5.2: The First Law of Thermodynamics

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Skills to Develop

To calculate changes in internal energy

The relationship between the energy change of a system and that of its surroundings is given by the first law of thermodynamics, which states that the energy of the universe is constant. Using Equation 18.1, we can express this law mathematically as follows:

\[E_{univ}=ΔE_{sys}+ΔE_{surr}=0 \tag{5.2.1a}\]

\[\Delta{E_{sys}}=−ΔE_{surr} \tag{5.2.1b}\]

where the subscripts univ, sys, and surr refer to the universe, the system, and the surroundings, respectively. Thus the change in energy of a system is identical in magnitude but opposite in sign to the change in energy of its surroundings.

Note

The tendency of all systems, chemical or otherwise, is to move toward the state with the lowest possible energy.

An important factor that determines the outcome of a chemical reaction is the tendency of all systems, chemical or otherwise, to move toward the lowest possible overall energy state. As a brick dropped from a rooftop falls, its potential energy is converted to kinetic energy; when it reaches ground level, it has achieved a state of lower potential energy. Anyone nearby will notice that energy is transferred to the surroundings as the noise of the impact reverberates and the dust rises when the brick hits the ground. Similarly, if a spark ignites a mixture of isooctane and oxygen in an internal combustion engine, carbon dioxide and water form spontaneously, while potential energy (in the form of the relative positions of atoms in the molecules) is released to the surroundings as heat and work. The internal energy content of the \(CO_2/H_2O\) product mixture is less than that of the isooctane \(O_2\) reactant mixture. The two cases differ, however, in the form in which the energy is released to the surroundings. In the case of the falling brick, the energy is transferred as work done on whatever happens to be in the path of the brick; in the case of burning isooctane, the energy can be released as solely heat (if the reaction is carried out in an open container) or as a mixture of heat and work (if the reaction is carried out in the cylinder of an internal combustion engine). Because heat and work are the only two ways in which energy can be transferred between a system and its surroundings, any change in the internal energy of the system is the sum of the heat transferred (q) and the work done (w):

\[ΔE_{sys} = q + w \tag{5.2.2}\]

Although \(q\) and \(w\) are not state functions on their own, their sum (\(ΔE_{sys}\)) is independent of the path taken and is therefore a state function. A major task for the designers of any machine that converts energy to work is to maximize the amount of work obtained and minimize the amount of energy released to the environment as heat. An example is the combustion of coal to produce electricity. Although the maximum amount of energy available from the process is fixed by the energy content of the reactants and the products, the fraction of that energy that can be used to perform useful work is not fixed. Because we focus almost exclusively on the changes in the energy of a system, we will not use “sys” as a subscript unless we need to distinguish explicitly between a system and its surroundings.

Note

Although \(q\) and \(w\) are not state functions, their sum (\(ΔE_{sys}\)) is independent of the path taken and therefore is a state function.

Example 5.2.1

A sample of an ideal gas in the cylinder of an engine is compressed from 400 mL to 50.0 mL during the compression stroke against a constant pressure of 8.00 atm. At the same time, 140 J of energy is transferred from the gas to the surroundings as heat. What is the total change in the internal energy (ΔE) of the gas in joules?

Given: initial volume, final volume, external pressure, and quantity of energy transferred as heat

Asked for: total change in internal energy

Strategy:

Determine the sign of \(q\) to use in Equation 5.2.2.
From Equation 5.2.2 calculate \(w\) from the values given. Substitute this value into Equation 5.2.2 to calculate \(ΔE\).

SOLUTION

A From Equation 5.2.2, we know that ΔE = q + w. We are given the magnitude of q (140 J) and need only determine its sign. Because energy is transferred from the system (the gas) to the surroundings, q is negative by convention.

B Because the gas is being compressed, we know that work is being done on the system, so \(w\) must be positive. From Equation 5.2.2,

\[w=-P_{\textrm{ext}}\Delta V=-8.00\textrm{ atm}(\textrm{0.0500 L} - \textrm{0.400 L})\left(\dfrac{\textrm{101.3 J}}{\mathrm{L\cdot atm}} \right)=284\textrm{ J}\]

Thus

ΔE = q + w = −140 J + 284 J = 144 J

In this case, although work is done on the gas, increasing its internal energy, heat flows from the system to the surroundings, decreasing its internal energy by 144 J. The work done and the heat transferred can have opposite signs.

Exercise 5.2.1

A sample of an ideal gas is allowed to expand from an initial volume of 0.200 L to a final volume of 3.50 L against a constant external pressure of 0.995 atm. At the same time, 117 J of heat is transferred from the surroundings to the gas. What is the total change in the internal energy (ΔE) of the gas in joules?

Answer: −216 J

Note

By convention, both heat flow and work have a negative sign when energy is transferred from a system to its surroundings and vice versa.

Enthalpy

To further understand the relationship between heat flow (q) and the resulting change in internal energy (ΔE), we can look at two sets of limiting conditions: reactions that occur at constant volume and reactions that occur at constant pressure. We will assume that PV work is the only kind of work possible for the system, so we can substitute its definition from Equation 18.5 into Equation 5.2.2 to obtain the following:

\[ΔE = q − PΔV \tag{5.2.2}\]

where the subscripts have been deleted.

If the reaction occurs in a closed vessel, the volume of the system is fixed, and ΔV is zero. Under these conditions, the heat flow (often given the symbol q_v to indicate constant volume) must equal ΔE:

\[\underset{\textrm{constant volume}}{q_{\textrm v}=\Delta E} \tag{5.2.3}\]

No \(PV\) work can be done, and the change in the internal energy of the system is equal to the amount of heat transferred from the system to the surroundings or vice versa.

Many chemical reactions are not, however, carried out in sealed containers at constant volume but in open containers at a more or less constant pressure of about 1 atm. The heat flow under these conditions is given the symbol qp to indicate constant pressure. Replacing q in Equation 18.8 by q_p and rearranging to solve for q_p,

\(\underset{\textrm{constant pressure}}{q_{\textrm p}=\Delta E+P\Delta V} \tag{5.2.4}\)

Thus, at constant pressure, the heat flow for any process is equal to the change in the internal energy of the system plus the PV work done, as we stated in Chapter 5.

Because conditions of constant pressure are so important in chemistry, a new state function called enthalpy (H) is defined as \(H = E + PV\). At constant pressure, the change in the enthalpy of a system is as follows:

\[ΔH = ΔE + Δ(PV) = ΔE + PΔV \tag{5.2.5}\]

Comparing the previous two equations shows that at constant pressure, the change in the enthalpy of a system is equal to the heat flow: \(ΔH = q_p\). This expression is consistent with our definition of enthalpy in Chapter 5, where we stated that enthalpy is the heat absorbed or produced during any process that occurs at constant pressure.

Note

At constant pressure, the change in the enthalpy of a system is equal to the heat flow: \(ΔH = q_p\).

Example 5.2.2

The molar enthalpy of fusion for ice at 0.0°C and a pressure of 1.00 atm is 6.01 kJ, and the molar volumes of ice and water at 0°C are 0.0197 L and 0.0180 L, respectively. Calculate ΔH and ΔE for the melting of ice at 0.0°C.

Given: enthalpy of fusion for ice, pressure, and molar volumes of ice and water

Asked for: ΔH and ΔE for ice melting at 0.0°C

Strategy:

A Determine the sign of q and set this value equal to ΔH.
B Calculate Δ(PV) from the information given.
C Determine ΔE by substituting the calculated values into Equation 18.11.

SOLUTION

A Because 6.01 kJ of heat is absorbed from the surroundings when 1 mol of ice melts, q = +6.01 kJ. When the process is carried out at constant pressure, q = q_p = ΔH = 6.01 kJ.

B To find ΔE using Equation 18.11, we need to calculate Δ(PV). The process is carried out at a constant pressure of 1.00 atm, so

\(\begin{align}\Delta(PV)&=P\Delta V=P(V_{\textrm f}-V)=(1.00\textrm{ atm})(\textrm{0.0180 L}-\textrm{0.0197 L})
\\ &=(-1.7\times10^{-3}\;\mathrm{L\cdot atm})(101.3\;\mathrm{J/L\cdot atm})=-0.0017\textrm{ J}\end{align}\)

C Substituting the calculated values of ΔH and PΔV into Equation 18.11,

ΔE = ΔH − PΔV = 6010 J − (−0.0017 J) = 6010 J = 6.01 kJ

Exercise 5.2.2

At 298 K and 1 atm, the conversion of graphite to diamond requires the input of 1.850 kJ of heat per mole of carbon. The molar volumes of graphite and diamond are 0.00534 L and 0.00342 L, respectively. Calculate ΔH and ΔE for the conversion of C (graphite) to C (diamond) under these conditions.

Answer: ΔH = 1.85 kJ/mol; ΔE = 1.85 kJ/mol

The Relationship between ΔH and ΔE

If ΔH for a reaction is known, we can use the change in the enthalpy of the system (Equation 18.11) to calculate its change in internal energy. When a reaction involves only solids, liquids, liquid solutions, or any combination of these, the volume does not change appreciably (ΔV = 0). Under these conditions, we can simplify Equation 18.11 to ΔH = ΔE. If gases are involved, however, ΔH and ΔE can differ significantly. We can calculate ΔE from the measured value of ΔH by using the right side of Equation 18.11 together with the ideal gas law, PV = nRT. Recognizing that Δ(PV) = Δ(nRT), we can rewrite Equation 18.11 as follows:

\[ΔH = ΔE + Δ(PV) = ΔE + Δ(nRT) \tag{5.2.6}\]

At constant temperature, Δ(nRT) = RTΔn, where Δn is the difference between the final and initial numbers of moles of gas. Thus

\[ΔE = ΔH − RTΔn \tag{5.2.7}\]

For reactions that result in a net production of gas, Δn > 0, so ΔE < ΔH. Conversely, endothermic reactions (ΔH > 0) that result in a net consumption of gas have Δn < 0 and ΔE > ΔH. The relationship between ΔH and ΔE for systems involving gases is illustrated in Example 5.2.3.

Note

For reactions that result in a net production of gas, ΔE < ΔH. For endothermic reactions that result in a net consumption of gas, ΔE > ΔH.

Example 5.2.3

The combustion of graphite to produce carbon dioxide is described by the equation C (graphite, s) + O₂(g) → CO₂(g). At 298 K and 1.0 atm, ΔH = −393.5 kJ/mol of graphite for this reaction, and the molar volume of graphite is 0.0053 L. What is ΔE for the reaction?

Given: balanced chemical equation, temperature, pressure, ΔH, and molar volume of reactant

Asked for: ΔE

Strategy:

Use the balanced chemical equation to calculate the change in the number of moles of gas during the reaction.
Substitute this value and the data given into Equation 18.13 to obtain ΔE.

SOLUTION

A In this reaction, 1 mol of gas (CO₂) is produced, and 1 mol of gas (O₂) is consumed. Thus Δn = 1 − 1 = 0.

B Substituting this calculated value and the given values into Equation 18.13,

ΔE=ΔH−RTΔn=(−393.5 kJ/mol)−[8.314 J/(mol⋅K)](298 K)(0)

=(−393.5 kJ/mol)−(0 J/mol)=−393.5 kJ/mol

To understand why only the change in the volume of the gases needs to be considered, notice that the molar volume of graphite is only 0.0053 L. A change in the number of moles of gas corresponds to a volume change of 22.4 L/mol of gas at standard temperature and pressure (STP), so the volume of gas consumed or produced in this case is (1)(22.4 L) = 22.4 L, which is much, much greater than the volume of 1 mol of a solid such as graphite.

Exercise 5.2.3

Calculate ΔE for the conversion of oxygen gas to ozone at 298 K:

\[3O_2(g) \rightarrow 2O_3(g).\]

The value of ΔH for the reaction is 285.4 kJ.

Answer: 288 kJ

As the exercise in Example 5.2.3 illustrates, the magnitudes of ΔH and ΔE for reactions that involve gases are generally rather similar, even when there is a net production or consumption of gases.

Summary

Enthalpy is a state function, and the change in enthalpy of a system is equal to the sum of the change in the internal energy of the system and the PV work done.

The first law of thermodynamics states that the energy of the universe is constant. The change in the internal energy of a system is the sum of the heat transferred and the work done. At constant pressure, heat flow (q) and internal energy (E) are related to the system’s enthalpy (H). The heat flow is equal to the change in the internal energy of the system plus the PV work done. When the volume of a system is constant, changes in its internal energy can be calculated by substituting the ideal gas law into the equation for ΔE.