Calculate the enthalpy of the reaction \(\mathrm{P}_{4}(\mathrm{~s})+10 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{PCl}_{5}(\mathrm{~s})\) from the reactions $$ \begin{gathered} \mathrm{P}_{4}(\mathrm{~s})+6 \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{PCl}_{3}(\mathrm{I}) \\ \Delta H^{\circ}=-1278.8 \mathrm{~kJ} \\ \mathrm{PCl}_{3}(\mathrm{l})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{PCl}_{5}(\mathrm{~s}) \\ \Delta H^{m}=-124 \mathrm{~kJ} \end{gathered} $$

Short Answer

Expert verified
-5611.2 kJ

Step by step solution

01

Identify the Known Reactions

The problem provides two known reactions with their respective enthalpy changes. The first reaction is the formation of PCl3 from P4 and Cl2 with an enthalpy change ΔH° = -1278.8 kJ, and the second reaction is the conversion of PCl3 and Cl2 into PCl5 with an enthalpy change ΔHm = -124 kJ.
02

Write the Target Reaction

The target reaction is the formation of PCl5 from P4 and Cl2:\[\mathrm{P}_{4}(\mathrm{~s}) + 10\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 4\mathrm{PCl}_{5}(\mathrm{~s})\]
03

Break Down the Target Reaction

In order to calculate the enthalpy of the target reaction, we need to combine the two known reactions in a way that yields the target reaction upon addition.Known Reaction 1 (multiplied by 4): \[4(\mathrm{P}_{4}(\mathrm{~s}) + 6\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 4\mathrm{PCl}_{3}(\mathrm{I}))\]Known Reaction 2 (multiplied by 4): \[4(\mathrm{PCl}_{3}(\mathrm{l}) + \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{PCl}_{5}(\mathrm{~s}))\]The coefficients are multiplied by 4 to match the four moles of PCl5 produced in the target reaction.
04

Calculate the Enthalpy Change for the Multiplied Reactions

Multiply the enthalpy changes of the given reactions by their coefficients to find the total enthalpy change:For Reaction 1 (multiplied by 4):\[4(-1278.8 \mathrm{~kJ}) = -5115.2 \mathrm{~kJ}\]For Reaction 2 (multiplied by 4):\[4(-124 \mathrm{~kJ}) = -496 \mathrm{~kJ}\]
05

Add the Modified Reactions

Add the reactions together, ensuring that the intermediates cancel out:Multiplying Reaction 1 by 4,\[4 \mathrm{P}_{4} + 24 \mathrm{Cl}_{2} \rightarrow 16 \mathrm{PCl}_{3}\]Multiplying Reaction 2 by 4,\[16 \mathrm{PCl}_{3} + 4 \mathrm{Cl}_{2} \rightarrow 4 \mathrm{PCl}_{5}\]Adding these modified reactions cancels out the 16 moles of PCl3, giving:\[4 \mathrm{P}_{4} + 24 \mathrm{Cl}_{2} + 4 \mathrm{Cl}_{2} \rightarrow 4 \mathrm{PCl}_{5}\].The total enthalpy change for this combined reaction is the sum of the enthalpy changes from each modified reaction.
06

Combine the Enthalpy Changes

Add the enthalpy changes of the modified reactions to calculate the overall enthalpy change of the target reaction:\[-5115.2 \mathrm{~kJ} + (-496 \mathrm{~kJ}) = -5611.2 \mathrm{~kJ}\]

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Key Concepts

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

Chemical Thermodynamics
Chemical thermodynamics studies the relationship between heat, work, and chemical reactions or with the physical changes of state that classify them. It is a bridge connecting thermodynamics and chemistry and is essential for understanding energy changes in chemical systems. Enthalpy, represented as \(H\), is a fundamental concept here; it describes the total heat content of a system at constant pressure.

The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed, only transformed. In chemical reactions, this principle is observed through the changes in enthalpy, where the heat released or absorbed can be calculated using the reaction's stoichiometry and the enthalpies of individual reactions. By considering the entropic and enthalpic factors, chemical thermodynamics allows us to predict whether a process will occur spontaneously. It is key in designing chemical reactions that are favorable from both an energetic and practical standpoint.
Enthalpy of Reaction
The enthalpy of reaction, commonly denoted as \(\Delta H\), is a measure of the heat change during a chemical reaction carried out at constant pressure. It is either absorbed or evolved, indicating endothermic or exothermic reactions, respectively. When presented with stepwise reactions, like the formation of PCl5 from P4 and Cl2, the Hess's law comes into play. According to Hess's law, the total enthalpy change in a chemical reaction is the same, no matter how the reaction occurs, as long as the initial and final states are the same.

Hence, by algebraically summing up the enthalpy changes (\(\Delta H\)) for each step provided, we can find the overall \(\Delta H\) for the reaction of interest. This principle is critical in calculating the enthalpy change for reactions where the direct method may be impracticable. By capturing the calorimetric data from related reactions, like those given in the exercise, and applying Hess's law, we deduce the enthalpy change for complex reactions.
Stoichiometry
Stoichiometry is the quantitative study of reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced in a given reaction based on the reaction coefficients indicated in the balanced chemical equation.

In practical terms, stoichiometry can tell us how much product we can expect from a certain amount of reactants, which is invaluable in both laboratory and industrial chemical synthesis. When calculating enthalpy changes, stoichiometry plays a pivotal role. For the accurate determination of enthalpy change, we utilize the stoichiometric coefficients to adjust the known enthalpies of formation or reaction. This is exemplified in the step by step solution where the known reactions are multiplied by their respective stoichiometric coefficients to align with the target reaction. This allows for precise cancelation of intermediates and provides the correct enthalpy change associated with the final balanced equation for the chemical process under investigation.

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Most popular questions from this chapter

Use the enthalpies of formation in Appendix \(2 \mathrm{~A}\) to calculate the standard enthalpy of the following reactions: (a) the replacement of deuterium by ordinary hydrogen in heavy water: \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{D}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{D}_{2}(\mathrm{~g})\) (b) the removal of sulfur from the hydrogen sulfide and sulfur dioxide in natural gas: \(2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})+\mathrm{SO}_{2}(\mathrm{~g}) \rightarrow 3 \mathrm{~S}(\mathrm{~s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (c) the oxidation of ammonia: \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

(a) Calculate the work for a system that releases \(346 \mathrm{~kJ}\) of heat in a process for which the decrease in internal energy is \(125 \mathrm{~kJ}\). (b) Is work done on or by the system during this process?

Calculate the lattice enthalpy of solid potassium bromide, \(\mathrm{KBr}(\mathrm{s}) \rightarrow \mathrm{K}^{+}(\mathrm{g})+\mathrm{Br}^{-}(\mathrm{g})\), from the following information: $$ \begin{aligned} &\Delta H_{i}{ }^{\circ}\left(\mathrm{KBr}, \text { s) }=-394 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\right. \\ &\Delta H_{i}{ }^{\circ}(\mathrm{K}, \mathrm{g})=+89.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \end{aligned} $$ First ionization energy of \(\mathrm{K}(\mathrm{g})=+425.0 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) \(\Delta H_{\text {vap }}{ }^{\circ}\left(\mathrm{Br}_{2}, \mathrm{l}\right)=+30.9 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) Br-Br bond dissociation cuthalpy \(=+192.9 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) Electron attachment to \(\mathrm{Br}(\mathrm{g})\) : $$ \mathrm{Br}(\mathrm{g})+\mathrm{e}^{-}(\mathrm{g}) \rightarrow \mathrm{Be}^{-}(\mathrm{g}), \quad \Delta H^{*}=-331.0 \mathrm{~kJ} $$

Two successive stages in the industrial manufacture of sulfuric acid are the combustion of sulfur and the oxidation of sulfur dioxide to sulfur trioxide. From the standard reaction enthalpies $$ \begin{gathered} \mathrm{S}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{SO}_{2}(\mathrm{~g}) \\ \Delta H^{\circ}=-296.83 \mathrm{~kJ} \\ 2 \mathrm{~S}(\mathrm{~s})+3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{SO}_{3}(\mathrm{~g}) \\ \Delta H^{\circ}=-791.44 \mathrm{~kJ} \end{gathered} $$ Calculate the reaction enthalpy for the oxidation of sulfur dioxide to sulfur trioxide in the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{SO}_{3}(\mathrm{~g})\).

Write the thermochemical equations that give the values of the standard enthalpies of formation for (a) \(\mathrm{CH}_{2} \mathrm{COOH}\) (I); (b) \(\mathrm{SO}_{2}\) (g); (c) \(\mathrm{CO}_{2}\) (g).

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