+---
+header-includes:
+ - \usepackage{mhchem}
+ - \usepackage{tabularx}
+columns: 2
+geometry: margin=2cm
+---
+
# Rates and Equilibria
-## Energy profile diagrams
+## Energy
+
+### Enthalphy
+
+$$\Delta H = H_{\text{products}} - H_{\text{reactants}}$$
+
+**Endothermic** (products > reactants, $\Delta H > 0$)
+**Exothermic** (reactants > products, $\Delta H < 0$)
+
+![](graphics/endothermic-profile.png){#id .class width=25%}
+![](graphics/exothermic-profile.png){#id .class width=25%}
+
+### Activation energy $E_A$
$$E_A = E_{\text{max}} - E_{\text{initial}}$$
- Energy always needed to initiate reaction (break bonds of reactants)
- Reactant particles must collide at correct angle, energy etc
- Most collisions are not fruitful
+- Energy must be greater than or equal to $E_A$
-![](graphics/endothermic-profile.png)
-![](graphics/exothermic-profile.png)
+### Kinetic energy
+
+- **Temperature** - measure of _avg_ kinetic energy of particles. Over time each particle will eventually have enough energy to overcome $E_A$
+- Note same distribution indicates same temperature
+- $\uparrow$ rate with $\uparrow T$ mainly caused by $\uparrow E_K \implies$ greater collision force
+![](graphics/ke-temperature.png)
+
+## Rates
**Ways to increase rate of reaction:**
2. Increase concentration/pressure
3. Increase temperature
-## Kinetic energy
-
-Temperature - measure of _avg_ kinetic energy of particles. Over time each particle will eventually have enough energy to overcome $E_A$.
-Note same distribution indicates same temperature.
-![](graphics/ke-temperature.png)
-
-## Catalysts
+### Catalysts
- alternate reaction pathway, with lower $E_A$
- increased rate of reaction
- involved in reaction but regenerated at end
+- does not alter $K_c$ or extent of reaction
+- attracts reaction products
+- removal/addition of catalyst does not push system out of equilibrium
**Homogenous** catalyst: same state as reactants and products, e.g. Cl* radicals.
**Hetrogenous** catalyst: different state, easily separated. Preferred for manufacturing.
![](graphics/catalyst-graph.png)
+
+- Many catalysts involve transition elements
+- **Solid catalysts** - particles around catalyst with high surface energy *adsorb* gas molecules, lowering $E_A$
+- **Haber process** (ammonia producition) - enzymes are catalysts for one reaction each. Adsorption (bonding on surface) forms ammonia \ce{NH3}.
+
+## Equilibrium systems
+
+*Equilibrium* - the stage at which quantities of reactants and products remain unchanged
+
+Reaction graphs - exponential/logarithmic curves for reaction rates with time (simultaneous curves forward/back)
+
+\begin{tabularx}{\columnwidth}{ | l | X |}
+ \hline
+ \parbox[c]{2.2cm}{\includegraphics[width=2cm]{graphics/rxn-complete.png} } & \textbf{Complete reaction} - all reactant becomes product \\
+ \hline
+ \parbox[c]{2.2cm}{\includegraphics[width=2cm]{graphics/rxn-incomplete.png} } & \textbf{Incomplete reaction} - goes both ways and reaches equilibrium \\
+ \hline
+\end{tabularx}
+
+- All reactions are equilibrium reactions, but extent of backwards reaction may be negligible
+- Double arrow indicates equilibrium reaction
+- At equilibrium, rate of forward reaction = rate of back reaction.
+- Approaching equilibrium, forward rate $>$ back rate
+
+### Equilibrium constant $K_c$
+
+For \ce{$\alpha$A + $\beta$B + $\dots$ <=> $\chi$X + $\psi$Y + $\dots$}:
+
+$$K_c = {{[\ce{X}]^\chi \cdot [\ce{Y}]^\psi \cdot \dots} \over {[\ce{A}]^\alpha \cdot [\ce{B}]^\beta \cdot \dots}}$$
+
+More generally, for reactants $n_i \ce{R}_i$ and products $m_i \ce{P}_i$:
+
+$$K_c = {{\prod\limits^{|P|}_{i=1} [P_i]^{m_i}} \over {\prod\limits^{|R|}_{i=1} [R_i]^{n_i}}} \> | \> i \in \mathbb{N}^*$$
+
+Indicates extent of reaction
+If value is high ($> 10^4$), then [products] > [reactants]
+If value is low ($< 10^4$), then [reactants] > [products]
+
+- **$K_c$ depends on direction that equation is written (L to R)**
+- If $K_c$ is small, equilibrium lies *to the left*
+- aka *equilibrium expression*
+- For reverse reaction, use $K_c^\prime = {1 \over K_c}$
+- For coefficients, use $K_c^\prime = K_c^n$
+
+## Reaction constant (quotient) $Q$
+
+Proportion of products/reactants at a give time (specific $K_C$). If $Q=K_c$, then reaction is at equilibrium.
+
+## Le Châtelier’s principle
+
+> Any change that affects the position of an equilibrium causes that equilibrium to shift, if possible, in such a way as to partially oppose the effect of that change.
+
+### Changing volume / pressure
+
+1. $\Delta V < 0 \implies [\Sigma \text{particles}] \uparrow$, therefore system reacts in direction that produces less particles
+2. $\Delta V > 0 \implies [\Sigma \text{particles}] \downarrow$, therefore system reacts in direction that produces more particles
+2. $n(\text{left}) = n(\text{right})$ (volume change does not disturb equilibrium)
+
+### Changing temperature
+
+Only method that alters $K_c$.
+
+Changing temperature changes kinetic energy. System's response depends on whether reaction is exothermic or endothermic.
+
+- Exothermic - increase temp decreases $K_c$
+- Endothermic - increase temp increases $K_c$
+
+Time-concentration graph: smooth change
+
+### Changing concentration
+
+- Decreasing "total" concentration of system causes a shift towards reaction which produces more particles
+
+## Yield
+
+$$\text{yield \%} = {{\text{actual mass obtained} \over \text{theoretical maximum mass}} \times 100}$$
+
+- Yield may be lower than expected due to equilibrium reaction (incomplete)
+- $\uparrow$ yield $\equiv$ forward rxn; $\downarrow$ yield $\equiv$ back rxn
+- *Rate-yield conflict*: rxn is slower at eq. point further to RHS
+- This is ameliorated by catalysts, high pressure and removal of product
+
+## Acid/base equilibria
+
+Strong acid: $\ce{HA -> H+ + A-}$
+Weak acid: $\ce{HA <=> H+ + A-}$
+
+For weak acids, dilution causes increase in % ionisation.
+$\therefore [\ce{HA}] \propto 1 \div \text{\% ionisation}$
+(see 2013 exam, m.c. q20)
+
+$$\text{\% ionisation} = {{[\ce{H+}] \over [\ce{HA}]} \times 100}$$
+
+When a weak acid is diluted:
+
+- amount of $\ce{H3O+}$ increases
+- equilibrium shifts right
+- overall $[\ce{H3O+}]$ decreases
+- therefore pH increases