Internal energy is an important thermodynamic quantity because it measures a system's capacity to do work on its surroundings. In chemistry, however, most reactions are carried out under conditions of constant pressure, typically the pressure of the surrounding atmosphere. Under these conditions, it is useful to define a new thermodynamic state function called enthalpy.
Enthalpy, \(H\), is defined as
\[ H \equiv U + pV \]
Because \(U\), \(p\), and \(V\) are all state functions, enthalpy is also a state function. Changes in enthalpy therefore depend only on the initial and final states of a system and not on the pathway followed.
Taking the differential of the definition gives
\[ dH = dU + d(pV) \]
\[ dH = dU + p\,dV + V\,dp \]
Consider a reversible process in which the only work performed is pressure-volume work. The First Law gives
\[ dU = dq + dw \]
\[ dU = dq - p\,dV \]
Substituting this expression into the differential form of the enthalpy equation gives
\[ dH = dq - p\,dV + p\,dV + V\,dp \]
\[ dH = dq + V\,dp \]
If the process occurs at constant pressure,
\[ dp = 0 \]
and therefore
\[ dH = dq \]
Integrating between the initial and final states yields
\[ \Delta H = q_p \]
where \(q_p\) represents the heat transferred under constant-pressure conditions.
Most laboratory reactions occur at approximately constant pressure. Because enthalpy changes are equal to heat flow under these conditions, enthalpy provides a convenient way to describe the energy changes associated with chemical and physical processes.
If a reaction releases heat at constant pressure, the enthalpy change is negative:
\[ \Delta H < 0 \]
Such reactions are called exothermic.
If a reaction absorbs heat at constant pressure, the enthalpy change is positive:
\[ \Delta H > 0 \]
Such reactions are called endothermic.
Since enthalpy changes are directly related to heat flow at constant pressure, it is natural to define a constant-pressure heat capacity:
\[ C_P = \left( \frac{\partial H} {\partial T} \right)_P \]
For small temperature changes,
\[ dH = C_P\,dT \]
and if \(C_P\) is approximately constant,
\[ \Delta H = C_P\Delta T \]
Big picture: Enthalpy is a state function that is particularly useful for describing chemical reactions and physical processes occurring at constant pressure. Under these conditions, changes in enthalpy are equal to the heat transferred, making enthalpy one of the most important quantities in thermochemistry.
One mole of an ideal gas expands reversibly and isothermally at \(298\ \mathrm{K}\) from an initial volume of \(10.0\ \mathrm{L}\) to a final volume of \(20.0\ \mathrm{L}\). Calculate \(q\), \(w\), \(\Delta U\), and \(\Delta H\).
Because the process is a reversible isothermal expansion of an ideal gas, the work is given by
\[ w = -nRT \ln \left( \frac{V_2}{V_1} \right) \]
Substituting the given values:
\[ w = -(1.00\ \mathrm{mol}) \left( 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}} \right) (298\ \mathrm{K}) \ln \left( \frac{20.0}{10.0} \right) \]
\[ w = -(2477.6\ \mathrm{J}) \ln(2) \]
\[ w = -1.72\times10^3\ \mathrm{J} \]
\[ w = -1.72\ \mathrm{kJ} \]
For an ideal gas undergoing an isothermal process,
\[ \Delta U = 0 \]
because the internal energy of an ideal gas depends only on temperature, and the temperature does not change during an isothermal process.
Applying the First Law,
\[ \Delta U = q + w \]
\[ 0 = q + (-1.72\ \mathrm{kJ}) \]
\[ q = +1.72\ \mathrm{kJ} \]
Since the gas does work on the surroundings, heat must flow into the system to maintain a constant temperature.
Finally, for an ideal gas undergoing an isothermal process,
\[ \Delta H = 0 \]
because enthalpy, like internal energy, depends only on temperature for an ideal gas.
| Quantity | Value |
|---|---|
| \(q\) | \(+1.72\ \mathrm{kJ}\) |
| \(w\) | \(-1.72\ \mathrm{kJ}\) |
| \(\Delta U\) | \(0\) |
| \(\Delta H\) | \(0\) |
Physical interpretation: The gas performs work on the surroundings as it expands. To maintain a constant temperature, exactly the same amount of energy must enter the system as heat. As a result, both the internal energy and the enthalpy remain unchanged.
A \(10.0\ \mathrm{g}\) sample of copper is heated reversibly at constant pressure from \(20.0^\circ\mathrm{C}\) to \(40.0^\circ\mathrm{C}\). Given that the molar heat capacity of copper is
\[ C_P = 24.44\ \mathrm{J\,mol^{-1}\,K^{-1}} \]
calculate \(q\), \(w\), \(\Delta U\), and \(\Delta H\).
First calculate the number of moles of copper:
\[ n = \frac{ 10.0\ \mathrm{g} } { 63.546\ \mathrm{g\,mol^{-1}} } \]
\[ n = 0.1574\ \mathrm{mol} \]
The temperature change is
\[ \Delta T = 40.0^\circ\mathrm{C} - 20.0^\circ\mathrm{C} \]
\[ \Delta T = 20.0\ \mathrm{K} \]
At constant pressure,
\[ q_P = \Delta H = nC_P\Delta T \]
Substituting the values:
\[ q = (0.1574\ \mathrm{mol}) \left( 24.44\ \mathrm{J\,mol^{-1}\,K^{-1}} \right) (20.0\ \mathrm{K}) \]
\[ q = 76.9\ \mathrm{J} \]
Therefore,
\[ \Delta H = +76.9\ \mathrm{J} \]
Because copper is a nearly incompressible solid, the volume change is extremely small. As a result, the pressure-volume work is negligible:
\[ w \approx 0 \]
Applying the First Law,
\[ \Delta U = q + w \]
\[ \Delta U \approx 76.9\ \mathrm{J} \]
| Quantity | Value |
|---|---|
| \(q\) | \(+76.9\ \mathrm{J}\) |
| \(w\) | \(\approx 0\) |
| \(\Delta U\) | \(+76.9\ \mathrm{J}\) |
| \(\Delta H\) | \(+76.9\ \mathrm{J}\) |
Physical interpretation: Unlike gases, solids undergo very little volume change when heated. Consequently, almost all of the energy added as heat increases the internal energy of the solid, making \(\Delta U\) and \(\Delta H\) nearly identical.
A \(1.50\ \mathrm{mol}\) sample of a monatomic ideal gas \(\left(C_V=\frac{3}{2}R\right)\) occupies a constant volume of \(10.0\ \mathrm{L}\). The gas is heated from \(15.0^\circ\mathrm{C}\) to \(45.0^\circ\mathrm{C}\). Calculate \(q\), \(w\), \(\Delta U\), and \(\Delta H\).
First determine the temperature change:
\[ \Delta T = 45.0^\circ\mathrm{C} - 15.0^\circ\mathrm{C} \]
\[ \Delta T = 30.0\ \mathrm{K} \]
Because the process occurs at constant volume,
\[ \Delta V = 0 \]
and therefore
\[ w = 0 \]
The First Law then simplifies to
\[ \Delta U = q \]
For a constant-volume process,
\[ \Delta U = nC_V\Delta T \]
Since
\[ C_V = \frac{3}{2}R = \frac{3}{2} (8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}) \]
\[ C_V = 12.47\ \mathrm{J\,mol^{-1}\,K^{-1}} \]
Substituting the values:
\[ \Delta U = (1.50\ \mathrm{mol}) (12.47\ \mathrm{J\,mol^{-1}\,K^{-1}}) (30.0\ \mathrm{K}) \]
\[ \Delta U = 561\ \mathrm{J} \]
Therefore,
\[ q = +561\ \mathrm{J} \]
\[ w = 0 \]
\[ \Delta U = +561\ \mathrm{J} \]
To determine the enthalpy change, use
\[ \Delta H = nC_P\Delta T \]
where
\[ C_P = C_V + R = \frac{5}{2}R \]
\[ C_P = 20.79\ \mathrm{J\,mol^{-1}\,K^{-1}} \]
\[ \Delta H = (1.50\ mol) (20.79\ J\ mol^{-1}\ K\ ^{-1}) (30.0\ K) \]
\[ \Delta H = 936\ \mathrm{J} \]
| Quantity | Value |
|---|---|
| \(q\) | \(+561\ \mathrm{J}\) |
| \(w\) | \(0\) |
| \(\Delta U\) | \(+561\ \mathrm{J}\) |
| \(\Delta H\) | \(+936\ \mathrm{J}\) |
Physical interpretation: Because the volume remains constant, no pressure-volume work is performed. All of the heat added to the system increases the internal energy of the gas. The enthalpy change is larger than the internal-energy change because enthalpy includes both internal energy and the \(pV\) contribution.
Calculate \(q\), \(w\), \(\Delta U\), and \(\Delta H\) for the randomly generated ideal-gas process. Choose the correct set of values.
| Summary (for ideal gas) | |||||
|---|---|---|---|---|---|
| Path | q | w | ΔU | ΔH | \[ C_p = C_V + R \] |
| Isothermal (ΔT = 0) | |||||
| Isobaric (Δp = 0) | |||||
| Isochoric (ΔV = 0) | |||||
Big picture: Enthalpy is the most useful thermodynamic quantity for chemistry because most reactions occur at approximately constant pressure. Under these conditions, changes in enthalpy are equal to heat flow, making enthalpy a convenient way to describe and measure energy changes in chemical and physical processes.