91Ó°ÊÓ

You wish to know the enthalpy change for the formation of liquid \(\mathrm{PCl}_{3}\) from the elements. $$ \mathrm{P}_{4}(\mathrm{s})+6 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{PCl}_{3}(\ell) \quad \Delta_{\mathrm{r}} H^{\circ}=? $$ The enthalpy change for the formation of \(\mathrm{PCl}_{5}\) from the elements can be determined experimentally, as can the enthalpy change for the reaction of \(\mathrm{PCl}_{3}(\ell)\) with more chlorine to give \(\mathrm{PCl}_{5}(\mathrm{s}):\) \(\begin{aligned} \mathrm{P}_{4}(\mathrm{s})+10 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{PCl}_{5}(\mathrm{s}) & \\ \Delta_{r} H^{\circ} &=-1774.0 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn} \\\ \mathrm{PCl}_{3}(\ell)+\mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{PCl}_{5}(\mathrm{s}) & \\ \Delta_{\mathrm{r}} H^{\circ} &=-123.8 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn} \end{aligned}\) Use these data to calculate the enthalpy change for the formation of 1.00 mol of \(\mathrm{PCl}_{3}(\ell)\) from phosphorus and chlorine.

Step by step solution

01

Write the Target Reaction

The target reaction is the formation of liquid \( \mathrm{PCl}_{3} \) from its elements: \( \mathrm{P}_{4} (\mathrm{s}) + 6 \mathrm{Cl}_{2} (\mathrm{g}) \rightarrow 4 \mathrm{PCl}_{3} (\text{l}) \). We need to find \( \Delta_{r}H^{\circ} \) for this reaction.

02

Write the Known Reactions

We have two known reactions:1. Formation of \( \mathrm{PCl}_{5} \): \( \mathrm{P}_{4} (\mathrm{s}) + 10 \mathrm{Cl}_{2} (\mathrm{g}) \rightarrow 4 \mathrm{PCl}_{5} (\mathrm{s}) \) with \( \Delta_{r} H^{\circ} = -1774.0 \text{ kJ/mol-rxn} \).2. Reaction of \( \mathrm{PCl}_{3} \) with \( \mathrm{Cl}_{2} \): \( \mathrm{PCl}_{3}(\ell) + \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{PCl}_{5}(\mathrm{s}) \) with \( \Delta_{r} H^{\circ} = -123.8 \text{ kJ/mol-rxn} \).

03

Reverse the Second Reaction

To match the target reaction, we invert the second reaction, which gives:\( \mathrm{PCl}_{5}(\mathrm{s}) \rightarrow \mathrm{PCl}_{3}(\ell) + \mathrm{Cl}_{2}(\mathrm{g}) \). Reversing changes the sign of enthalpy change: \( \Delta_{r} H^{\circ} = +123.8 \text{ kJ/mol-rxn} \).

04

Apply Hess's Law

Using Hess's Law, the enthalpy change for the target reaction can be found by manipulating the known reactions. We align and add:\( 1. \) \( \mathrm{P}_{4} (\mathrm{s}) + 10 \mathrm{Cl}_{2} (\mathrm{g}) \rightarrow 4 \mathrm{PCl}_{5} (\mathrm{s}) \) \( \quad \Delta_{r} H^{\circ} = -1774.0 \text{ kJ} \)\( 2. \) \( 4 \mathrm{PCl}_{5}(\mathrm{s}) \rightarrow 4 \mathrm{PCl}_{3}(\ell) + 4 \mathrm{Cl}_{2}(\mathrm{g}) \) \( \quad \Delta_{r} H^{\circ} = 4 \times 123.8 \text{ kJ} = 495.2 \text{ kJ} \)The net reaction is: \( \mathrm{P}_{4} (\mathrm{s}) + 6 \mathrm{Cl}_{2} (\mathrm{g}) \rightarrow 4 \mathrm{PCl}_{3} (\ell) \).

05

Calculate the Target Enthalpy Change

Add the enthalpy changes from Step 4:\( \Delta_{r} H^{\circ} = (-1774.0 \text{ kJ}) + (495.2 \text{ kJ}) = -1278.8 \text{ kJ} \) for 4 mol of \( \mathrm{PCl}_{3}(\ell) \).For 1 mol of \( \mathrm{PCl}_{3}(\ell) \), divide the value by 4: \( \Delta_{r} H^{\circ} = \frac{-1278.8}{4} = -319.7 \text{ kJ/mol} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hess's Law

Hess's Law is an important principle in chemistry that allows us to calculate the enthalpy change of a reaction that cannot be measured directly. It states that the total enthalpy change for a chemical reaction is the same, no matter how the reaction is carried out in steps. This means that the enthalpy change depends only on the initial and final states, not the pathway taken.

In the exercise provided, we applied Hess's Law by using two given reactions to determine the enthalpy change for the formation of liquid \( \mathrm{PCl}_3 \). We reversed a reaction and combined it with another to align with the reaction of interest. By adding the enthalpy changes of each step, we determined the enthalpy change for the overall process.

Using Hess's Law offers a flexible approach to enthalpy calculations, particularly useful when direct measurement is not possible. It's like piecing together a puzzle where known segments help complete the picture.

Enthalpy of Formation

The enthalpy of formation is the heat change that occurs when one mole of a compound forms from its constituent elements in their standard states. It is a crucial concept as it allows chemists to understand the energy aspects of forming compounds.

In the context of the exercise, the target is to find the enthalpy change for the formation of \( \mathrm{PCl}_3(\ell) \) from elemental phosphorus and chlorine. This means you're determining how much energy is involved in forming a mole of \( \mathrm{PCl}_3 \) from \( \mathrm{P}_4(s) \) and \( \mathrm{Cl}_2(g) \).

This helps us comprehend the energy efficiency and feasibility of a chemical process. When the enthalpy of formation is negative, as in this case, it indicates that the process releases energy, which is typically more spontaneous.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products through the breaking and forming of bonds, which involve energy changes. Each reaction is characterized by an enthalpy change \( (\Delta H) \), which tells us whether the reaction absorbs or releases energy.

In this exercise, we dealt with two particular reactions: one forming \( \mathrm{PCl}_5 \) and another converting \( \mathrm{PCl}_5 \) back to \( \mathrm{PCl}_3 \). By examining these reactions, we were able to determine the enthalpy for the desired reaction through manipulation and application of Hess's Law.

Understanding the energy changes in reactions is essential for predicting reaction behavior and the conditions needed to drive reactions. Reactions that release energy, such as many combustion reactions, are termed exothermic, while those that absorb energy are endothermic. These concepts are fundamental in explaining how and why reactions occur.