Question

Given the following data $$\begin{array}{cl}\mathrm{P}_{4}(s)+6 \mathrm{Cl}_{2}(g) \longrightarrow 4 \mathrm{PCl}_{3}(g) & \Delta H=-1225.6 \mathrm{kJ} \\ \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) & \Delta H=-2967.3 \mathrm{kJ} \\\ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(g) & \Delta H=-84.2 \mathrm{kJ} \\ \mathrm{PCl}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Cl}_{3} \mathrm{PO}(g) & \Delta H=-285.7 \mathrm{kJ} \end{array}$$ 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 (ΔH) for the reaction \(P_{4}O_{10}(s) + 6PCl_{5}(g) \rightarrow 10Cl_{3}PO(g)\) is -6329.5 kJ.

Step-by-step solution

Identify the target reaction

: The target reaction we want to find the ΔH for is: \(P_{4}O_{10}(s) + 6PCl_{5}(g) → 10Cl_{3}PO(g)\)

Manipulate the given reactions

: To obtain the target reaction, we will need to multiply and invert the given reactions as necessary. First, observe that the fourth reaction directly produces \(Cl_{3}PO(g)\), so we will multiply the entire reaction and its ΔH by 10: \(10 PCl_{3}(g) + 5O_{2}(g) → 10 Cl_{3}PO(g)\), ΔH = -285.7 kJ × 10 = -2857 kJ Second, note that the third reaction must be multiplied by 6 to obtain 6PCl5: \(6( PCl_{3}(g) + Cl_{2}(g) → PCl_{5}(g))\), ΔH = -84.2 kJ × 6 = -505.2 kJ Lastly, the second reaction produces \(P_{4}O_{10}(s)\). It can be used as is: \(P_{4}(s)+5O_{2}(g) → P_{4}O_{10}(s)\), ΔH = -2967.3 kJ

Combine the modified reactions

: Now, we will combine the modified reactions from Step 2: 10 PCl3 + 5 O2 → 10 Cl3PO, ΔH = -2857 kJ (Reaction A) 6 (PCl3 + Cl2 → PCl5), ΔH = -505.2 kJ (Reaction B) P4 + 5 O2 → P4O10, ΔH = -2967.3 kJ (Reaction C) First, add Reaction A and B: 10 PCl3 + 5 O2 + 6 (PCl3 + Cl2 → PCl5) → 10 Cl3PO + 6 PCl5, ΔH = -3362.2 kJ Now, notice that there are 5 O2 on both sides of the reaction, cancel them out: 10 PCl3 + 6 (PCl3 + Cl2 → PCl5) → 10 Cl3PO + 6 PCl5, ΔH = -3362.2 kJ Next, add Reaction C to the resulting reaction: (P4 + 5 O2 → P4O10) + (10 PCl3 + 6 (PCl3 + Cl2 → PCl5) → 10 Cl3PO + 6 PCl5), ΔH = -3362.2 kJ + (-2967.3 kJ)

Calculate the ΔH for the target reaction

: Now, calculate the overall ΔH for the target reaction by adding the ΔH values of the combined reactions: ΔH = (-3362.2 kJ) + (-2967.3 kJ) = -6329.5 kJ So, the ΔH for the reaction \(P_{4}O_{10}(s) + 6PCl_{5}(g) → 10Cl_{3}PO(g)\) is -6329.5 kJ.

Key concepts

Thermochemical Equations

Thermochemical equations are a critical component of understanding chemical reactions and energy changes. At the most basic level, these equations reflect the law of conservation of energy, highlighting how energy is transformed during chemical reactions. More specifically, a thermochemical equation presents a chemical reaction along with the enthalpy change (ΔH), effectively combining a stoichiometric equation with thermodynamic information.

In the context of our exercise, each reaction is supplied with a corresponding ΔH value that reflects the heat released or absorbed. For instance, the reaction \( P_{4}(s) + 6Cl_{2}(g) \rightarrow 4PCl_{3}(g) \) comes with a ΔH of -1225.6 kJ. The negative sign indicates that the reaction is exothermic: it releases heat into the surroundings. In contrast, a positive ΔH would suggest an endothermic reaction, where heat is absorbed from the surroundings.

Understanding these equations is fundamental since they allow us to calculate the overall energy change of a sequence of reactions, as is necessary when using Hess's Law. This understanding helps us recognize that the thermodynamic quantities are directly related to the stoichiometry of the reaction: if we multiply both sides of the chemical equation by a factor, the ΔH must also be multiplied by the same factor. For example, when we multiply the fourth reaction in the exercise by 10 to match the stoichiometry required, the ΔH value also gets multiplied by 10.

Hess's Law

Hess's Law, named after Russian chemist Germain Henri Hess, is a central concept in thermodynamics and provides a practical approach to calculate the enthalpy change of a reaction that may be difficult to measure directly. The law states that if a chemical equation can be expressed as the sum of a series of steps, then the enthalpy change for the overall process is simply the sum of the enthalpy changes for the individual steps.

This principle is grounded in the idea that enthalpy is a state function, meaning it depends only on the initial and final state of the system, not on the path taken. Therefore, whether a reaction occurs in one step or multiple steps, the total ΔH will be the same. In the exercise provided, we use Hess’s Law to determine the ΔH for a target reaction by manipulating and combining given reactions, ensuring the law of conservation of mass is observed.

To apply Hess's Law to the exercise at hand, we manipulate the given thermochemical equations in such a way that, when added together, they result in the target equation. The manipulated equations included changing the coefficients, reversing the reactions, and then summing the corresponding ΔH values. These manipulations are vital to achieving the same molecular species on both sides of the equation, allowing us to sum the ΔH values accurately.

Enthalpy (ΔH)

Enthalpy, represented by the symbol ΔH, is a measure of the total heat content in a chemical system under constant pressure. It is an essential concept in understanding energy transfer during chemical reactions. The ΔH can be positive or negative, reflecting endothermic and exothermic processes, respectively. Endothermic reactions absorb heat from the surroundings, resulting in a temperature decrease unless the system is supplied with heat. On the contrary, exothermic reactions release heat, often raising the temperature of the environment unless the heat is removed.

In our exercise, each chemical equation specifies a ΔH value, allowing us to determine the heat change involved with each reaction. By applying Hess’s Law, we can extract the ΔH for a more complex reaction that isn't directly provided by the experimental data. It’s important to point out that enthalpy changes are often reported in kilojoules (kJ) and are dependent on the quantities of the reactants and products, which is why we keep close attention to stoichiometry in our calculations.

Lastly, recognizing that the ΔH for a reversed reaction is equal in magnitude but opposite in sign to the ΔH for the forward reaction is essential for correctly manipulating the equations. In our exercise, we were able to rightfully deduce the ΔH of the targeted complex reaction with careful adjustments and combinations of the provided thermochemical equations based on the principles of enthalpy.