Question

Given two processes: (i) \(1 / 2 \mathrm{P}_{4}(\mathrm{~s})+3 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{PCl}_{3}(1) ; \Delta H\) \(=-635 \mathrm{~kJ}\) (ii) \(\mathrm{PCl}_{3}(\mathrm{l})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{PCl}_{5}(\mathrm{~s}) ; \Delta H\) \(=-137 \mathrm{~kJ}\) The value of \(\Delta_{\mathrm{f}} H\) of \(\mathrm{PCl}_{5}(\mathrm{~s})\) is (a) \(454.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(-454.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-772 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(-498 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Short answer

The value of Δ_fH for PCl_5(s) is -772 kJ/mol, making the correct answer (c) -772 kJ/mol^{-1}.

Step-by-step solution

Understanding the Problem

We are given two reaction equations with their associated enthalpy changes (ΔH). We need to use these to calculate the standard enthalpy of formation (Δ_fH) for PCl_5(s). Enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states.

Combining the Two Equations

The target molecule PCl_5(s) is formed in the second equation. To find its enthalpy of formation, we combine both equations in a way that reflects the formation of PCl_5(s) from its elemental forms in their standard states. We achieve this by adding the two given equations.

Calculating the Enthalpy of Formation

By adding the enthalpy changes of both reactions, we get the overall enthalpy change for the formation of PCl_5(s) from the elemental PCl_3(l) and Cl_2(g). Since PCl_3(l) is formed from P_4(s) and Cl_2(g) in the first reaction, we effectively have the formation of PCl_5(s) from phosphorous and chlorine, the elements in their standard states.

Key concepts

Physical Chemistry and Enthalpy

Physical chemistry involves the study of how matter behaves on a molecular and atomic level, and how chemical reactions occur. From the energetic standpoint, a key concept in physical chemistry is enthalpy (\( \triangle H \)), which is the heat content of a system. When we delve into thermochemistry, a branch of physical chemistry, we look at how heat energy is absorbed or released during chemical reactions.

Enthalpy changes in reactions are crucial because they indicate whether a reaction is exothermic (releases heat) or endothermic (absorbs heat). For instance, in the given equation, the formation of PCl_3 from P_4 and Cl_2 is associated with a negative enthalpy change, indicating that the process is exothermic and heat is being released. Understanding the concept of enthalpy is essential not just for solving problems but also for predicting and controlling chemical processes.

Thermochemistry and Enthalpy of Formation

Thermochemistry is the study of the heat changes that accompany chemical reactions. An important aspect of thermochemistry is the standard enthalpy of formation, denoted as \( \triangle_{f}H \). This is the change in enthalpy when one mole of a compound is formed from its elements in their standard states at 1 atm pressure and at a specified temperature, usually 25°C.

The standard enthalpy of formation for any element in its most stable form is zero, as elements are considered to be formed from themselves. When solving problems related to the enthalpy of formation, it is essential to remember that it corresponds to the formation of one mole of a compound from its constituent elements. The exercise provided requires combining two reactions to find the net enthalpy change for the formation of PCl_5, which serves as an excellent example for learners to understand this concept in a practical context.

Chemical Reactions and Energy Transfers

Chemical reactions are transformations where reactants convert into products through breaking and forming of chemical bonds. Enthalpy change is a vital indicator of these energy transfers that occur during reactions. It's key to note that the sum of enthalpies for individual steps in a reaction pathway can lead to the total enthalpy change for the overall reaction, as illustrated by Hess's Law. This principle is what allows us to add the enthalpy changes from multiple steps to find the enthalpy of formation for a more complex compound like PCl_5.

In the exercise, solving for the enthalpy of formation of PCl_5 involves combining energy changes from multiple steps. This reinforces the concept that energy conservation is fundamental to understanding chemical reactions. The answer will be negative, as indicated by the given options, since the formation of this stable compound from its elements releases energy, making it an exothermic process.