Question

Given the following data $$ \begin{aligned} \mathrm{P}_{4}(s)+6 \mathrm{Cl}_{2}(g) & \longrightarrow 4 \mathrm{PCl}_{3}(g) & & \Delta H=-1225.6 \mathrm{~kJ} \\ \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) & \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) & & \Delta H=-2967.3 \mathrm{~kJ} \\ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(g) & & \Delta H=-84.2 \mathrm{~kJ} \\ \mathrm{PCl}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Cl}_{3} \mathrm{PO}(g) & & \Delta H=-285.7 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the reaction $$ \mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{PCl}_{5}(g) \longrightarrow 10 \mathrm{Cl}_{3} \mathrm{PO}(g) $$

Short answer

The enthalpy change for the target reaction \(\mathrm{P}_{4}\mathrm{O}_{10}(s) + 6\mathrm{PCl}_{5}(g) \longrightarrow 10\mathrm{Cl}_{3}\mathrm{PO}(g)\) is \(\Delta H = -1620.5 \mathrm{~kJ}\).

Step-by-step solution

Identify the target reaction

The target reaction is given as: \[ \mathrm{P}_{4}\mathrm{O}_{10}(s) + 6 \mathrm{PCl}_{5}(g) \longrightarrow 10 \mathrm{Cl}_{3}\mathrm{PO}(g). \]

Manipulate the given reactions

In order to get to the desired reaction, we must find a way to combine the given reactions. To do this, we will manipulate the given reactions accordingly: 1. Reverse the second reaction so that P4O10 is on the left side of the reaction. After reversing, it becomes: \[ \mathrm{P}_{4}\mathrm{O}_{10}(s) \longrightarrow \mathrm{P}_{4}(s) + 5\mathrm{O}_{2}(g), \quad \Delta H = +2967.3 \mathrm{~kJ}. \] Note that the enthalpy change also changes its sign since the reaction is reversed. 2. Multiply the third reaction by 6 so that 6 PCl5 will be on the left side: \[ 6\mathrm{PCl}_{3}(g) + 6\mathrm{Cl}_{2}(g) \longrightarrow 6\mathrm{PCl}_{5}(g), \quad \Delta H = -84.2 \cdot 6 \mathrm{~kJ}. \] 3. Multiply the fourth reaction by 10 so that there will be 10 Cl3PO on the right side of the reaction: \[ 10\mathrm{PCl}_{3}(g) + 5\mathrm{O}_{2}(g) \longrightarrow 10\mathrm{Cl}_{3}\mathrm{PO}(g), \quad \Delta H = -285.7 \cdot 10 \mathrm{~kJ}. \]

Combine the manipulated reactions

Add the manipulated reactions together to form the target reaction: \[ \begin{aligned} \cancel{\mathrm{P}_{4}\mathrm{O}_{10}(s)} &\longrightarrow \cancel{\mathrm{P}_{4}(s)} + \cancel{5\mathrm{O}_{2}(g)} \\ \cancel{\mathrm{P}_{4}(s)} + 6\cancel{\mathrm{Cl}_{2}(g)} &\longrightarrow 4\cancel{\mathrm{PCl}_{3}(g)} \\ 6\cancel{\mathrm{PCl}_{3}(g)} + 6\cancel{\mathrm{Cl}_{2}(g)} &\longrightarrow 6\mathrm{PCl}_{5}(g) \\ 10\cancel{\mathrm{PCl}_{3}(g)} + \cancel{5\mathrm{O}_{2}(g)} &\longrightarrow 10\mathrm{Cl}_{3}\mathrm{PO}(g) \\ \hline \mathrm{P}_{4}\mathrm{O}_{10}(s) + 6\mathrm{PCl}_{5}(g) &\longrightarrow 10\mathrm{Cl}_{3}\mathrm{PO}(g) \end{aligned} \]

Calculate the enthalpy change of the target reaction

Add up the enthalpy changes of the manipulated reactions to get the enthalpy change for the target reaction: \[ \Delta H_\text{target} = (+2967.3) + (-1225.6) + (-84.2 \cdot 6) + (-285.7 \cdot 10) \mathrm{~kJ}. \] Calculate the numerical value: \[ \Delta H_\text{target} = 2967.3 - 1225.6 - 505.2 - 2857 \mathrm{~kJ} = -1620.5 \mathrm{~kJ}. \] So the enthalpy change for the target reaction is \(\Delta H=-1620.5 \mathrm{~kJ}\).

Key concepts

Hess's Law

Hess's Law is a fundamental principle in chemistry that helps us understand how energy changes in chemical reactions. It states that the total change in enthalpy, or heat content, in a reaction is the same regardless of the pathway taken, as long as the initial and final conditions are identical. This principle ensures that even if a reaction happens in multiple steps, the total enthalpy change will remain constant.​

In practical terms, Hess's Law allows us to calculate the enthalpy change of a reaction that might not be straightforward to measure directly. By combining known reactions in a sequence that reaches our target reaction, we can easily add up their enthalpy changes to find the desired result. This makes Hess's Law incredibly useful in thermochemistry, where measuring every single reaction separately can be challenging. Knowing this helps you in solving determinations of energy required or released in complex reactions.

For the target reaction given, using Hess's Law involves manipulating the provided reactions so they add up correctly to the desired equation. Each step's enthalpy change helps build to this total enthalpy change.

Thermochemical Equations

Thermochemical equations are more than just typical chemical equations. They incorporate the enthalpy change as part of the equation, giving us valuable information about the energy transformation occurring during the reaction. The enthalpy change, denoted as \(\Delta H\), can be positive, indicating an endothermic reaction where energy is absorbed, or negative, indicating an exothermic reaction where energy is released.

When analyzing thermochemical reactions, it's important to pay attention to the equation's stoichiometry, as well as the physical states of the reactants and products (solid, liquid, gas). These influence the enthalpy changes, along with temperature and pressure conditions.

In the problem, each provided reaction has its \(\Delta H\) value, depicting its energy dynamics. By following the steps to rearrange these equations—reversing, multiplying, and combining—we determine the enthalpy change for the complex reaction targeted. Through these thermochemical equations, we gain insights into the energy profiles of the compounds involved.

Chemical Reactions

Chemical reactions involve rearranging atoms to form new substances and are at the core of chemistry. They are represented as equations showing the transformation from reactants to products. This transformation involves the breaking and forming of chemical bonds, which in turn affects the enthalpy change.

It's essential to understand that chemical reactions are governed by the law of conservation of mass, where mass is neither created nor destroyed. Hence, balancing chemical equations is crucial as it ensures this principle holds true by maintaining equal numbers of each type of atom on both sides of the equation.

In thermochemical contexts, changes in energy accompany these molecular transformations. Reviewing the reactions given, each displays different interactions and transitions, from \(\mathrm{PCl}_3\) forming \(\mathrm{PCl}_5\), to the combustion of \(\mathrm{P}_4\). Recognizing these reactions' properties helps in determining the involved energy changes and adjusting them to achieve the desired reaction pathway and \(\Delta H\).

Energy Conservation

Energy conservation is a pillar of both physics and chemistry, affirming that energy cannot be created or destroyed, only converted from one form to another. In the realm of chemical reactions, this translates to the First Law of Thermodynamics, dictating that the total energy of an isolated system remains constant.

Every chemical reaction involves energy changes due to bond-breaking and bond-forming processes. In an exothermic reaction, energy is released to the surroundings due to the formation of stronger bonds in the products. Conversely, in endothermic reactions, energy is absorbed, leading to weaker product bonds compared to reactants.

In our exercises, energy conservation emerges through the application's sequence of manipulating chemical equations to meet Hess's Law. By understanding each reaction's \(\Delta H\) and collectively summing these within the boundaries of energy conservation, we calculate the overall energy change with confidence. Mastery of these laws ensures a consistent approach to evaluating the thermodynamics of reactions without violating the fundamental energy conservation principles.