Question

Calculate the enthalpy change for the reaction $$ \mathrm{P}_{4} \mathrm{O}_{6}(s)+2 \mathrm{O}_{2}(g) \cdots \mathrm{P}_{4} \mathrm{O}_{10}(s) $$ given the following enthalpies of reaction: $$ \begin{array}{ll} \mathrm{P}_{4}(\mathrm{~s})+3 \mathrm{O}_{2}(g) \rightarrow \mathrm{P}_{4} \mathrm{O}_{6}(s) & \Delta H=-1640.1 \mathrm{~kJ} \\ \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g)-\cdots & \mathrm{P}_{4} \mathrm{O}_{1} 0(s) & \Delta H=-2940.1 \mathrm{~kJ} \end{array} $$

Short answer

The enthalpy change for the reaction \( \mathrm{P}_{4} \mathrm{O}_{6}(s)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) \) is \(-1300.0 ~\text{kJ}\).

Step-by-step solution

Reverse the First Reaction

The desired reaction includes P4O6(s) in the reactants, but in the given reactions, it appears as a product in the first reaction. To obtain the desired reaction, we need to reverse the first reaction as follows: $$ \mathrm{P}_{4} \mathrm{O}_{6}(s) \rightarrow \mathrm{P}_{4}(s) + 3 \mathrm{O}_{2}(g) $$

Calculate the Enthalpy Change for the Reversed First Reaction

When a reaction is reversed, the sign of its enthalpy change is flipped. Thus, the enthalpy change for the reversed first reaction is: $$ \Delta H_{\text{reversed 1}} = -(-1640.1~\text{kJ}) = 1640.1~\text{kJ} $$

Add the Reversed First Reaction to the Second Reaction

Now we can add the reversed first reaction and the second reaction to obtain the desired reaction: $$ \begin{array}{l} \mathrm{P}_{4} \mathrm{O}_{6}(s) \rightarrow \mathrm{P}_{4}(s) + 3 \mathrm{O}_{2}(g) \\ + \\ \mathrm{P}_{4}(s) + 5 \mathrm{O}_{2}(g) \rightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) \\ \hline \mathrm{P}_{4} \mathrm{O}_{6}(s) + 2 \mathrm{O}_{2}(g) \rightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) \end{array} $$

Calculate the Enthalpy Change for the Required Reaction

According to Hess's Law, the enthalpy change for the required reaction equals the sum of the enthalpy changes for the reversed first reaction and the second reaction: $$ \Delta H_{\text{required}} = \Delta H_{\text{reversed 1}} + \Delta H_{\text{2}} = 1640.1 ~\text{kJ} + (-2940.1) ~\text{kJ} = -1300.0 ~\text{kJ} $$

Final Answer

The enthalpy change for the given reaction is -1300.0 kJ.

Key concepts

Hess's Law

Understanding Hess's Law is critical for mastering the calculation of enthalpy changes in chemical reactions. It serves as a foundation for predicting the heat flow during chemical transformations. Hess's Law states that the total enthalpy change for a chemical reaction is the same, regardless of the number of steps taken to complete the reaction or the path taken.

Hess's Law allows us to use known enthalpies of reactions to determine unknown ones. You practically act as an accountant for energy, making sure that all the heat transactions even out in the end. In a step-by-step process, you could reverse, multiply, or divide given equations and adjust their enthalpies accordingly to fit the desired chemical equation.

For exampe, in the solution to the textbook problem, the first reaction is reversed, which means we flip the sign of the enthalpy: \( \Delta H_{\text{reversed 1}} = -(-1640.1~\text{kJ}) = 1640.1~\text{kJ} \). Then, we add this to the enthalpy change for the second reaction, and we obtain the total enthalpy change for the desired reaction. This approach neatly illustrates Hess's Law in action, simplifying the calculation of enthalpy change that might otherwise seem daunting.

Chemical Thermodynamics

Chemical thermodynamics involves the study of energy changes accompanying chemical and physical processes. It's the science that deals with the principles of energy conversion and the direction in which reactions can occur.

Enthalpy, denoted by \(H\), is a measure of the total energy of a thermodynamic system. It includes both internal energy, which is the energy required to create a system, and the energy needed to make room for it by displacing its environment and establishing its volume and pressure.

Enthalpy change, \(\Delta H\), tells us whether a reaction is exothermic (releasing energy to the surroundings, \(\Delta H < 0\) ) or endothermic (absorbing energy from the surroundings, \(\Delta H > 0\) ). This is crucial information because it helps scientists and engineers design chemical processes, understand stability of compounds, and predict whether reactions will occur spontaneously.

The enthalpy change calculated in our exercise also determines the amount of heat exchanged with the surroundings during the reaction. This has important practical applications in fields like energy production, material synthesis, and environmental science.

Stoichiometry

Stoichiometry is a section of chemistry that involves calculating the relative quantities of reactants and products in chemical reactions. It is derived from the Greek words 'stoicheion' (element) and 'metron' (measure), effectively meaning 'the measure of elements'. Stoichiometry is based on the law of conservation of mass where the mass of the reactants equals the mass of the products.

For stoichiometric calculations, it is essential to understand the mole concept, balanced equations, and molar relationships. In our textbook example, stoichiometry ensures that the atoms balance out on both sides of the equation when adding the reversed first reaction to the second reaction. Moreover, it ensures that the coefficients of the equations match the needs of the chemical reaction we are analyzing.

Stoichiometry isn't just theoretical; it's used in the lab to predict how much product will form from a given amount of reactants, which is essential in industrial applications for maximizing efficiency and minimizing waste. In the educational setting, it strengthens a student's ability to think critically and logically about the material and energy balances in chemical reactions.