Problem 88 - Chemistry The Central Science

Question

Consider the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \longrightarrow 2 \mathrm{HI}(g) .(\mathbf{a})\) Use the bond enthalpies in Table 5.4 to estimate \(\Delta H\) for this reaction, ignoring the fact that iodine is in the solid state. (b) Without doing a calculation, predict whether your estimate in part (a) is more negative or less negative than the true reaction enthalpy. (c) Use the enthalpies of formation in Appendix \(C\) to determine the true reaction enthalpy.

Step-by-Step Solution

Answer

The short answer to the given problem is: (a) The estimated enthalpy change for the reaction ignoring iodine's solid-state is \(\Delta H = 4 \, \text{kJ/mol}\). (b) The estimate is expected to be less negative than the true reaction enthalpy since it ignores the solid-state of iodine. (c) The true reaction enthalpy using enthalpies of formation is \(\Delta H_\text{true} = -52.96 \, \text{kJ/mol}\), confirming that the estimate is less negative than the true reaction enthalpy.

Step 1: Identify the bonds in the reaction

Begin by identifying the bonds involved in the given reaction: H2(g) + I2(s) -> 2HI(g) The bonds involved in this reaction are: 1. H-H bond (in H2) 2. I-I bond (in I2) 3. Two H-I bonds (in 2HI)

Step 2: Find bond enthalpies for each bond type

Next, use Table 5.4 (as mentioned in the exercise) to find the bond enthalpies for each bond type involved in the reaction: 1. H-H bond: 436 kJ/mol 2. I-I bond: 150 kJ/mol 3. H-I bond: 295 kJ/mol

Step 3: Calculate the reaction enthalpy using bond enthalpies

The enthalpy change for the reaction is the difference between the sum of the bond enthalpies of the products and the sum of the bond enthalpies of the reactants: \(\Delta H = \sum(\text{Bond enthalpies of products}) - \sum(\text{Bond enthalpies of reactants})\) \(\Delta H = 2 \times 295 \, \text{kJ/mol} - (436 + 150) \, \text{kJ/mol}\) \(\Delta H = 590 \, \text{kJ/mol} - 586 \, \text{kJ/mol}\) \(\Delta H = 4 \, \text{kJ/mol}\)

Step 4: Predict whether the estimate is more or less negative than the true reaction enthalpy

The calculation in step 3 ignored that iodine is in the solid state, so it is expected that this will affect the true value of the reaction enthalpy. Since it takes energy to go from a solid to a gaseous state, the true reaction enthalpy is expected to be more negative than the estimate.

Step 5: Calculate the true reaction enthalpy using enthalpies of formation

Use the enthalpies of formation from Appendix C to determine the true reaction enthalpy. We need the enthalpies of formation for all species involved in the reaction: 1. Enthalpy of formation for H2(g): 0 kJ/mol (by definition, since it's an element in its standard state) 2. Enthalpy of formation for I2(s): 0 kJ/mol (by definition, since it's an element in its standard state) 3. Enthalpy of formation for HI(g): -26.48 kJ/mol Now calculate the true reaction enthalpy using the formula: \(\Delta H_\text{true} = \sum \Delta H_\text{products} - \sum \Delta H_\text{reactants}\) \(\Delta H_\text{true} = 2 \times (-26.48 \, \text{kJ/mol}) - (0 + 0) \, \text{kJ/mol}\) \(\Delta H_\text{true} = -52.96 \, \text{kJ/mol}\) The true reaction enthalpy is more negative than the estimate, confirming our prediction from step 4.

Key Concepts

Bond Enthalpies

Understanding bond enthalpies is essential when studying chemical reactions. Bond enthalpy, also known as bond dissociation energy, is the amount of energy required to break one mole of a bond in a molecule to form separated atoms, usually in the gas phase. It is a measure of bond strength in a chemical bond.

When a chemical reaction occurs, bonds are broken in the reactants and new bonds are formed in the products. To estimate the overall energy change, or reaction enthalpy (Delta H), we account for the energy needed to break bonds and the energy released upon forming new bonds. The bond enthalpy values provided in reference tables are average quantities because bond energies can vary depending on the molecular environment.

To calculate the reaction enthalpy using bond enthalpies, we use the formula: Delta H = sum(Bond enthalpies of products) - sum(Bond enthalpies of reactants).By applying this principle to the reaction of hydrogen (H_{2}) and iodine (I_{2}) forming hydrogen iodide (HI), we can estimate the reaction enthalpy by subtracting the sum of the bond enthalpies of reactants from the sum of the bond enthalpies of products.

Enthalpies of Formation

The enthalpies of formation, or standard heats of formation, are another crucial concept linked to reaction enthalpy calculations. The standard enthalpy of formation (Delta H_f^{o}) of a compound is the change in enthalpy when one mole of the compound is formed from its elements in their standard states.

For elements in their standard states, such as O_{2}(g), H_{2}(g), and I_{2}(s) at 25°C and 1 atmosphere pressure, the enthalpy of formation is zero by convention. This simplifies calculations as formation from the standard state does not require any energy input or release.

The true reaction enthalpy can be more accurately determined using enthalpies of formation because it considers the state of reactants and products (solid, liquid, gas) and the conditions under which they react. The calculation follows the formula: Delta H_{true} = sum Delta H_f^{o}_{products} - sum Delta H_f^{o}_{reactants}.For the reactants and products, we find their respective Delta H_f^{o} values from a reference, such as Appendix C in chemistry textbooks. For the cited reaction, the true enthalpy change is found to be significantly more negative when considering the enthalpies of formation.

Chemical Bonds

Chemical bonds are the glue that holds atoms together in molecules. They result from the attraction between atoms that share or transfer valence electrons. The three major types of chemical bonds are ionic, covalent, and metallic. In our context, we are interested in covalent bonds, which involve the sharing of electron pairs between atoms.

The strength of a chemical bond is related to its bond enthalpy; stronger bonds require more energy to break. Factors such as bond length, bond order, and the presence of electronegative elements can affect bond strength. For example, the H-H bond is a single covalent bond and has a bond enthalpy of 436 kJ/mol, while the H-I bond, also a single bond but between hydrogen and the more electronegative iodine, has a bond enthalpy of 295 kJ/mol.

Understanding the types of chemical bonds and their associated energies helps predict the behavior of compounds during chemical reactions. This knowledge is used to calculate the energy changes that occur, as seen in our exercise, where the bond enthalpies of H-H, I-I, and H-I were essential in estimating the reaction enthalpy.