First Law of Thermodynamics

---

Work and Heat

James P. Joule (1818-1889)

• Joule was one of the pioneers of modern thermodynamics.
• Among his experiments, Joule attempted to measure the effect of work on the temperature of water.

---

The First Law of Thermodynamics

The capacity of a system to do work is increased by heating the system or doing work on it.

\[\;\]

• The first law translates to \[\Delta {U}={q}+{w}\] \[{dU}={dq}+{dw}\]

---

Heat

• Heat is the flow of energy that, in the absence of other changes, will change the temperature of the system.
• Endothermic: heat flows into the system (\(q_{system}>0\))
• Exothermic: heat flows out of the system (\(q_{system}<0\))
• An infinitesimal amount of heat flowing can be related to the temperature change as \[{dq}={CdT}\] where \(C\) is the heat capacity

Example 3.1

How much energy is needed to raise the temperature of \(5.0g\) of water from \(21.0^{o}C\) to \(25.0^{o}C\)?

\[\;\]

Side Note: Partial Derivative

• The ideal gas law tells us that \[{P(V.n.T)}=\frac{nRT}{V}\]
• How do we calculate the infinitesimal change in the pressure (\(dP\)) if all the variable can change?
• The infinitesimal change in any function can be calculated as \[{df}=\sum_{i}\left( \frac{\delta f}{\delta x_{i}} \right )_{x_{j}\neq {i}}{dx}_{i}\]

---

Work

• Work can take various forms, some of which we have already spoken about
  

  Type of Work Displacement Resistance

• Expansion: \[{dw}=-P_{ext}{dV}\]

• Electrical: \[{dw}=-W{dQ}\]

• Extension: \[{dw}={TdL}\]

• Stretching: \[{dw}={sdA}\]

Example 3.2

What is the work done by \(1.00{mol}\) of an ideal gas expanding from a volume of \(22.4L\) to a volume of \(44.8L\) against a constant external pressure of \(0.500{atm}\)?

---

Reversible and Irreversible Pathways

Reversible Gas Expansions

Irreversible Gas Expansions

Isochoric Expansion

Isobaric Expansion

Adiabatic Pathways

---

Calorimetry

• The techniques of calorimetry can be used to measure \(q\) for a chemical reaction directly
• Calorimetry is based on the idea that if heat leaves the system it must enter the surroundings
• We measure the temperature change in the surroundngs, usually water

Bomb Calorimetry

• Bomb calorimetry is used mostly to measure the heat evolved in combustion reactions

“Water Equivalent” of a Bomb Calorimeter

• Bomb calorimeters are calibrated by carrying out a reaction with a known \(\Delta U_{rxn}\) - commonly the combustion of benzoic acid
• This calibration reaction allows the researcher to calculate the “water equivalent” (\(W\)) of the calorimeter
  • \(n\) is the number of moles of benzoic acid used
  • \(\Delta U_c\) is the internal energy of compustion of benzoic acid (\(3225.7 \frac{kJ}{mol}\) at \(26^oC\))
  • \(e_{wire}\) is the energy released in the combustion of the fuse wire
  • \(e_{other}\) accounts for any other corrections

\[W=\frac{n\Delta{U_c}+e_{wire}+e_{other}}{\Delta T}\]

Using the “Water Equivalent”

• The “water equivalent” can be used to determine the \(\Delta U_c\) for an unknown reaction \[\Delta U_c = \frac{W\Delta T-e_{wire}-e_{other}}{n_{sample}}\]

\[\;\]

• The enthalpy of combustion can be found from the internal energy of combustion \[ \begin{align} \Delta H &= \Delta U + \Delta (PV) \\ &= \Delta U + RT \Delta{n_{gas}} \\ \end{align} \]
• For benzoic acid \[C_6H_5COOH(s)+7.5O_2(g) \rightarrow 7CO_2(g)+3H_2O(l)\] so \(\Delta n_{gas} = -0.5mol\)

Example 3.8

A student burned a \(0.7842g\) sample of benzoic acid (\(C_7H_6O_2\)) in a bomb calorimeter initially at \(25.0^oC\) and saw a temperature increase of \(2.02^oC\). She then burned a \(0.5348g\) sample of naphthalene (\(C_10H_8\)) (again from an initial temperature of \(25^oC\)) and saw a temperature increase of \(2.24^oC\). From this data, calculate \(\Delta H_c\) for naphthalene (assuming \(e_{wire}\) and \(e_{other}\) are unimportant).

---

Temperature Dependence of Enthalpy

• At constant pressure, \(dH=C_pdT\)
• For a temperature change we get \[\Delta H = \int_{T_1}^{T_2} C_pdT\]
• If \(C_p\) is independent of temperature this reduces to \[\Delta H = C_p \Delta T\]

---

Varying Heat Capacity

• A common empirical model can fit heat capacities over broad temperature ranges \[C_p(T) = a + bT + \frac{c}{T^2}\]
• Plugging this into our enthalpy change equation yields \[ \begin{align} \Delta H &= \int_{T_1}^{T_2}\left(a+bT+\frac{c}{T^2}\right)dT \\ &= a(T_2-T_1)+\frac{b}{2}(T_2^2-T_1^2)-\frac{c}{3}\left(\frac{1}{T_2^3}-\frac{1}{T_1^3} \right) \\ \end{align} \]
• For chemical reactions \[\Delta H_{rxn}(T_2)=\Delta H_{rxn}(T_1)+\Delta C_p\Delta T\]

Empirical Parameters for the Temperature Dependence of \(C_p\)

Substance	\(a(J{mol}^{-1}K^{-1})\)	\(b(J{mol}^{-1}K^{-2})\)	\(c(J{mol}^{-1}K)\)
\(C(gr)\)	\(16.86\)	\(4.77 \cdot 10^{-3}\)	\(-8.54 \cdot 10^5\)
\(CO_2(g)\)	\(44.22\)	\(8.79 \cdot 10^{-3}\)	\(-8.62 \cdot 10^5\)
\(H_2O(l)\)	\(75.29\)	\(0\)	\(0\)
\(N_2(g)\)	\(28.58\)	\(3.77 \cdot 10^{-3}\)	\(-5.0 \cdot 10^4\)
\(Pb(s)\)	\(22.13\)	\(1.172 \cdot 10^{-2}\)	\(9.6 \cdot 10^4\)

\[\;\]

Example 3.9

What is the molar enthalpy change for a temperature increase from \(273K\) to \(353K\) for \(Pb(s)\)?

\[\;\]

Example 3.10

The enthalpy of formation of \(NH_3(g)\) is \(-46.11\frac{kJ}{mol}\) at \(25^oC\). Calculate the enthalpy of formation at \(100^oC\). Assuming heat capacities are independent of temperature.

Substance	\(C_p(J{mol}^{-1}K^{-1})\)
\(N_2(g)\)	\(29.12\)
\(H_2(g)\)	\(28.82\)
\(NH_3(g)\)	\(35.06\)

---

Reaction Enthalpies

• Reaction enthalpies are difficult to tabulate
• We can us Hess’ Law to simplify this
• We can use tabulated enthalpies to calculate the enthalpies of other reactions

Example 3.11

Find \(\Delta H_{rxn}\) for the reaction \[2CO(g)+O_2(g) \rightarrow 2CO_2(g)\] given that \[ \begin{align} C(gr)+\frac{1}{2}O_2(g) \rightarrow CO(g) & &\Delta H = -110.53kJ \\ C(gr)+O_2(g) \rightarrow CO_2(g) & &\Delta H = -393.51kJ \\ \end{align} \]

---

Standard Enthalpy of Formation

• One problem with many thermodynamic state variables is that it is possible to measure changes but not absolute values
• It is therefore necessary to define a zero
  • the standard enthalpy of formation of a pure element in its standard state is zero

The standard state of a substance is the most stable form of that substance at 1 atmosphere pressure and the specified temperature.

---

Standard Formation Reactions

• We measure and tabulate the standard formation reactions for compounds

• These reactions involve one of the substance in its standard state from the elements in their standard states

• \({NaCl}{(s)}\) :

\[ \begin{align} {Na(s)}+\frac{1}{2}{Cl}_{2}{(g)} \rightarrow {NaCl(s)} & & \Delta H_f^o = -411.2 \frac{kJ}{mol} \\ \end{align} \]

• \(C_3H_8{(g)}\) : \[ \begin{align} C(gr)+4H_2(g) \rightarrow C_3H_8(g) & & \Delta H_f^o = -103.8 \frac{kJ}{mol} \end{align} \]

• We can use Hess’ Law with standard heats of formation

---

Alternation Use of Heats of Formation

• Because standard formation reactions begin with elements in their standard states we can simplify Hess’ Law
• The reaction enthalpy can be calculated from the standard formation reaction enthalpy by \[\Delta H_{rxn}= \sum_{products }\upsilon \cdot \Delta H_f^o - \sum_{reactants} \nu \cdot \Delta H_f^o\] where \(\nu\) is the stoichiometric coefficient of a species in the balanced chemical reaction

---

Ionization Reactions

---

Putting the \(1^{st}\) Law to Work

---