Ch. 11 Thermodynamics

Definitions

system: part of the universe under study
- open system: can transfer energy and matter to/from surroundings
- closed system: energy can be transfered, matter cannot
- isolated system: no transfer of energy or matter to/from surroundings
surroundings: everything outside the system
state functions: aspect of chemical system, eg pressure, temperature, moles; changes depend only on initial and final states
- internal energy: $E$
- enthalpy: $H$
- entropy: $S$
- Gibbs free energy: $G$
extensive properties: properties that change as the amount of sample changes $($ $\Delta H$ , $\Delta S$ , $\Delta G$ , $\Delta E$ $)$
intensive properties: properties that only depend on type not amount
standard state: pressure is $\pu{1 atm}$ , temperature is $\pu{25^\circ C}$ , compound amount is $\pu{1 mol}$
exothermic: system loses energy to surroundings $(\Delta H<0)$
endothermic: system absorbs energy from surroundings $(\Delta H>0)$

First Law of Thermodynamics

conservation of energy: energy cannot be created nor destroyed; only transferred
specific heat ( $c$ ): amount of energy needed to raise $\pu{1g}$ $1 g$ of a substance by $\pu{1^\circ C}$ $1^{\circ} C$ , $c_\ce{H2O}=\pu{4.184J/g^\circ C}$ $c_{H X_{2} O} = 4.184 J / g^{\circ} C$
- Dulong and Petit law: $\text{specific heat}\times\text{molar mass}\approx\pu{25 J/mol^\circ C}$
$q=mc\Delta T$ where $q$ is heat energy transferred, $m$ is mass, $c$ is specific heat, $\Delta T$ is change in temperature $(\Delta C$ or $K)$

Example

A $\pu{26.00g}$ sample of a $\pu{85.25^\circ C}$ metal is added to an insulated cup with $\pu{75g}$ of $\pu{24.00^\circ C}$ water. The final temperature of the water and metal is $\pu{28.34^\circ C}$ . What is the specific heat of the metal?

By conservation of energy, $q_\text{water}=-q_\text{metal}$ . Thus from the formula,
$m_\text{water}c_\text{water}\Delta T_\text{water}=-m_\text{metal}c_\text{metal}\Delta T_\text{metal}$
$(\pu{75.00g})(\pu{4.184J//g^\circ C})(\pu{28.34^\circ C}-\pu{24.00^\circ C})=-(\pu{26.00g})(c_\text{metal})(\pu{28.34^\circ C}-\pu{85.25^\circ C})$
$c_\text{metal}=\frac{(75.00)(4.184)(4.34)}{(26.00)(56.91)}\pu{J//g^\circ C}=\pu{0.9204 J//g^\circ C}$

standard heat of a reaction ( $\Delta H^\circ$ , units: $\pu{kJ}$ ): heat produced when the number of moles specified in a chemical equation reacts
Hess's Law: same mathematical operations on chemical equation apply to $\Delta H_\text{react}^\circ$ $Δ H_{react}^{\circ}$
- i.e. multiplying by a constant and adding two equations
- in particular, $\Delta H_\text{forward react}^\circ=-\Delta H_\text{reverse react}^\circ$

Example

Find $\Delta H^\circ$ for the reaction below, given the following reactions and subsequent $\Delta H^\circ$ values:
$\ce{H2SO4(l) -> SO3(g) + H2O(g)}$

\ce{H2S(g) + 2O2(g) -> H2SO4(l)}

\Delta H=\pu{-235.5 kJ}

\ce{H2S(g) + 2O2(g) → SO3(g) + H2O(l)}

\Delta H=\pu{-207 kJ}

\ce{H2O(l) → H2O(g)}

\Delta H=\pu{44kJ}

Since $\ce{H2SO4}$ must be on the left, we start with the reverse of the first equation

\ce{H2SO4(l) -> H2S(g) + 2O2(g)}

\Delta H=\pu{235.5 kJ}

Next, we must get rid of $\ce{H2S}$ and $\ce{2O2}$ , so add the second equation

\ce{H2SO4(l) -> SO3(g) + H2O(l)}

\Delta H=\pu{28.5 kJ}

Finally, we remove the $\ce{H2O(l)}$ and convert it to $\ce{H2O(g)}$ , so add the third equation

\ce{H2SO4(l) -> SO3(g) + H2O(g)}

\Delta H=\pu{72.5 kJ}

So, the final $\Delta H^\circ$ is $\pu{72.5 kJ}$

Heat of formation: tabulated values used to calculate $\Delta H_\text{react}^\circ$ $Δ H_{react}^{\circ}$
- $\Delta H_f^\circ(\text{any element})=0$
- $\Delta H_\text{react}^\circ=\sum(\Delta H_f^\circ\times\text{coef})_\text{products}-\sum(\Delta H_f^\circ\times\text{coef})_\text{reactants}$ $Δ H_{react}^{\circ} = \sum (Δ H_{f}^{\circ} \times coef)_{products} - \sum (Δ H_{f}^{\circ} \times coef)_{reactants}$
  - products minus reactants

Example

Calculate the heat of combustion of $\ce{CH4}$ given the following heats of formation:
$\Delta H_f^\circ(\ce{CH4(g)})=\pu{-74.8 kJ/mol}$
$\Delta H_f^\circ(\ce{CO2(g)})=\pu{-110.5 kJ/mol}$
$\Delta H_f^\circ(\ce{H2O(g)})=\pu{-241.8 kJ/mol}$

The combustion of $\ce{CH4}$ is given by the equation
$\ce{CH4 + 2O2 -> CO2 + 2H2O}$
Since $\Delta H_f^\circ(\ce{O2})=0$ ,
$\Delta H_\text{react}^\circ=[(\pu{2 mol})(\pu{-241.8 kJ//mol})+(\pu{1 mol})(\pu{-110.5 kJ//mol})]-[(\pu{1 mol})(\pu{-74.8 kJ//mol})]=\pu{-519.3 kJ}$

Second Law of Thermodynamics

entropy ( $S$ $S$ ): measure of how chaotic a system is (number of ways a system can arrange the particles within the system)
- particle increase increases entropy
- volume increase increases entropy
- state change (from solid $\rightarrow$ liquid $\rightarrow$ gas) increases entropy
standard entropy ( $S^\circ$ $S^{\circ}$ , units: $\pu{J/mol ^\circ C}$ $J / m o l^{\circ} C$ ): entropy based on mole
- calculated the same way as $\Delta H^\circ$

Gibbs Free-Energy

free-energy change $(\Delta G^\circ)$ $(Δ G^{\circ})$ : maximum amount of energy available from any chemical reaction
- $\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$
- negative $\Delta G^\circ$ means thermodynamically favorable
  - reaction that occurs without additional energy input; spontaneous
  - $\Delta G_\text{react}^\circ$ calculated the same way as $\Delta H^\circ$ and $\Delta S^\circ$ , but temperature-specific (standard state is $\Delta G_{298}^\circ$ )
- nonstandard state free-energy change is simply $\Delta G=\Delta G^\circ+RT\ln Q$ , where $Q$ $Q$ is the reaction quotient
  - at standard state, $Q=1$ so $\Delta G=\Delta G^\circ$
  - at equilibrium, $\Delta G=0$ and $Q=K$ ( $K$ is equilibrium constant), so the equilibrium condition is $\Delta G^\circ=-RT\ln K$