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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?

Short Answer

Expert verified
The enthalpy change for the given process is \(532.7 \, \mathrm{kJ}\). The estimated internal energy change (\(\Delta U\)) is also \(532.7 \, \mathrm{kJ}\) based on the assumptions of constant pressure and small changes in volume. The value of \(\Delta U\) is only an estimate because precise values of pressure and \(\Delta V\) were not provided in the question.

Step by step solution

01

Calculate the total heat absorbed

We can find the total heat absorbed (\(\Delta H\)) by summing up the heat absorbed in each step. This gives us: \(\Delta H = 155.3 \, \mathrm{kJ} + 320.1 \, \mathrm{kJ} + 57.3 \, \mathrm{kJ} = 532.7 \, \mathrm{kJ}\).
02

Estimate the internal energy change

We can estimate the internal energy change (\(\Delta U\)) using the equation \(\Delta U = \Delta H - P\Delta V\). But since both the pressure (P) and the change in volume (\(\Delta V\)) are not given in the problem, we can make an assumption that the pressure is relatively constant and that the change in volume is relatively small due both to the conversion happening in the solid state and the process happening at constant pressure. Given these assumptions, \(\Delta U \approx \Delta H = 532.7 \, \mathrm{kJ}\).
03

Explain why \(\Delta U\) is only estimated

The value of \(\Delta U\) is only an estimate because we do not have precise information of the pressure and the change in volume for the process. The formula used is an approximation under assumptions of constant pressure and small changes in volume. These assumptions will not always hold true and thus the value of \(\Delta U\) calculated is only an estimate.

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Key Concepts

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

Sodium Carbonate
Sodium carbonate is a widely used chemical compound, often found in its decahydrate form, commonly known as washing soda. It gets this name due to its use in laundry detergents, where it aids in removing dirt and greasing. Sodium carbonate has several forms based on the number of water molecules attached to its crystalline structure. These forms include decahydrate, heptahydrate, and monohydrate. These terms refer to having ten, seven, and one water molecule respectively bonded in the crystalline structure.
Heat treatment causes sodium carbonate to lose water molecules, transforming from decahydrate to soda ash, the anhydrous form. Soda ash has various industrial applications from glass manufacturing to softening water.
The process of heating sodium carbonate and its gradual transformation to less hydrated forms involves the absorption of heat at each step, known as an endothermic reaction. Understanding this process helps in calculating the enthalpy change ( ∆H ) as heat is absorbed.
Internal Energy
Internal energy ( ∆U ) refers to the total energy contained within a system. It is a measure of both the kinetic and potential energies at the molecular level. When a reaction like the heating of sodium carbonate takes place, internal energy reflects the energy changes within that reaction.
In chemical thermodynamics, the relationship between enthalpy change ( ∆H ) and internal energy is given by the equation: ∆U = ∆H - P∆V . This equation considers both enthalpy and pressure-volume work done by the system during a reaction. When assumptions about the constancy of pressure and volume are made, approximations about ∆U can be derived from ∆H .
  • In the specific process of converting washing soda to soda ash, ∆U was estimated based on the summed enthalpy changes, representing the total heat absorbed.
  • Estimations are necessary because exact pressure-volume changes during solid-state reactions aren't easily measurable.
Heat Absorption
Heat absorption is a critical component when understanding the transformation of substances like sodium carbonate. During endothermic processes, heat is absorbed into the system from its surroundings, causing changes in the structure or state of the material.
In our case, the heat absorbed during each phase of heating sodium carbonate causes water molecules to be driven off from the structure, leading to its conversion into less hydrated forms such as soda ash.
The total heat absorbed ( ∆H ) in the sequence of transformations within sodium carbonate was calculated by adding the heat quantities specific to each dehydration step: 155.3 kJ for decahydrate, 320.1 kJ for heptahydrate, and 57.3 kJ for monohydrate.
This value of ∆H reflects the complex series of energy changes needed to convert the substance through its various phases, and understanding it helps in predicting and controlling chemical reactions.

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Most popular questions from this chapter

In a student experiment to confirm Hess's law, the reaction $$\mathrm{NH}_{3}(\text { concd aq })+\mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq})$$ was carried out in two different ways. First, \(8.00 \mathrm{mL}\) of concentrated \(\mathrm{NH}_{3}(\text { aq })\) was added to \(100.0 \mathrm{mL}\) of 1.00 M HCl in a calorimeter. [The NH \(_{3}(\) aq) was slightly in excess.] The reactants were initially at \(23.8^{\circ} \mathrm{C},\) and the final temperature after neutralization was \(35.8^{\circ} \mathrm{C} .\) In the second experiment, air was bubbled through \(100.0 \mathrm{mL}\) of concentrated \(\mathrm{NH}_{3}(\mathrm{aq})\) sweeping out \(\mathrm{NH}_{3}(\mathrm{g})\) (see sketch). The \(\mathrm{NH}_{3}(\mathrm{g})\) was neutralized in \(100.0 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). The temperature of the concentrated \(\mathrm{NH}_{3}(\text { aq })\) fell from 19.3 to \(13.2^{\circ} \mathrm{C} .\) At the same time, the temperature of the 1.00 M HCl rose from 23.8 to 42.9 ^ C as it was neutralized by \(\mathrm{NH}_{3}(\mathrm{g}) .\) Assume that all solutions have densities of \(1.00 \mathrm{g} / \mathrm{mL}\) and specific heats of \(4.18 \mathrm{Jg}^{-1 \circ} \mathrm{C}^{-1}\) (a) Write the two equations and \(\Delta H\) values for the processes occurring in the second experiment. Show that the sum of these two equations is the same as the equation for the reaction in the first experiment. (b) Show that, within the limits of experimental error, \(\Delta H\) for the overall reaction is the same in the two experiments, thereby confirming Hess's law.

Can a chemical compound have a standard enthalpy of formation of zero? If so, how likely is this to occur? Explain.

Calculate the quantity of work associated with a \(3.5 \mathrm{L}\) expansion of a gas \((\Delta V)\) against a pressure of \(748 \space\mathrm{mmHg}\) in the units (a) atm \(\mathrm{L} ;\) (b) joules (J); (c) calories (cal).

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 to determine \(\Delta H^{\circ}\) for the reaction $$\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}), \text { given that }$$ $$\begin{array}{l} \text { C(graphite) }+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g}) \\ &\left.\qquad \Delta H^{\circ}=-110.54 \mathrm{k} \mathrm{J}\right] \end{array}$$ $$\begin{aligned} &\text { C(graphite) }+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})\\\ &&\Delta H^{\circ}=-393.51 \mathrm{kJ} \end{aligned}$$

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