Question

In ionic compounds with certain metals, hydrogen exists as the hydride ion, \(\mathrm{H}^{-}\). Determine the electron affinity of hydrogen; that is, \(\Delta H\) for the process \(\mathrm{H}(\mathrm{g})+e^{-} \rightarrow \mathrm{H}^{-}(\mathrm{g}) .\) To do so, use data from Section \(12-7 ;\) the bond energy of \(\mathrm{H}_{2}(\mathrm{g})\) from table 10.3 \(-812 \mathrm{kJmol}^{-1} \mathrm{NaH}\) for the lattice energy of \(\mathrm{NaH}(\mathrm{s})\) and \(-57 \mathrm{kJmol}^{-1}\) NaH for the enthalpy of formation of \(\mathrm{NaH}(\mathrm{s})\)

Short answer

The electron affinity of hydrogen in this case is +465 \, kJ \, mol^{-1}

Step-by-step solution

Identify the given energy parameters

We're given four pieces of data: the bond energy of Hydrogen \(-812 \, kJ \, mol^{-1}\), the lattice energy of Sodium Hydride, the enthalpy of formation of Sodium Hydride \(-57\, kJ \, mol^{-1}\), and the ionization energy of Sodium which can be found from the data table.

Write the corresponding thermodynamic equations

Next, write down the thermodynamic equations for the processes that correspond to these energy parameters. \n (1) \( \mathrm{H}_2 (g) \rightarrow 2H(g)\) with a ΔH of +812 kJ mol^(-1)\n (2) \(\mathrm{Na}(g) \rightarrow \mathrm{Na^+}(g) + e^-\) The ΔH for this process is +496 kJ mol^(-1) which is the ionization energy of Na and can be found in the data table as mentioned earlier\n (3) \(\mathrm{Na^+}(g) + H^- (g) \rightarrow \mathrm{NaH}(s)\) The ΔH for this process is the opposite of the lattice energy of NaH, therefore it's -900 kJ mol^(-1)\n (4) \(\mathrm{H}(g) + \mathrm{Na}(g) \rightarrow \mathrm{NaH}(s)\) The ΔH for this process is the enthalpy of formation of NaH, -57 kJ mol^(-1)

Applying Hess's Law

Combine equations (1), (2) and (3) so that they add up to equation (4) while respecting the law of conservation of energy. Be sure to adjust the associated ΔH values for each step accordingly. When you add all these equations up the result should be \[ ΔH = ΔH_1 + ΔH_2 + ΔH_3 = (+812) + (+496) + (-900) = +408 kJ mol^{-1} \]. Compare this sum to the actual enthalpy of reaction, which is -57 kJ mol^{-1}. The difference between these values will give the electron affinity of Hydrogen.

Calculate electron affinity

Finally, subtract the actual enthalpy of reaction from the calculated value to solve for electron affinity: Electron affinity = \[ ΔH_{calculated} - ΔH_{actual} = 408 -(-57) = +465 kJ mol^{-1} \]

Key concepts

Hydride Ion

In chemistry, the hydride ion is represented as \( \text{H}^- \). It's an anion of hydrogen, formed when hydrogen gains an extra electron. This process creates a negatively charged ion due to the surplus of electrons compared to protons. Hydride ions are particularly interesting because they are involved in various chemical reactions, particularly with certain metals to form compounds. Such ions are critical in the formation of ionic compounds, where they pair with metal cations to create stable ionic bonds. The formation of a hydride ion involves an electron capture by a hydrogen atom, which links directly to the concept of electron affinity. Electron affinity measures how easily an atom can gain an electron. For hydrogen, this is represented by the equation \( \text{H(g)} + e^- \rightarrow \text{H}^-(g) \), highlighting the formation of the hydride ion in gaseous phase.

Hess's Law

Hess's Law is an important principle in thermodynamics. It states that the total enthalpy change for a chemical reaction is the same, no matter how many stages the reaction is carried out in. This means you can add up the enthalpy changes of individual reaction steps to obtain the overall enthalpy change. Practically, this allows chemists to calculate enthalpy changes for reactions that might be difficult to measure directly. Hess's Law is a reflection of the conservation of energy, which states that energy cannot be created or destroyed.In the context of the exercise, Hess's Law is used to determine the overall energy change when sodium hydride \( \text{NaH} \) is formed. By reorganizing and combining known reaction steps, the total enthalpy change can be derived. This calculation is instrumental in finding the electron affinity of hydrogen.

Thermodynamic Equations

Thermodynamic equations are mathematical expressions used to represent the relationships between different thermodynamic properties. They describe the energy changes that occur during chemical reactions, such as bond formation or breaking.In this exercise, several thermodynamic equations are employed:- Breaking the bond in \( \text{H}_2 \) leading to the formation of individual hydrogen atoms: \( \text{H}_2 \rightarrow 2\text{H} \). The energy required for this is 812 kJ/mol.- Ionizing \( \text{Na} \) to form \( \text{Na}^+ \) and an electron: \( \text{Na} \rightarrow \text{Na}^+ + e^- \), which requires 496 kJ/mol.- Forming \( \text{NaH} \) from gas-phase \( \text{Na}^+ \) and \( \text{H}^- \), where the energy involved corresponds to the lattice energy.These thermodynamic equations allow us to compute the energy changes needed for the formation of sodium hydride from its elements, helping us understand the reaction dynamics fully.

Lattice Energy

Lattice energy is a measure of the energy released when ions combine to form an ionic solid. It is one of the key energies involved in the formation of compounds like sodium hydride \( \text{NaH} \).The lattice energy can be thought of as the strength of the forces holding the ions together in the crystal lattice. A high lattice energy indicates a very stable compound, as it means more energy is required to break the compound into its respective ions.In the exercise, lattice energy plays a pivotal role in calculating the electron affinity of hydrogen. By understanding the energy released or required to form \( \text{NaH} \) from \( \text{Na}^+ \) and \( \text{H}^- \), it becomes easier to apply Hess's Law efficiently and accurately. The lattice energy for NaH is negative, signaling that energy is released, thus highlighting the exothermic nature of the compound's formation.