Question

Use bond energies (Table \(8.5 ),\) values of electron affinities (Table 7.7\()\) , and the ionization energy of hydrogen \((1312 \mathrm{kJ} / \mathrm{mol})\) to estimate \(\Delta H\) for each of the following reactions. a. \(\mathrm{HF}(g) \rightarrow \mathrm{H}^{+}(g)+\mathrm{F}^{-}(g)\) b. \(\mathrm{HCl}(g) \rightarrow \mathrm{H}^{+}(g)+\mathrm{Cl}^{-}(g)\) c. \(\mathrm{HI}(g) \rightarrow \mathrm{H}^{+}(g)+\mathrm{I}^{-}(g)\) d. \(\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{H}^{+}(g)+\mathrm{OH}^{-}(g)\)

Short answer

The estimated enthalpy changes for the given reactions are: a. HF(g) → H+(g) + F−(g); ∆H = 2205 kJ/mol b. HCl(g) → H+(g) + Cl−(g); ∆H = 2092 kJ/mol c. HI(g) → H+(g) + I−(g); ∆H = 1905 kJ/mol d. H2O(g) → H+(g) + OH−(g); ∆H = 1916 kJ/mol

Step-by-step solution

Write the General Formula for ∆H

We can use the following general formula to estimate the enthalpy change of the given reactions: ∆H = Sum of enthalpies of bonds broken - Sum of enthalpies of bonds formed 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)

Determine the Enthalpies of Bonds Broken and Formed

For each reaction, we must determine the enthalpies of the bonds that are broken and the bonds that are formed. We will find the appropriate bond energies (Table 8.5), ionization energies, and electron affinity values (Table 7.7) for each reaction component from the given information. a. For HF(g), the bond between H and F is broken. Bond energy of HF = 565 kJ/mol Ionization energy of H = 1312 kJ/mol Electron affinity of F = -328 kJ/mol b. For HCl(g), the bond between H and Cl is broken. Bond energy of HCl = 431 kJ/mol Ionization energy of H = 1312 kJ/mol Electron affinity of Cl = -349 kJ/mol c. For HI(g), the bond between H and I is broken. Bond energy of HI = 298 kJ/mol Ionization energy of H = 1312 kJ/mol Electron affinity of I = -295 kJ/mol d. For H2O(g), the bond between H and O is broken. Bond energy of O-H = 463 kJ/mol Ionization energy of H = 1312 kJ/mol Electron affinity of O = -141 kJ/mol

Calculate ∆H for Each Reaction

Use the general formula to estimate the enthalpy change for each reaction. a. ∆H = (Bond energy of HF + Ionization energy of H) - Electron affinity of F ∆H = (565 + 1312) - (-328) ∆H = 2205 kJ/mol b. ∆H = (Bond energy of HCl + Ionization energy of H) - Electron affinity of Cl ∆H = (431 + 1312) - (-349) ∆H = 2092 kJ/mol c. ∆H = (Bond energy of HI + Ionization energy of H) - Electron affinity of I ∆H = (298 + 1312) - (-295) ∆H = 1905 kJ/mol d. ∆H = (Bond energy of O-H + Ionization energy of H) - Electron affinity of O ∆H = (463 + 1312) - (-141) ∆H = 1916 kJ/mol

Conclusion

The estimated enthalpy changes for the given reactions are: a. HF(g) → H+(g) + F−(g); ∆H = 2205 kJ/mol b. HCl(g) → H+(g) + Cl−(g); ∆H = 2092 kJ/mol c. HI(g) → H+(g) + I−(g); ∆H = 1905 kJ/mol d. H2O(g) → H+(g) + OH−(g); ∆H = 1916 kJ/mol

Key concepts

Bond Energies

Bond energies are crucial when it comes to understanding reaction enthalpies. In chemistry, bond energy is the amount of energy needed to break one mole of a bond. This is a significant factor in reactions because breaking bonds requires energy. Conversely, forming bonds releases energy.
For the reactions provided in the original exercise, calculating the energy needed to break specific bonds helps estimate the enthalpy change, \(\Delta H\). Remember: The higher the bond energy, the stronger the bond. For example, the hydrogen-fluoride bond (HF) has a relatively high bond energy of 565 kJ/mol, which signifies a strong bond that requires more energy to break.
Understanding these bonds gives us insight into the stability and reactivity of molecules. Bond energies are usually presented in units of kJ/mol and can be found in tables, such as the one referenced in Table 8.5 of the textbook.

Ionization Energy

Ionization energy represents the amount of energy required to remove one mole of electrons from one mole of gaseous atoms or ions. It is an essential concept in determining \(\Delta H\) for reactions that involve ion formation.
In our context, we consider the ionization energy of hydrogen, specifically for converting H(g) to H⁺(g).
This process requires a substantial 1312 kJ/mol, a fact that contributes significantly to the enthalpy changes in reactions involving hydrogen. Ionization energy is a key factor in understanding why some elements form ions more readily than others. It also affects the reactivity and bonding behavior of elements, particularly in gaseous states.
Generally, ionization energy increases across a period, as the nuclear charge becomes stronger, attracting electrons more tightly and making them harder to remove.

Electron Affinity

Electron affinity measures the energy change when an electron is added to a neutral atom in the gaseous state to form an anion. This concept is the complement to ionization energy and plays a critical role in the estimation of reaction enthalpy.
In the reactions given, we calculate the electron affinity for converting gaseous atoms like F, Cl, and others to their anionic forms, such as F⁻.
This process might release energy, resulting in negative values for electron affinity, like F at -328 kJ/mol and Cl at -349 kJ/mol.
This negative value indicates that when these elements gain an electron, the system loses energy, often indicating a stable anion is formed. Electron affinity can vary widely among different elements, influencing their chemical behavior, particularly in reactions involving electron transfer.

Enthalpy Estimation

Estimating the enthalpy change, \(\Delta H\), for a reaction is vital for understanding the energetics of chemical processes. The enthalpy change indicates whether a reaction is endothermic (absorbs energy) or exothermic (releases energy).
Using the values of bond energies, ionization energy, and electron affinity, we can approximate \(\Delta H\) for reactions, considering both energy required for breaking bonds and energy released in formulating new bonds.
As per the formula: \(\Delta H = \text{{Sum of enthalpies of bonds broken}} - \text{{Sum of enthalpies of bonds formed}}\). Each component's contribution must be calculated precisely to derive accurate estimates.
This estimation is fundamental in practical applications, ranging from predicting the feasibility of reactions to designing processes for industrial synthesis. Breaking down a reaction step-by-step helps predict its overall energy changes and potential behavior.