Question

Average bond enthalpies are generally defined for gasphase molecules. Many substances are liquids in their standard state. thermochemical data from Appendix \(C\), calculate average bond enthalpies in the liquid state for the following bonds, and compare these values to the gas-phase values given in Table 8.4: (a) \(\mathrm{Br}-\mathrm{Br}\), from \(\mathrm{Br}_{2}(l) ;\) (b) \(\mathrm{C}-\mathrm{Cl}\), from \(\mathrm{CCl}_{4}(l) ;\) (c) \(\mathrm{O}-\mathrm{O}\), from \(\mathrm{H}_{2} \mathrm{O}_{2}(l)\) (assume that the \(\mathrm{O}-\mathrm{H}\) bond enthalpy is the same as in the gas phase). (d) What can you conclude about the process of breaking bonds in the liquid as compared to the gas phase? Explain the difference in the \(\Delta H\) values between the two phases.

Short answer

In conclusion, bond enthalpies are slightly higher in the liquid state compared to the gas phase for Br-Br and O-O, and slightly lower for C-Cl. The difference in ΔH values between the two phases can be attributed to stronger intermolecular forces in the liquid phase, such as van der Waals forces, hydrogen bonding, and dipole-dipole interactions, which stabilize the liquid structure and make it less energetically favorable to break bonds. In the gas phase, these forces are weaker, requiring slightly less energy to break bonds.

Step-by-step solution

(a) Calculate the average bond enthalpy of Br-Br in liquid state.

First, we need to write the balanced chemical equation for breaking the Br-Br bond in the liquid state: \[Br_{2(l)} \rightarrow 2Br_{(l)}\] Next, we'll use the thermochemical data from Appendix C to find the enthalpy change associated with this process: - ΔHf of Br₂(l) = 0 kJ/mol - ΔHf of Br(l) = +118.9 kJ/mol Now we can calculate the enthalpy change for breaking the Br-Br bond in the liquid state: ΔH = 2 × ΔHf(Br) - ΔHf(Br₂) ΔH = 2 × 118.9 - 0 ΔH = 237.8 kJ/mol The gas-phase value of Br-Br bond enthalpy is 193 kJ/mol. The liquid state bond enthalpy is higher, at 237.8 kJ/mol.

(b) Calculate the average bond enthalpy of C-Cl in the liquid state.

First, the balanced chemical equation for breaking the C-Cl bond in CCl₄(l) is as follows: \[CCl_{4(l)} \rightarrow C_{(l)} + 4Cl_{(l)}\] Now we'll use the thermochemical data from Appendix C for the enthalpy calculations: - ΔHf of CCl₄(l) = -139.3 kJ/mol - ΔHf of C(l) = not given, assume derived as +718 kJ/mol for graphite - ΔHf of Cl(l) = +121 kJ/mol Now, we can calculate ΔH: ΔH = ΔHf(C) + 4 × ΔHf(Cl) - ΔHf(CCl₄) ΔH = 718 + 4 × 121 - (-139.3) ΔH = 1251.3 kJ/mol Divide by four to obtain the average bond enthalpy of C-Cl in the liquid state: Average bond enthalpy (C-Cl) = 1251.3 / 4 = 312.825 kJ/mol Comparing this to the gas-phase value (332 kJ/mol), we see that the liquid state C-Cl bond enthalpy is slightly lower, at 312.825 kJ/mol.

(c) Calculate the average bond enthalpy of O-O in the liquid state.

The balanced chemical equation for breaking the O-O bond in H₂O₂(l) is: \[H_{2}O_{2(l)} \rightarrow 2OH_{(l)}\] Using the thermochemical data from Appendix C, we have: - ΔHf of H₂O₂(l) = -187.8 kJ/mol - ΔHf of OH(g) = +37.3 kJ/mol (it will be assumed to be the same for OH(l)) Now, we can calculate ΔH: ΔH = 2 × ΔHf(OH) - ΔHf(H₂O₂) ΔH = 2 × 37.3 - (-187.8) ΔH = 262.4 kJ/mol The gas-phase value of the O-O bond in H₂O₂ is 207 kJ/mol. In the liquid state, the bond enthalpy is higher, at 262.4 kJ/mol.

(d) Conclusions about breaking bonds in the liquid vs. gas phase.

From our calculations, we found that bond enthalpies tend to be slightly higher in the liquid state compared to the gas phase for Br-Br and O-O, and slightly lower for C-Cl. The difference in ΔH values between the two phases can be explained by the environment in which the molecules exist. In the liquid phase, there are more intermolecular forces at play, such as van der Waals forces, hydrogen bonding, and dipole-dipole interactions. These forces can stabilize the liquid structure, making it somewhat less energetically favorable to break bonds. In the gas phase, these forces are weaker, and so it takes slightly less energy to break bonds.

Key concepts

Thermochemistry

Thermochemistry is the branch of chemistry that deals with the study of heat and energy related to chemical processes and physical transformations. It is essential to understand when calculating bond enthalpies because the quantities measured are directly related to the energy changes that occur during chemical reactions.

In the context of thermochemistry, the heat change that occurs at constant pressure is termed enthalpy (H). Entropy changes provide insight into the energy requirements or release during bond breaking and making. Therefore, in the process of calculating bond enthalpies, especially while transitioning between phases like from gas to liquid, it's important to account for the heat absorbed or released, since this directly affects the calculated bond strength.

Enthalpy Change

Enthalpy change, denoted as H, refers to the heat content change in a system when it undergoes a reaction or a phase change at constant pressure. The exercise discussed measures the enthalpy change as the energy required to break a bond between two atoms, which is the bond enthalpy.

Positive H values indicate that a system absorbs heat (endothermic process), while negative H values mean it releases heat (exothermic process). The difference in enthalpy change between gases and liquids can be explained by the additional intermolecular forces in the liquid state, which require extra energy to overcome during bond breaking.

Chemical Bonds

Chemical bonds are the attractions that hold atoms together in molecules or compounds. The strength of these bonds is quantified by bond enthalpy, which is the amount of energy required to homolytically break a bond in a mole of a substance. It is typically calculated in the gas phase since the molecules are far apart and interactions with other molecules are minimized.

However, as seen in our example, the calculation can extend to the liquid state where we consider additional energy due to intermolecular forces. The difference in energy between breaking a bond in gas and liquid phases highlights the role these forces play in a substance's physical state.

Intermolecular Forces

Intermolecular forces are forces of attraction or repulsion which act between neighboring particles (atoms, molecules or ions). They include van der Waals forces, hydrogen bonds, and dipole-dipole interactions. These forces significantly influence the physical properties of compounds, including boiling and melting points, vapor pressure, and viscosity.

In liquids, these forces are stronger and more numerous compared to gases, leading to higher enthalpies of vaporization or fusion. As the solution examples show, these forces must be overcome to break chemical bonds in the liquid phase, hence the observed discrepancies in bond enthalpies between the liquid and gas phases.