Question

The standard heats of formation of ${\mathrm{H}}_2\mathrm{O}(g),{\mathrm{H}}_2\mathrm{S}(g),{\mathrm{H}}_2\mathrm{Se}(g)$, and ${\mathrm{H}}_2\mathrm{Te}(g)$ are $-241.8,-20.17,+29.7,$ and $+99.6\mathrm{kJ}/$ mol, respectively. The enthalpies necessary to convert the elements in their standard states to one mole of gaseous atoms are $248,277,227,$ and $197\mathrm{kJ}/\mathrm{mol}$ of atoms for $\mathrm{O},\mathrm{S}$, Se, and Te, respectively. The enthalpy for dissociation of ${\mathrm{H}}_2$ is $436\mathrm{kJ}/\mathrm{mol}$. Calculate the average $\mathrm{H}-\mathrm{O},\mathrm{H}-\mathrm{S},\mathrm{H}-\mathrm{Se}$, and $\mathrm{H}-$ Te bond enthalpies, and comment on their trend.

Short answer

The calculated average bond enthalpies for H-O, H-S, H-Se, and H-Te are -291.9, -257.585, -207.65, and -157.7 kJ/mol, respectively. The bond enthalpies decrease as we move from H-O to H-Te, which can be attributed to the increasing size of the central atom and decreasing electronegativity as we move down the periodic table (from O to Te). The weaker the bond strength, the lower the bond enthalpy, making it easier to break the bond.

Step-by-step solution

Understand the reactions needed for bond enthalpy calculations

We define ΔHf as the heat of formation, ΔHat as the enthalpy required to convert the element in its standard state to gaseous atoms, and ΔH_diss as enthalpy for dissociation of a molecule. Using Hess' Law, we will set up an equation for each bond enthalpy. For example, consider the calculation of H-O bond enthalpy: We need to treat half of these reactions: $1/2H_2(g)+1/2O_2\to H-O$ Consider two reactions: 1. The formation of H2O (for which the enthalpy change is ΔHf): $H_2(g)+1/2O_2(g)\to H_{2}O(g)$ 2. The dissociation of H2 and O2 into their gaseous atoms (for which the enthalpy change is ΔH_diss and half of ΔHat): $1/2H_2(g)+1/2O_2\to H(g)+1/2O(g)$ Using these reactions, we can find the average H-O bond enthalpy.

Calculate the H-O bond enthalpy

According to Hess' Law, the enthalpy change of the desired reaction can be found by combining the equations and their corresponding enthalpy changes. From the given data, $\mathrm{\Delta }{H}_f=-241.8kJ/mol$ (for gaseous H2O), $\mathrm{\Delta }{H}_{diss}=436kJ/mol$ (for H2), and the enthalpy needed to convert O2 to 1/2O(g) can be obtained from the given data as $1/2\times 248=124kJ/mol$. Now, let's find the enthalpy for the desired reaction (H-O bond): ΔH (H-O) = ΔHf - ΔH_diss - ΔHat(O) ΔH (H-O) = -241.8 - 1/2(436) - 124 ΔH (H-O) = -241.8 - 218 - 124 = -583.8 kJ/mol However, this is the bond enthalpy for two H-O bonds in a water molecule. Thus, Average H-O bond enthalpy = (-583.8 / 2) kJ/mol = -291.9 kJ/mol

Calculate the H-S, H-Se, and H-Te bond enthalpies

Using the same method outlined in Steps 1 and 2, we can calculate the bond enthalpies for H-S, H-Se, and H-Te from the given values of ΔHf, ΔHat, and ΔH_diss. For H-S: ΔH (H-S) = ΔHf - ΔH_diss - ΔHat(S) ΔH (H-S) = -20.17 - 1/2(436) - 277 ΔH (H-S) = -20.17 - 218 - 277 = -515.17 kJ/mol Average H-S bond enthalpy = (-515.17 / 2) kJ/mol = -257.585 kJ/mol For H-Se: ΔH (H-Se) = ΔHf - ΔH_diss - ΔHat(Se) ΔH (H-Se) = 29.7 - 1/2(436) - 227 ΔH (H-Se) = 29.7 - 218 - 227 = -415.3 kJ/mol Average H-Se bond enthalpy = (-415.3 / 2) kJ/mol = -207.65 kJ/mol For H-Te: ΔH (H-Te) = ΔHf - ΔH_diss - ΔHat(Te) ΔH (H-Te) = 99.6 - 1/2(436) - 197 ΔH (H-Te) = 99.6 - 218 - 197 = -315.4 kJ/mol Average H-Te bond enthalpy = (-315.4 / 2) kJ/mol = -157.7 kJ/mol

Analyze the trend in bond enthalpies

Now we have calculated the average bond enthalpies for H-O, H-S, H-Se, and H-Te. We find that their values are -291.9, -257.585, -207.65, and -157.7 kJ/mol, respectively. It is observed that the bond enthalpies decrease as we move from H-O to H-Te. This trend can be attributed to the increasing size of the central atom and decreasing electronegativity as we move down the periodic table (from O to Te). The weaker the bond strength, the lower the bond enthalpy, making it easier to break the bond.

Key concepts

Hess' Law

Hess' Law is a powerful tool in thermodynamics that allows us to determine the enthalpy change of a complex chemical reaction by breaking it down into simpler steps. It is based on the principle that the total enthalpy change of a reaction is the same, regardless of whether it occurs in one step or multiple steps.

In this exercise, Hess' Law facilitates the calculation of bond enthalpies by setting up equations for each bond formation as a combination of known reactions. For example, to find the H-O bond enthalpy, we consider the formation of water from its elements and the dissociation into gaseous atoms. By rearranging these steps and applying Hess' Law, we can accurately calculate the average bond enthalpy. This makes Hess' Law extremely valuable for indirect determination of energy changes, especially when direct measurement is challenging.

Heat of Formation

The heat of formation, denoted as $\mathrm{\Delta }{H}_f$, is the enthalpy change when one mole of a compound forms from its elements in their standard states. It is a critical value used to determine energy changes in chemical reactions.

In this exercise, the heat of formation values for water, hydrogen sulfide, hydrogen selenide, and hydrogen telluride serve as starting points for evaluating the enthalpy changes necessary to calculate bond enthalpies. Since these reactions involve the formation of compounds from elemental gases, $\mathrm{\Delta }{H}_f$ values are essential in determining the total energy change associated with forming specific bonds. Understanding heat of formation helps us appreciate how energy-efficient or demanding the synthesis of molecules can be.

Enthalpy Change

Enthalpy change, represented by the symbol $\mathrm{\Delta }H$, is the heat absorbed or released in a chemical reaction at constant pressure. It provides insight into the energy dynamics of chemical processes. Enthalpy changes can be exothermic (release heat) or endothermic (absorb heat).

In bond enthalpy calculations, such as those in this exercise, a clear understanding of enthalpy change is crucial. For instance, to compute the bond enthalpy for H-O, we calculate the enthalpy change during water formation, splitting of hydrogen, and oxygen into gaseous atoms. Summing these values while applying Hess' Law gives us the desired bond enthalpy. This approach demonstrates how manipulation of different kinds of enthalpy changes leads to insight into bond energies and reaction energetics.

Periodic Table Trends

Periodic table trends reveal patterns in the properties of elements that help predict chemical behavior. As a group is descended, characteristics like atomic size increase, and electronegativity typically decreases. These trends have real impacts on bond enthalpy.

In this specific exercise, we observe a trend in bond enthalpies from H-O to H-Te. The bond strengths diminish, evidenced by lower bond enthalpies, as one moves down the group from oxygen to tellurium. This is because larger atomic radii and reduced electronegativity generally weaken the bond strength, making bonds easier to break. Understanding periodic trends allows us to predict and explain observed bond energy data and provides a systematic framework for anticipating chemical behaviors.