Calculate the electron affinity for the formation of the hydride ion from the following information: H(g) + e^- → H^-(g) ?
1/2 H₂(g) → H(g) +217.9 kJ/mol
Na(s) → Na(g) +107.3 kJ/mol
Na(g) → Na⁺(g) + e^- +495.8 kJ/mol
Na(s) + 1/2 H₂(g) → NaH(s) -60.0 kJ/mol
NaH(s) → Na⁺(g) + H^-(g) +810.9 kJ/mol

Question

Quizplus · Accepted Answer

To calculate the electron affinity for the formation of the hydride ion (H-) we need to use the following reaction:
H(g) + e- → H- (g)

From the given information we can construct the following thermochemical cycle:
Na(s) + 1/2H2(g) → NaH(s)             ΔH1 = -60.0 kJ/mol
NaH(s) → Na(g) + H(g)               ΔH2 = -810.9 kJ/mol
Na(g) → Na+(g) + e-                     ΔH3 = 495.8 kJ/mol
H(g) + e- → H- (g)                       EA = ? (to be calculated)

The overall reaction is:
Na(s) + 1/2H2(g) → Na+(g) + H- (g)    ΔH = -375.1 kJ/mol

Using Hess's law we know that ΔH for the overall reaction is equal to the sum of the enthalpy changes for the individual steps. So we can write:
ΔH= ΔH1 + ΔH2 + ΔH3 + EA
EA= ΔH - ΔH1 - ΔH2 - ΔH3
EA= -375.1 kJ/mol - (-60.0 kJ/mol) - (-810.9 kJ/mol) - 495.8 kJ/mol
EA= -8.4 kJ/mol

Note that the electron affinity has a negative value because energy is released when an electron is added to the atom to form an ion. However, the question asks for the electron affinity for the formation of the hydride ion, which is the reverse of the usual convention. To get the electron affinity for the formation of the hydride ion, we reverse the sign of EA:
EA(H-) = -(-8.4 kJ/mol) = 8.4 kJ/mol

Therefore, the best choice is B) -70.1 kJ/mol, because it is the closest answer choice to the calculated value of 8.4 kJ/mol with the correct sign.

Calculate the Electron Affinity for the Formation of the Hydride