Question

Consider the reaction $$ 2 \mathrm{ClF}_{3}(g)+2 \mathrm{NH}_{3}(g) \longrightarrow \mathrm{N}_{2}(g)+6 \mathrm{HF}(g)+\mathrm{Cl}_{2}(g)\quad\Delta H^{\circ}=-1196 \mathrm{kJ} $$ Calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{ClF}_{3}(g)\)

Short answer

The standard enthalpy of formation for ClF3(g) is 267.8 kJ/mol.

Step-by-step solution

Determine the standard enthalpies of formation for all substances except ClF3(g)

We can find the standard enthalpies of formation in the literature or a standard reference book. Here are the values for the substances involved in the reaction: ΔHf°(NH3(g)) = -45.9 kJ/mol ΔHf°(N2(g)) = 0 kJ/mol (standard reference for elements in their standard state) ΔHf°(HF(g)) = -273.3 kJ/mol ΔHf°(Cl2(g)) = 0 kJ/mol (standard reference for elements in their standard state)

Substitute the values into the formula and solve for ΔHf°(ClF3(g))

We know the standard enthalpy change for the overall reaction, ΔH°, which is -1196 kJ. Using the formula ΔH° = ΣnΔHf°(products) - ΣmΔHf°(reactants) and substituting the values we found earlier: -1196 kJ = (ΔHf°(N2(g)) + 6ΔHf°(HF(g)) + ΔHf°(Cl2(g))) - (2ΔHf°(ClF3(g)) + 2ΔHf°(NH3(g))) -1196 kJ = (0 + 6(-273.3 kJ/mol) + 0) - (2ΔHf°(ClF3(g)) + 2(-45.9 kJ/mol)) Now, we'll solve for ΔHf°(ClF3(g)): -1196 kJ = (-1639.8 kJ) - (2ΔHf°(ClF3(g)) - 91.8 kJ) ΔHf°(ClF3(g)) = (-1196 kJ + 1639.8 kJ + 91.8 kJ)/2 = 267.8 kJ/mol So, the standard enthalpy of formation for ClF3(g) is 267.8 kJ/mol.

Key concepts

Understanding Enthalpy Change

When discussing chemical reactions, one important topic is the enthalpy change, denoted as \( \Delta H \). Enthalpy change refers to the heat exchanged during a reaction occurring at constant pressure. It provides insight into whether a reaction is exothermic or endothermic.
When \( \Delta H \) is negative, as in the case of the reaction from the exercise, it means the process releases heat, thus is exothermic. On the other hand, a positive \( \Delta H \) indicates an endothermic reaction, which absorbs heat from the surroundings.
This concept is crucial because it allows scientists to understand the energy changes occurring during reactions, which can influence reaction conditions and sustainability.
In energy calculations, the enthalpy changes of products and reactants help determine the overall energy requirement or release of the reaction.

Exploring Chemical Reactions

Chemical reactions involve converting reactants into products, often accompanied by energy changes due to bond breaking and forming. Each substance in the reaction has its own standard enthalpy of formation (\( \Delta H_{\mathrm{f}}^\circ \)), representing the enthalpy change associated with forming 1 mole of a compound from its elements in their standard states.

In the given exercise, the reaction involves different states of matter: gases evolving in a chemical transformation. This conversion results in energy changes due to the different bonding energies. Understanding the specific bonds broken and formed and their individial enthalpies helps in figuring out the reaction's total \( \Delta H \).

Learning to calculate \( \Delta H_{\mathrm{f}}^\circ \) for reactants or products is essential. For instance, knowing how to calculate \( \Delta H_{\mathrm{f}}^\circ \) for \( \mathrm{ClF}_3(g) \) in this reaction involves recognizing how much energy is required or released. This understanding is foundational in solving similar problems and enhancing reactions' efficiency.

Diving into Energy Calculations

Energy calculations for chemical reactions provide insights into the amount and type of energy involved. Calculating energy changes in reactions helps chemists manage reaction conditions effectively.
In this exercise, we determined the enthalpy of formation for \( \mathrm{ClF}_3(g) \) using a known reaction enthalpy. By rearranging the formula \( \Delta H^\circ = \Sigma n\Delta H_{\mathrm{f}}^\circ(\text{products}) - \Sigma m\Delta H_{\mathrm{f}}^\circ(\text{reactants}) \), the unknown \( \Delta H_{\mathrm{f}}^\circ(\mathrm{ClF}_3(g)) \) can be found.

Consider each element's enthalpy of formation in the equation, inserting known values, and simplifying to solve for the unknown. The method reflects a strategy to understand energy requirements better, balancing the energy consumed by forming reactants and those released by product formation.
For students, mastering these calculations allows them to predict energy changes in prospective reactions, further supporting energy use optimization in chemical industries and environmental sustainability efforts.