Problem 92 - Chemistry Matter and Change

Question

Use standard enthalpies of formation from Table \(R-11\) on page 975 to calculate \(\Delta H_{\text { ran for the following }}^{\circ}\) reaction. \(\mathrm{P}_{4} \mathrm{O}_{6}(\mathrm{s})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{P}_{4} \mathrm{O}_{10}(\mathrm{s})\)

Step-by-Step Solution

Answer

The standard enthalpy change for the given reaction, \(\Delta H_{\text{ran}}^{\circ}\), is -1340.2 kJ/mol, calculated using the standard enthalpies of formation \(\Delta H_{f}^{\circ}\) from Table R-11.

Step 1: Find the standard enthalpies of formation from Table R-11

Look up the standard enthalpies of formation, \(\Delta H_{f}^{\circ}\), for the reactants and products in Table R-11: \[\Delta H_{f}^{\circ}(P_4O_6(s)) = -1640.1 \, \text{kJ/mol}\] \[\Delta H_{f}^{\circ}(O_2(g)) = 0 \, \text{kJ/mol}\] \[\Delta H_{f}^{\circ}(P_4O_{10}(s)) = -2980.3 \, \text{kJ/mol}\]

Step 2: Apply the equation to calculate the standard enthalpy change for the reaction

Now that we have the standard enthalpies of formation of the reactants and the product, we can apply the equation mentioned in the analysis: \[\Delta H_{\text{ran}}^{\circ} = \sum \Delta H_{f}^{\circ}(\text{products}) - \sum \Delta H_{f}^{\circ}(\text{reactants})\] We can substitute the enthalpy values found in step 1, considering that there are 2 moles of \(O_2(g)\) in the reaction: \[\Delta H_{\text{ran}}^{\circ} = \Delta H_{f}^{\circ}(P_4O_{10}(s)) - (\Delta H_{f}^{\circ}(P_4O_6(s)) + 2\Delta H_{f}^{\circ}(O_2(g)))\] \[\Delta H_{\text{ran}}^{\circ} = (-2980.3 \, \text{kJ/mol}) - (-1640.1\, \text{kJ/mol} + 2(0\, \text{kJ/mol}))\]

Step 3: Calculate the standard enthalpy change for the reaction

Now we just need to carry out the arithmetic: \[\Delta H_{\text{ran}}^{\circ} = -2980.3\, \text{kJ/mol} + 1640.1\, \text{kJ/mol}\] \[\Delta H_{\text{ran}}^{\circ} = -1340.2\, \text{kJ/mol}\] The standard enthalpy change for the reaction, \(\Delta H_{\text{ran}}^{\circ}\), is -1340.2 kJ/mol.

Key Concepts

Enthalpies of Formation

Enthalpies of formation are essential values in thermochemistry. They tell us how much energy is absorbed or released when one mole of a compound forms from its constituent elements at standard conditions. Standard conditions usually mean a pressure of 1 atmosphere and a temperature of 25°C (298 K). Each element has a standard state, which is the pure form of the element that is most stable under these conditions. For example, the enthalpy of formation for \( \mathrm{O}_2(g) \) is zero because it's in its elemental form. In contrast, \( \mathrm{P}_4\mathrm{O}_6(s) \) and \( \mathrm{P}_4\mathrm{O}_{10}(s) \) require certain amounts of energy to form from elemental phosphorus and oxygen under those conditions. These enthalpies can be found in tables, like Table R-11 in chemistry textbooks, and they serve as the building blocks for calculating the energy changes in reactions.

Chemical Reactions

Chemical reactions involve the breaking and forming of bonds to transform reactants into products. These processes either absorb or release energy. Endothermic reactions absorb energy from the surroundings, leading to a decrease in temperature of the immediate environment. In contrast, exothermic reactions release energy, often in the form of heat, causing an increase in temperature. The net energy change of a reaction is quantified by its enthalpy change \( ( \Delta H ) \), which can be calculated using the enthalpies of formation for the substances involved. A reaction is exothermic if it releases more energy in forming the products than it absorbs in breaking the reactants. Conversely, it's endothermic if it absorbs more energy in breaking the bonds of the reactants than it releases in forming the products. In the supplied reaction, the calculation shows that the reaction of \( \mathrm{P}_4\mathrm{O}_6 \) and \( \mathrm{O}_2 \) is exothermic with a standard enthalpy change of \( -1340.2\, \text{kJ/mol} \). This indicates that energy is released as the reactants transform into the product \( \mathrm{P}_4\mathrm{O}_{10}(s) \).

Thermochemistry

Thermochemistry is the branch of chemistry concerned with the heat involved in chemical processes. It helps us understand how energy flows in reactions, which is crucial for both academic studies and industrial applications. Thermochemistry involves various concepts, including heat, work, and enthalpy. One key aspect is determining whether a chemical process can occur spontaneously. This is closely related to enthalpy changes, where spontaneous reactions typically exude energy and have a negative enthalpy change directionally. In practice, we use standard enthalpies of formation to calculate the overall energy change in reactions. These calculations help predict reaction feasibilities and are vital for designing processes in fields like material science and engineering. Understanding the principles of thermochemistry and its tools allows us to control chemical reactions better and develop new technologies.