Question

Study the following thermodynamic data given by E. H. P. Cordfunke, A. S. Booji and M. Y. Furkalionk. (i) \(\mathrm{DyCl}_{3}(\mathrm{~s}) \rightarrow \mathrm{DyCl}_{3}(\) aq., in \(4.0 \mathrm{M}-\mathrm{HCl}) ;\) \(\Delta H^{0}=-180.06 \mathrm{kJmol}^{-1}\) (ii) \(\mathrm{Dy}(\mathrm{s})+3 \mathrm{HCl}(\mathrm{aq}, 4.0 \mathrm{M}) \rightarrow \mathrm{DyCl}_{3}\) \((\mathrm{aq}\), in \(4.0 \mathrm{M}-\mathrm{HCl})+3 / 2 \mathrm{H}_{2}(\mathrm{~g})\) \(\Delta H^{\circ}=-699.43 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (iii) \(1 / 2 \mathrm{H}_{2}(\mathrm{~g})+1 / 2 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{HCl}_{\text { }}(\mathrm{aq},\), \(4.0 \mathrm{M}) ; \Delta H^{\circ}=-158.31 \mathrm{~kJ} \mathrm{~mol}^{-1}\) What is \(\Delta H_{\mathrm{f}}^{0}\) of \(\mathrm{DyCl}_{3}(\mathrm{~s})\) from these data? (a) \(-248.58 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(-994.30 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-3977.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(-1469.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Short answer

-994.30 kJ mol^{-1}

Step-by-step solution

- Write the target equation for \( \Delta H_{\mathrm{f}}^{0} \) of \( \mathrm{DyCl}_{3}(\mathrm{~s}) \)

You want to find the standard heat of formation (\( \Delta H_{\mathrm{f}}^{0} \)) of \( \mathrm{DyCl}_{3}(\mathrm{~s}) \). This refers to the enthalpy change when one mole of a substance is formed from its elements in their standard states. The general reaction for the formation of \( \mathrm{DyCl}_{3}(\mathrm{~s}) \) is:\[ \mathrm{Dy}(\mathrm{s}) + \frac{3}{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{DyCl}_{3}(\mathrm{~s}) \]

- Use Hess's Law

We can use Hess's Law to combine the given reactions to form the target equation from Step 1. Hess's Law states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change for the final reaction is the sum of the enthalpy changes of the individual reactions.

- Modify the given reactions

Modify the given reactions to match the elements in the target reaction:\[\text{(ii)} \quad \mathrm{Dy}(\mathrm{s}) + 3 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{DyCl}_{3}(\mathrm{aq}) + \frac{3}{2} \mathrm{H}_{2}(\mathrm{g}) \quad \Delta H^{\circ} = -699.43 \, \mathrm{kJ} \, \mathrm{mol}^{-1}\]\[\text{(iii)} \quad \frac{3}{2} \left( \frac{1}{2} \mathrm{H}_{2}(\mathrm{g}) + \frac{1}{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{HCl}(\mathrm{aq}) \right) \quad \Delta H^{\circ} = 3 \times (-158.31) \, \mathrm{kJ} \, \mathrm{mol}^{-1}\]\[\text{(i)} \quad \mathrm{DyCl}_{3}(\mathrm{s}) \rightarrow \mathrm{DyCl}_{3}(\mathrm{aq}) \quad \Delta H^{\circ} = -180.06 \, \mathrm{kJ} \, \mathrm{mol}^{-1}\]We reverse reaction (i) to get \( \Delta H^{\circ} \) for the opposite process, and multiply reaction (iii) by 3 to get the amount of \( \mathrm{HCl} \) used in reaction (ii).

- Combine the modified reactions

Combine the equations, ensuring that the species that are not found in the target equation cancel out:\[\begin{align*}-\Delta H^{\circ}_\text{(i)} & : \mathrm{DyCl}_{3}(\mathrm{aq}) \rightarrow \mathrm{DyCl}_{3}(\mathrm{s}) & (+180.06 \, \mathrm{kJ} \, \mathrm{mol}^{-1}) \Delta H^{\circ}_\text{(ii)} & : \mathrm{Dy}(\mathrm{s}) + 3 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{DyCl}_{3}(\mathrm{aq}) + \frac{3}{2} \mathrm{H}_{2}(\mathrm{g}) & (-699.43 \, \mathrm{kJ} \, \mathrm{mol}^{-1}) \(3 \times \Delta H^{\circ}_\text{(iii)}\) & : 3(\frac{1}{2} \mathrm{H}_{2}(\mathrm{g}) + \frac{1}{2} \mathrm{Cl}_{2}(\mathrm{g})) \rightarrow 3 \mathrm{HCl}(\mathrm{aq}) & (-3 \times 158.31 \, \mathrm{kJ} \, \mathrm{mol}^{-1}) \end{align*}\]After combining, the \( \mathrm{HCl}(\mathrm{aq}) \) and \( \mathrm{DyCl}_{3}(\mathrm{aq}) \) drop out, leaving the formation of \( \mathrm{DyCl}_{3}(\mathrm{s}) \) from its elements.

- Calculate total enthalpy change for the target reaction

Add the enthalpy changes of the individual reactions after considering their stoichiometry and the direction of the reaction:\[ \Delta H_{\text{total}} = +180.06 + (-699.43) + (3 \times -158.31) \, \mathrm{kJ} \, \mathrm{mol}^{-1} \]

- Perform the calculations

Carry out the arithmetic:\[ \Delta H_{\text{total}} = 180.06 - 699.43 - (3 \times 158.31) \]\[ \Delta H_{\text{total}} = 180.06 - 699.43 - 474.93 \]\[ \Delta H_{\text{total}} = -994.3 \, \mathrm{kJ} \, \mathrm{mol}^{-1} \]This will be the \( \Delta H_{\mathrm{f}}^{0} \) of \( \mathrm{DyCl}_{3}(\mathrm{s}) \).

Key concepts

Hess's Law

Hess's Law is a principle that asserts the total energy change for a chemical reaction is the same regardless of the number of steps the reaction is carried out in. This law is named after Germain Hess, a Russian chemist, and is fundamental in solving thermodynamics problems in chemistry.

Hess's Law is based on the first law of thermodynamics, which states that energy cannot be created or destroyed, making it possible to calculate the enthalpy change for a reaction even if it cannot be measured directly. Instead, chemists use chemical equations of known enthalpy changes related to the reaction of interest.

By manipulating these equations and combining their respective enthalpy changes—keeping in mind the stoichiometric coefficients and the directions of the reactions—we can deduce the enthalpy change for the overall reaction. This process critically relies on the idea that the enthalpy change for the reverse reaction is equal in magnitude but opposite in sign to the forward reaction. Moreover, if one multiplies both sides of a chemical equation by a constant, the enthalpy change for that reaction is also multiplied by the same constant.

In any thermodynamics problem involving Hess's Law, it's crucial to write down all known reactions and enthalpy changes, manipulate these equations to resemble the target equation, and then sum the enthalpy changes to find the enthalpy of the overall reaction.

Enthalpy Change

Enthalpy change, denoted by the symbol \( \Delta H \) and measured in kJ/mol, refers to the heat absorbed or released during a chemical reaction at constant pressure. It is a central concept in thermodynamics and chemistry that gauges the amount of energy exchanged with the surroundings during a chemical process.

The enthalpy change can be either positive or negative. A positive \( \Delta H \) indicates an endothermic reaction, where heat is absorbed from the surroundings, while a negative \( \Delta H \) indicates an exothermic reaction, in which heat is released to the surroundings. Understanding enthalpy change is essential for predicting how reactions will respond to changes in temperature, pressure, and concentration.

Students often encounter the concept of enthalpy change in the context of standard enthalpy changes, such as the standard enthalpy of formation. Calculating these values provides insights into the stability of substances and their potential to react under standard conditions. When solving enthalpy problems, it's important to pay attention to the state of matter of the reactants and products, as enthalpy change is dependent on these states.

Standard Heat of Formation

The standard heat of formation, or standard enthalpy of formation, symbolized as \( \Delta H_{f}^{0} \), is the change in enthalpy when one mole of a compound is formed from its elements in their standard states at 1 atm and 298.15 K (25°C). The standard state refers to the most stable physical form of an element or compound under standard conditions.

By convention, the standard enthalpy of formation for an element in its standard state is zero. This serves as a reference point for calculating the \( \Delta H_{f}^{0} \) of compounds. These values are quintessential for chemists and engineers because they allow for the calculation of enthalpy changes in reactions where direct measurements are not feasible.

To find the standard enthalpy of formation for a compound, like \( \mathrm{DyCl}_{3}(\mathrm{~s}) \) in our exercise, we use known enthalpy changes from other reactions involving its formation. By applying Hess's Law, we can systematically determine the \( \Delta H_{f}^{0} \) from these reactions. This ability to calculate enthalpy changes based on the formation of compounds from their elements is indispensable for predicting and understanding chemical behavior in various scientific and industrial processes.