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Under the entry \(\mathrm{H}_{2} \mathrm{SO}_{4},\) a reference source lists many values for the standard enthalpy of formation. For example, for pure \(\mathrm{H}_{2} \mathrm{SO}_{4}(1), \Delta H_{\mathrm{f}}^{\circ}=-814.0 \mathrm{kJ} / \mathrm{mol}\) for a solution with \(1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) \(-841.8 ;\) with \(10 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-880.5 ;\) with \(50 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) \(-886.8 ;\) with \(100 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-887.7 ;\) with \(500 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) \(-890.5 ;\) with \(1000 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-892.3 ;\) with \(10,000 \mathrm{mol}\) \(\mathrm{H}_{2} \mathrm{O},-900.8 ;\) and with \(100,000 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-907.3\) (a) Explain why these values are not all the same. (b) The value of \(\Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\right]\) in an infinitely dilute solution is \(-909.3 \mathrm{kJ} / \mathrm{mol} .\) What data from this chapter can you cite to confirm this value? Explain. (c) If \(500.0 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is prepared from pure \(\mathrm{H}_{2} \mathrm{SO}_{4}(1),\) what is the approximate change in temperature that should be observed? Assume that the \(\mathrm{H}_{2} \mathrm{SO}_{4}(1)\) and \(\mathrm{H}_{2} \mathrm{O}(1)\) are at the same temperature initially and that the specific heat of the \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is about \(4.2 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1}\).

Short Answer

Expert verified
As the number of water moles increases in the formation of \(H_2SO_4\), the enthalpy change (∆H) decreases. Hence, ∆Hf of \(H_2SO_4\) in different quantities of water are not the same. The ∆Hf of -909.3 kJ/mol for infinite dilution can be confirmed from the data trend. The temperature change for formation of 500.0 mL, 1.00 M \(H_2SO_4\) from pure \(H_2SO_4\) can be found using the heat exchange equation and comes out to be -21.8°C.

Step by step solution

01

Understanding Enthalpy Change

The enthalpy of formation is the heat change that results when one mole of a compound is formed from its elements. The differing enthalpy values for \(H_2SO_4\) come from it being in different solutions. As the ion becomes more and more dilute, the interactions with the water around it change, leading to different enthalpies of formation.
02

Confirming ∆Hf of Dilute \(H_2SO_4\)

The heat of formation of an infinitely dilute solution is the energy released or absorbed when 1 mole of a substance is dispersed infinitely throughout a solution. This value can be confirmed from the chapter data or from experiments by logic that as the number of water moles increases, the ∆Hf value decreases, approaching -909.3 kJ/mol.
03

Calculate Temperature Change

First, calculate the number of moles of \(H_2SO_4\) in 500.0 mL of 1.00 M solution, which is 0.500 mol. The heat added to the water can be found using ∆H = moles × ∆Hf, which is -909.3 kJ/mol × 0.500 mol = -454.65 kJ = -454650 J (since 1 kJ = 1000 J). Now, use the formula for heat exchange, q = mass × specific heat × ∆T. Substituting the known values, -454,650 J = 500 g × 4.2 J/g°C × ∆T, calculate ∆T.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dilution Effect
When we talk about the dilution effect, we're referring to how the concentration of a solution can influence various properties, such as its enthalpy of formation. As a solution becomes more dilute, with increasing amounts of solvent, the interactions between solute molecules change. This is because there are more solvent molecules available to interact with each solute molecule.

In the case of \(\mathrm{H}_{2} \mathrm{SO}_{4}\), as more water is added, the solution becomes less concentrated. The sulfuric acid molecules have more water molecules surrounding them, leading to changes in how these molecules interact energetically. This results in different values for the enthalpy of formation, which reflects the energy required or released during these interactions.
  • In highly concentrated solutions, the enthalpy of formation may differ largely due to strong solute-solute interactions.
  • As dilution increases, solute-solvent interactions become more dominant, altering the overall energy dynamics.
This is why data shows decreasing enthalpy of formation values as the solution becomes more diluted. Ultimately, the effects of dilution become minimal when the solution approaches infinite dilution.
Enthalpy Change
Enthalpy change, or \(\Delta H\), is essentially the heat change that occurs at constant pressure. It is a crucial concept in understanding chemical reactions and energy changes in processes. Specifically, the enthalpy of formation is the change in enthalpy when one mole of a substance is created from its constituent elements in their standard states.

For a substance like \(\mathrm{H}_{2} \mathrm{SO}_{4}\), different values of enthalpy of formation depend on the physical state and concentration of the substance in the solution. When sulfuric acid is prepared in water, the solution's enthalpy change depends on how much water is present.
  • Harsher conditions, such as initial concentrations, cause a different magnitude of energy change during formation compared to infinite dilution.
  • In a completely diluted state, the values tend to stabilize, depicting very minute changes in \(\Delta H\) as water molecules are in excess.
This behavior highlights the importance of considering the concentration and conditions in any chemical calculation or prediction involving enthalpy.
Heat of Formation
The heat of formation refers to the enthalpy change specifically when one mole of a compound forms from its individual elements in their stable states. It's a foundational concept in thermochemistry that allows us to compare the stability of different compounds and predict reaction behavior.

In our particular case of \(\mathrm{H}_{2} \mathrm{SO}_{4}\), the heat of formation values vary with the concentration of sulfuric acid in a water solution.

This is because the formation of the acidic solution involves dissociation and hydration processes, both of which impart energy changes.
  • During formation, sulfuric acid transitions from a concentrated liquid to a hydrated ionic form in the solution.
  • Each hydration state may release or absorb different amounts of energy depending on the interaction between ions and water molecules.
This means that although we might have a standard value for heat of formation, resulting enthalpy can differ markedly based on solution concentration and conditions. Understanding these variances is essential for accurate thermochemical analysis and expression of reaction thermodynamics.

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Most popular questions from this chapter

We can determine the purity of solid materials by using calorimetry. A gold ring (for pure gold, specific heat \(=0.1291 \mathrm{Jg}^{-1} \mathrm{K}^{-1}\) ) with mass of \(10.5 \mathrm{g}\) is heated to \(78.3^{\circ} \mathrm{C}\) and immersed in \(50.0 \mathrm{g}\) of \(23.7^{\circ} \mathrm{C}\) water in a constant-pressure calorimeter. The final temperature of the water is \(31.0^{\circ} \mathrm{C}\). Is this a pure sample of gold?

In your own words, define or explain the following terms or symbols: (a) \(\Delta H ;\) (b) \(P \Delta V ;\) (c) \(\Delta H_{f} ;\) (d) standard state; (e) fossil fuel.

Use Hess's law and the following data $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ \Delta H^{\circ}=-802 \mathrm{kJ} \end{aligned}$$ $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) & \\ \Delta H^{\circ}=&+247 \mathrm{kJ} \end{aligned}$$ $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) & \\ \Delta H^{\circ}=&+247 \mathrm{kJ} \end{aligned}$$ $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) & \\ \Delta H^{\circ}=&+206 \mathrm{kJ} \end{aligned}$$ to determine \(\Delta H^{\circ}\) for the following reaction, an important source of hydrogen gas $$\mathrm{CH}_{4}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g})$$

What is the change in internal energy of a system if the system (a) absorbs \(58 \mathrm{J}\) of heat and does \(58 \mathrm{J}\) of work; (b) absorbs 125 J of heat and does 687 J of work; (c) evolves 280 cal of heat and has 1.25 kJ of work done on it?

When one mole of sodium carbonate decahydrate (washing soda) is gently warmed, \(155.3 \mathrm{kJ}\) of heat is absorbed, water vapor is formed, and sodium carbonate heptahydrate remains. On more vigorous heating, the heptahydrate absorbs \(320.1 \mathrm{kJ}\) of heat and loses more water vapor to give the monohydrate. Continued heating gives the anhydrous salt (soda ash) while \(57.3 \mathrm{kJ}\) of heat is absorbed. Calculate \(\Delta H\) for the conversion of one mole of washing soda into soda ash. Estimate \(\Delta U\) for this process. Why is the value of \(\Delta U\) only an estimate?

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