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Question: Use bond energies (Table 13.6), values of electron affinities (Table 12.8), and the ionization energy of hydrogen (1312 kJ/mol) to estimate𝛥H for each of the following reactions.

a. HF(g)H+(g) + F-(g)

b. HCl(g)H+(g)+ Cl-(g)

c. HI(g)H+(g) + I-(g)

d. H2O(g)H+(g) + OH-(g)

(Electron affinity of OH(g) -180. kJ/mol.)

Short Answer

Expert verified

The ionization energy of Hydrogen is 1312 KJ/mol.

The electron affinity of OH(g) -180 kJ/mol.

Required data for Bond energy and electron affinities are given in table 13.6 & table 12.8.

Step by step solution

01

Step 1:Describing the reactions, ionization energy, and electron affinity

In the gaseous state, the hydrogen atom is ionized where it loses an electron from valence orbit. The ionized form is a proton (H+).

H(g)H+(g) + e-(g)(1)

The energy required to ionize the atom is the ionization energy. Therefore, the required ionization energy is 1312 KJ/mol.

Also, for F, Cl and Br atoms tend to gain electrons which further form F-, Cl- and Br- respectively.

Hence,

F + e- F-(2)

Cl + e- Cl-

Br + e-Br-

OH + e-OH-

02

Step 2:Summing up the above reactions to get required reactions

By adding above reactions respectively with (1) will get reactions

HF(g) H+(g) + F-(g)

HCl(g)H+(g)+ Cl-(g)

HI(g) H+(g) + I-(g)

H2O(g)H+(g) + OH-(g)

03

Determining 𝛥H for the above reactions

The enthalpy of reaction (𝛥H)is

𝛥H =𝞢𝛥H(products) -𝞢𝛥H(reactants)

Here

𝛥H = [𝛥H(I.E of H) +𝛥H(E.A of X)] - [𝛥H(HX)]

Therefore,

For (6)

𝛥H = [𝛥H (I.E of H) +𝛥H (E.A of F)] - [𝛥H(HF)]

= {1312) + (-327.8)}– {- 565}

= 1549.2 KJ/mol

For (7)

𝛥H = [𝛥H (I.E of H) +𝛥H (E.A of Cl)] - [𝛥H(HCl)]

={1312 + (-348.7)} – {-427}

= 1390.3 KJ/mol

For (8)

𝛥H = {𝛥H (I.E of H) +𝛥H (E.A of Br)] - {𝛥H(HBr)]

= {1312 + (-324.5)} - {-363}

= 1350.5 KJ/mol

For (9)

𝛥H = {𝛥H (I.E of H) +𝛥H (E.A of OH)] - {𝛥H(HOH)]

= (1312 + (-180)) – (-286)

=1418 KJ/mol

04

Determining the final values

The enthalpy of reaction(𝛥H) is1549.2, 1390.3, 1350.5, and1418 KJ/mol.

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