Question

For reactions in condensed phases (liquids and solids), the difference between \(\Delta H\) and \(\Delta U\) is usually quite small. This statement holds for reactions carried out under atmospheric conditions. For certain geochemical processes, however, the external pressure may be so great that \(\Delta H\) and \(\Delta U\) can differ by a significant amount. A well-known example is the slow conversion of graphite to diamond under Earth's surface. Calculate \(\Delta H-\Delta U\) for the conversion of 1 mole of graphite to 1 mole of diamond at a pressure of 50,000 atm. The densities of graphite and diamond are \(2.25 \mathrm{~g} / \mathrm{cm}^{3}\) and \(3.52 \mathrm{~g} / \mathrm{cm}^{3},\) respectively.

Short answer

\(\Delta H - \Delta U \approx 9.78 \, \text{kJ/mol}\) for the conversion.

Step-by-step solution

Calculate Molar Volume of Graphite

First, determine the molar volume of graphite using its density. The molecular weight of carbon is 12.01 g/mol. Using the equation for volume, \( V = \frac{m}{d} \), we find the volume of 1 mole of graphite:\[ V_{\text{graphite}} = \frac{12.01 \, \text{g/mol}}{2.25 \, \text{g/cm}^3} \approx 5.34 \, \text{cm}^3/\text{mol} \]

Calculate Molar Volume of Diamond

Next, we calculate the molar volume of diamond using its density. Again, using the molecular weight of carbon, calculate the volume:\[ V_{\text{diamond}} = \frac{12.01 \, \text{g/mol}}{3.52 \, \text{g/cm}^3} \approx 3.41 \, \text{cm}^3/\text{mol} \]

Calculate the Change in Molar Volume

Find the change in molar volume for the reaction transitioning graphite to diamond. This is the difference between the molar volumes:\[ \Delta V = V_{\text{diamond}} - V_{\text{graphite}} = 3.41 \, \text{cm}^3/\text{mol} - 5.34 \, \text{cm}^3/\text{mol} \approx -1.93 \, \text{cm}^3/\text{mol} \]

Convert Pressure to Appropriate Units

Note that pressure is given in atmospheres but we commonly use pascals in calculations involving pressure-volume work. The conversion factor is:\[ 1 \, \text{atm} = 101325 \, \text{Pa} \]So, for 50,000 atm:\[ P = 50000 \, \text{atm} \times 101325 \, \text{Pa/atm} = 5.06625 \times 10^9 \, \text{Pa} \]

Calculate Work Done in Conversion

The work done, \( \Delta P \Delta V \), can now be calculated. Use the formula:\[ W = -P \cdot \Delta V = -(5.06625 \times 10^9 \, \text{Pa}) \times (-1.93 \times 10^{-6} \, \text{m}^3/\text{mol}) \]Note the unit conversion: 1 cm\(^3\) = 1 \(\times 10^{-6}\) m\(^3\). This results in:\[ W \approx 9.78 \, \text{kJ/mol} \]

Relate Enthalpy to Internal Energy

The relationship between enthalpy change and internal energy change is given by:\[ \Delta H = \Delta U + \Delta (PV) \]Where \( \Delta (PV) = P \Delta V = W \). Thus:\[ \Delta H - \Delta U = W \approx 9.78 \, \text{kJ/mol} \]

Key concepts

Enthalpy

Enthalpy, often represented by the symbol \( H \), is a crucial concept in thermodynamics. It is a measure of the total heat content of a system. Enthalpy considers both the internal energy of a system and the product of its pressure and volume. In a chemical reaction, the change in enthalpy, \( \Delta H \), indicates the amount of heat absorbed or released at constant pressure.
For reactions in condensed phases (like the conversion of graphite to diamond), the enthalpy change can often be small. However, under extreme pressures, such as those found deep within the Earth, the difference between enthalpy change and internal energy change can be significant.
To understand the relationship between these changes during the graphite-to-diamond conversion, we look at the equation \( \Delta H = \Delta U + P\Delta V \). The term \( P\Delta V \) represents the work done by the system due to pressure-volume changes, playing a key role in the enthalpy change under these conditions.

Internal Energy

Internal energy, denoted as \( U \), is another fundamental concept in thermodynamics. It refers to the total energy contained within a chemical system, arising from kinetic and potential energies of molecules. Unlike enthalpy, internal energy does not consider the external pressure and volume that the system occupies, making it more intrinsic to the nature of the materials in the system.
In chemical reactions, the change in internal energy, \( \Delta U \), reflects the energy difference between products and reactants. For the graphite to diamond conversion, despite the large external pressure, the \( \Delta U \) might be small because it primarily accounts for the intrinsic energy change within the carbon atoms as they rearrange from one form to another.
Understanding \( \Delta U \) helps us grasp the intrinsic energy shifts, complementing our comprehension of enthalpy in terms of the reaction's heat exchange overall.

Pressure-Volume Work

Pressure-volume work, often abbreviated as PV work, is a term in thermodynamics that represents the work done by or against a system as its volume changes under pressure. Calculated as \( W = -P \cdot \Delta V \), this value becomes significant in processes under high pressure.
For instance, during the conversion of graphite to diamond, a high external pressure of 50,000 atm is applied. This immense pressure leads to a measurable change in the volume of the substance, from the looser structure of graphite to the extremely dense diamond form. It is this pressured transition that makes the difference in work for \( \Delta H \) versus \( \Delta U \) notable.
Thus, recognizing how pressure-volume work influences energy states can clarify how substances behave differently under varying pressures.

Molar Volume

Molar volume is a term that refers to the volume occupied by one mole of a substance. This quantity helps understand the space each molecule takes, crucial in exploring reactions under different conditions, such as high pressure.
In the original exercise, we calculated the molar volume of graphite and diamond to determine how they differ in structure. Graphite, having a regular molar volume of approximately 5.34 cm\(^3\)/mol, contrasts with diamond's much denser form at approximately 3.41 cm\(^3\)/mol.
When analyzing the change in molar volume \( \Delta V \) as part of a reaction such as graphite to diamond, it's critical as it directly influences the pressure-volume work element in the reaction's enthalpy calculation. A decrease in molar volume signifies the compact nature of the resulting product, diamond, explaining its distinct physical attributes compared to graphite.