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Chapter 12: Problem 88

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

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

Step by step solution

01

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.
02

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)
03

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.
04

Calculate electron affinity

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

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