Question

You wish to know the enthalpy change for the formation of liquid PCl grom the elements. $$\mathrm{P}_{4}(\mathrm{s})+6 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{PCl}_{3}(\ell) \quad \Delta H^{\circ}=?$$ 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 \(\mathrm{PCl}_{5}(\mathrm{s}):\) $$\begin{array}{cc}\mathrm{P}_{4}(\mathrm{s})+10 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{PCl}_{5}(\mathrm{s}) & \Delta H^{\circ}=-1774.0 \mathrm{kJ} \\\\\mathrm{PCl}_{3}(\ell)+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{PCl}_{5}(\mathrm{s}) & \Delta H^{\circ}=-123.8 \mathrm{kJ}\end{array}$$ Use these data to calculate the enthalpy change for the formation of 1.00 mol of \(\mathrm{PCl}_{3}(\ell)\) from phosphorus and chlorine.

Short answer

The enthalpy change for forming 1 mol of \( \mathrm{PCl}_3(\ell) \) is \(-319.7 \, \mathrm{kJ/mol}\).

Step-by-step solution

Write the target reaction

The target reaction for which we need to find the enthalpy change is: \[ \mathrm{P}_4(\mathrm{s}) + 6\mathrm{Cl}_2(\mathrm{g}) \longrightarrow 4\mathrm{PCl}_3(\ell) \] We will use Hess's Law to find the enthalpy change for this reaction.

Identify the given reactions

We have two given reactions with their enthalpy changes: \[ \mathrm{P}_4(\mathrm{s}) + 10 \mathrm{Cl}_2(\mathrm{g}) \longrightarrow 4 \mathrm{PCl}_5(\mathrm{s}) \quad \Delta H^{\circ} = -1774.0 \, \mathrm{kJ} \] \[ \mathrm{PCl}_3(\ell) + \mathrm{Cl}_2(\mathrm{g}) \longrightarrow \mathrm{PCl}_5(\mathrm{s}) \quad \Delta H^{\circ} = -123.8 \, \mathrm{kJ} \]

Reverse and adjust the second reaction

To use Hess's Law, we need the decomposition of \(\mathrm{PCl}_5(\mathrm{s})\) to \(\mathrm{PCl}_3(\ell)\) and \(\mathrm{Cl}_2(\mathrm{g})\): \[ \mathrm{PCl}_5(\mathrm{s}) \longrightarrow \mathrm{PCl}_3(\ell) + \mathrm{Cl}_2(\mathrm{g}) \] For this, we reverse the second reaction, changing the sign of its enthalpy: \( \Delta H^{\circ} = +123.8 \, \mathrm{kJ}\).

Combine the reactions

Combine the modified second reaction with the first reaction: \[ \mathrm{P}_4(\mathrm{s}) + 10 \mathrm{Cl}_2(\mathrm{g}) \longrightarrow 4 \mathrm{PCl}_5(\mathrm{s}) \] \[ 4 \mathrm{PCl}_5(\mathrm{s}) \longrightarrow 4 \mathrm{PCl}_3(\ell) + 4 \mathrm{Cl}_2(\mathrm{g}) \] Add these reactions to find: \[ \mathrm{P}_4(\mathrm{s}) + 6 \mathrm{Cl}_2(\mathrm{g}) \longrightarrow 4 \mathrm{PCl}_3(\ell) \] The sum of their enthalpies: \( -1774.0 \, \mathrm{kJ} + 4 \times 123.8 \, \mathrm{kJ} = -1278.8 \, \mathrm{kJ}\).

Calculate for one mole of PCl3(l)

The calculated enthalpy change \( -1278.8 \, \mathrm{kJ} \) is for the formation of 4 moles \( \mathrm{PCl}_3(\ell) \). Therefore, for 1 mole: \( \frac{-1278.8}{4} = -319.7 \, \mathrm{kJ/mol} \).

Key concepts

Enthalpy Change

Understanding enthalpy change is crucial in studying chemical reactions. Enthalpy, represented by the symbol \( H \), is a measure of the total energy of a thermodynamic system. It includes both internal energy and the energy required to make room for it by displacing its environment. In chemical reactions, we often look at enthalpy change, \( \Delta H \), which is the difference in enthalpy between the products and the reactants.

- When a reaction releases heat, the change is negative (exothermic reactions).
- When a reaction absorbs heat, the change is positive (endothermic reactions).

In the given exercise, the reaction \( \mathrm{P}_4(\mathrm{s}) + 6 \mathrm{Cl}_2(\mathrm{g}) \longrightarrow 4 \mathrm{PCl}_3(\ell) \) involves calculating the enthalpy change using other known reactions. You manipulate these equations to find the overall enthalpy. Remember, when a reaction is reversed, \( \Delta H \) changes sign. Moreover, when a reaction's coefficients are multiplied, \( \Delta H \) is also multiplied by the same factor.

Thermochemistry

Thermochemistry is the part of thermodynamics that deals with heat changes accompanying chemical reactions. It focuses on energy transfer, in the form of heat, during chemical processes. Thermochemistry principles help us predict whether a reaction will occur and how much energy will be released or absorbed.

Hess's Law is a foundational principle in thermochemistry that states that the total enthalpy change for a reaction is the same, regardless of the number of steps the reaction is carried out in. This means you can find the enthalpy change for a complex reaction by adding the enthalpy changes of individual steps that lead to the overall reaction.

This law was used effectively in the exercise to determine the enthalpy change of forming \( \mathrm{PCl}_3(\ell) \) by combining known reactions. By reversing and adjusting given reactions, we are allowed to calculate the desired \( \Delta H \) value, simplifying the investigation of complex chemical processes.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products, involving breaking and forming chemical bonds. They are often described by chemical equations that show the reactants and products, along with their physical states.

- The reactions in the exercise involve phosphorus \( \mathrm{P}_4(\mathrm{s}) \) and chlorine \( \mathrm{Cl}_2(\mathrm{g}) \) forming phosphorus trichloride \( \mathrm{PCl}_3(\ell) \).
- Intermediate reactions, such as the formation of phosphorus pentachloride \( \mathrm{PCl}_5(\mathrm{s}) \), play a crucial role in understanding the complete enthalpy change.
- Balancing chemical reactions is important to ensure mass and energy conservation in any thermochemical calculation.

By utilizing the principles of chemical reactions and thermochemistry laws, such as Hess's Law, we can efficiently calculate enthalpy changes for reactions that may not be easily measurable directly.