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Construct a table with the headings \(q, w, \Delta E,\) and \(\Delta H .\) For each of the following processes, deduce whether each of the quantities listed is positive \((+)\) negative \((-),\) or zero (0) . (a) Freezing of benzene. (b) Compression of an ideal gas at constant temperature. (c) Reaction of sodium with water. (d) Boiling liquid ammonia. (e) Heating a gas at constant volume. (f) Melting of ice.

Short Answer

Expert verified
The table would result as follows (where + is positive, - is negative and 0 is zero): (a) Freezing of benzene: q=-, w=0, \(\Delta E\)=-, \(\Delta H\)=-. (b) Compression of an ideal gas at constant temperature: q=-, w=+, \(\Delta E\)=0, \(\Delta H\)=0. (c) Reaction of sodium with water: q=+, w=0, \(\Delta E\)=+, \(\Delta H\)=+. (d) Boiling liquid ammonia: q=+, w=0, \(\Delta E\)=+, \(\Delta H\)=+. (e) Heating a gas at constant volume: q=+, w=0, \(\Delta E\)=+, \(\Delta H\)=+. (f) Melting of ice: q=+, w=0, \(\Delta E\)=+, \(\Delta H\)=+.

Step by step solution

01

- Analyzing the freezing of benzene

On freezing, a substance gives out heat (exothermic process) so \(q\) is negative. Since it's a spontaneous process, no work is done - \(w\) is zero. The change in internal energy \(\Delta E\) is negative because the kinetic energy decreases as particles slow down in a solid state. The enthalpy change \(\Delta H\) would also be negative as it is an exothermic process.
02

- Compression of an ideal gas at constant temperature

When a gas is compressed at constant temperature (isothermally), heat is given out so \(q\) is negative. Since work is done in compression, \(w\) is positive. The change in internal energy \(\Delta E\) is zero due to constant temperature. The enthalpy change \(\Delta H\) would be zero as enthalpy is a function of pressure and temperature only.
03

- Reaction of sodium with water

Sodium reacts violently, releasing a lot of heat, thus \(q\) is positive. It's a spontaneous reaction with no external work done, so \(w\) is zero. The change in internal energy \(\Delta E\) is positive as energy is absorbed during reaction. \(\Delta H\) is positive as it's an exothermic process.
04

- Boiling liquid ammonia

Boiling is an endothermic process, the liquid absorbs heat from surroundings making \(q\) positive. No work is done in this process, so \(w\) is zero. The change in internal energy, \(\Delta E\), is positive as heat is absorbed. As an endothermic process, the enthalpy change, \(\Delta H\), would also be positive.
05

- Heating a gas at constant volume

In this case, heat is supplied, so \(q\) is positive. Since the volume is constant, no work is done: \(w\) is zero. The change in internal energy \(\Delta E\) is positive because heat is absorbed. Finally, \(\Delta H\) is positive: in a constant-volume process, the change in enthalpy is equivalent to the heat absorbed or released.
06

- Melting of ice

Melting requires absorbing heat, making \(q\) positive. No work is done in the melting, so \(w\) is zero. The change in internal energy \(\Delta E\) is positive as heat is absorbed. As an endothermic process, the enthalpy change, \(\Delta H\), would also be positive.

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Key Concepts

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

Enthalpy Change
The concept of enthalpy change is central to the field of thermochemistry, involving the measurement of heat exchanged in a chemical or physical process at constant pressure. Enthalpy change, symbolized as \(\Delta H\), is a type of energy exchanged that includes both heat and work done against the atmosphere.

When a substance undergoes a transformation, such as when benzene freezes, the process can either release heat (exothermic, \(\Delta H < 0\)) or absorb heat (endothermic, \(\Delta H > 0\)). In the freezing of benzene, molecules lose kinetic energy and stabilize into a solid form, releasing heat to the surroundings, thus exhibiting a negative enthalpy change. On the contrary, processes like boiling ammonia or melting ice require an input of heat, indicating a positive enthalpy change.
Internal Energy
The internal energy of a system, often noted as \(\Delta E\) or \(\Delta U\), represents the totality of kinetic and potential energies of the particles within that system. It's the energy stored inside the system. Changes in a system’s temperature, volume, or chemical composition can alter its internal energy.

During phase transitions, such as melting or boiling, a system absorbs energy, increasing the kinetic energy of its particles and thus, its internal energy (e.g., \(\Delta E > 0\) during boiling of ammonia). In contrast, freezing and exothermic reactions release energy, decreasing the system's internal energy (e.g., \(\Delta E < 0\) for freezing benzene).
Heat Transfer
Heat transfer is the movement of thermal energy from one thing to another due to a temperature difference. Heat (\(q\)) can move via conduction, convection, or radiation and can be quantified in terms of energy units, such as joules or calories.

In thermodynamic processes, if heat is absorbed by the system from the surroundings, the heat transfer value is positive (\(q > 0\)), like when heating a gas at constant volume. Conversely, if the system loses heat to the surroundings, heat transfer is negative (\(q < 0\)), exemplified by the compression of an ideal gas where it releases heat. Understanding the direction of heat flow is crucial for mastering concepts in thermochemistry.
Phase Transitions
Phase transitions occur when a substance changes from one state of matter to another, such as from solid to liquid (melting) or liquid to gas (boiling). These processes are governed by energy changes and can be categorized as endothermic or exothermic.

During endothermic transitions, like melting ice or boiling ammonia, the substance absorbs energy from the surroundings, leading to a positive heat transfer (\(q > 0\)) and a rise in the internal energy (\(\Delta E > 0\)). Conversely, during exothermic changes like freezing of benzene, the substance releases energy, resulting in a negative heat transfer (\(q < 0\)) and a drop in the internal energy (\(\Delta E < 0\)).

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Most popular questions from this chapter

A gas expands and does \(P-V\) work on the surroundings equal to \(325 \mathrm{~J}\). At the same time, it absorbs \(127 \mathrm{~J}\) of heat from the surroundings. Calculate the change in energy of the gas.

Predict the value of \(\Delta H_{\mathrm{f}}^{\circ}\) (greater than, less than, or equal to zero) for these elements at \(25^{\circ} \mathrm{C}\) : (a) \(\mathrm{Br}_{2}(g)\) and \(\mathrm{Br}_{2}(l)\) (b) \(\mathrm{I}_{2}(g)\) and \(\mathrm{I}_{2}(s)\)

From the following data, $$ \begin{array}{c} \mathrm{C} \text { (graphite) }+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}(g) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-393.5 \mathrm{~kJ} / \mathrm{mol} \\\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-285.8 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-3119.6 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ calculate the enthalpy change for the reaction $$ 2 \mathrm{C}(\text { graphite })+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g) $$

(a) A snowmaking machine contains a mixture of compressed air and water vapor at about 20 atm. When the mixture is sprayed into the atmosphere it expands so rapidly that, as a good approximation, no heat exchange occurs between the system (air and water) and its surroundings. (In thermodynamics, such a process is called an adiabatic process.) Do a first law of thermodynamics analysis to show how snow is formed under these conditions. (b) If you have ever pumped air into a bicycle tire, you probably noticed a warming effect at the valve stem. The action of the pump compresses the air inside the pump and the tire. The process is rapid enough to be treated as an adiabatic process. Apply the first law of thermodynamics to account for the warming effect. (c) A driver's manual states that the stopping distance quadruples as the speed doubles; that is, if it takes \(30 \mathrm{ft}\) to stop a car traveling at \(25 \mathrm{mph}\) then it would take \(120 \mathrm{ft}\) to stop a car moving at 50 mph. Justify this statement by using the first law of thermodynamics. Assume that when a car is stopped, its kinetic energy \(\left(\frac{1}{2} m u^{2}\right)\) is totally converted to heat.

The work done to compress a gas is \(74 \mathrm{~J}\). As a result, \(26 \mathrm{~J}\) of heat is given off to the surroundings. Calculate the change in energy of the gas.

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