2}(\Delta H^{\circ}_{\text{f, } \mathrm{O}_{2}(g)}) - (\Delta H^{\circ}_{\text{f, } \mathrm{AgNO}_{3}(s)}) \]

Simplify the Enthalpy Equation

The standard enthalpy of formation of \( \mathrm{O}_{2}(g) \) is \( 0 \, \text{kJ/mol} \) because it is an element in its standard state. Therefore, the equation simplifies to:\[ 78.67 \, \text{kJ/mol} = \Delta H^{\circ}_{\text{f, } \mathrm{AgNO}_{2}(s)} - (-123.02 \, \text{kJ/mol}) \]

Solve for the Enthalpy of Formation of \( \mathrm{AgNO}_{2}(s) \)

Rearrange the equation to solve for \( \Delta H^{\circ}_{\text{f, } \mathrm{AgNO}_{2}(s)} \):\[ \Delta H^{\circ}_{\text{f, } \mathrm{AgNO}_{2}(s)} = 78.67 \, \text{kJ/mol} + 123.02 \, \text{kJ/mol} \]Calculate:\[ \Delta H^{\circ}_{\text{f, } \mathrm{AgNO}_{2}(s)} = 201.69 \, \text{kJ/mol} \]

Key concepts

Hess's Law

Hess's Law is an important principle in thermochemistry. It states that the total enthalpy change for a chemical reaction is the same regardless of the pathway taken.
This means that enthalpy is a state function and is path-independent. To put it simply, whether a reaction occurs in one step or multiple steps, the overall change in enthalpy is the same.
This allows us to calculate enthalpy changes that are not easy to measure directly, using other reactions whose enthalpy changes are known. In the context of the provided exercise, Hess's Law helps us determine the unknown enthalpy of formation for \( \mathrm{AgNO}_{2}(s) \) using the known enthalpies of the related substances.
It acts as a tool that lets us sum up the individual enthalpy changes to find the total change for the overall reaction.

Enthalpy of Formation

The enthalpy of formation, often denoted as \( \Delta H^{\circ}_{\text{f}} \), refers to the change in enthalpy when one mole of a compound is formed from its elements in their standard states.
It is a measure of the energy released or consumed during the creation of a compound. Standard enthalpies of formation are typically given in units of \( \text{kJ/mol} \) and are measured under standard conditions (1 atm pressure and 25°C).
For example, the standard enthalpy of formation for \( \mathrm{O}_{2}(g) \) is zero because oxygen is in its elemental form. In calculating the standard enthalpy of formation for \( \mathrm{AgNO}_{2}(s) \), the understanding of enthalpy of formation allows assigning values to the reactants and products in the reaction,
leading to the application of Hess's Law to find the enthalpy of formation for \( \mathrm{AgNO}_{2}(s) \). This calculation helps chemists understand the energy dynamics involved in forming or decomposing substances.

Chemical Reactions

Chemical reactions involve the transformation of substances, where reactants are converted into products. During these processes, bonds are broken and formed, resulting in energy changes.
In thermodynamics, these changes often involve enthalpy, a measure of heat content. The exercise provided presents a decomposition reaction of silver nitrate: - \( \mathrm{AgNO}_{3}(s) \rightarrow \mathrm{AgNO}_{2}(s) + \frac{1}{2}\mathrm{O}_{2}(g) \) In this reaction, silver nitrate (\( \mathrm{AgNO}_{3}(s) \)) breaks down into silver nitrite and oxygen gas.
The enthalpy change for this reaction demonstrates how energy is either absorbed or released during the breaking and forming of chemical bonds. Understanding chemical reactions is essential for grasping how energy is transferred and transformed during these processes. In the given problem, the calculation of enthalpy of formation for different components of the reaction helps us understand the energy changes involved in decomposing and forming new compounds.