REACTIVITY 2
How Much, How Fast & How Far?
Mole Ratio • Collision Theory • Le Châtelier • Rates • Equilibrium
2.1 How Much? (Stoichiometry)
1. The Mole Ratio
Chemical equations are recipes. The coefficients tell you the ratio of moles, not grams.
At the same temperature and pressure, Mole ratio = Volume ratio.
10 dm3 N2 reacts with 30 dm3 H2
2. Yields & Limiting Reactant
Examiner Tip: Calculate $\frac{\text{Moles}}{\text{Coefficient}}$ for each reactant. The smallest number is the limiting reactant.
Measures Reaction Efficiency
3. Atom Economy (Green Chemistry)
Measures waste, not just yield.
2.2 How Fast? (Kinetics)
1. Collision Theory
For a reaction to occur, particles must collide with:
2. Maxwell-Boltzmann Distribution
Shows the distribution of kinetic energy among particles.
- Curve flattens and peak shifts right.
- Total Area stays constant (same number of particles).
- Key Result: Significant increase in the area past $E_a$ (more successful collisions).
3. Factors Affecting Rate
| Factor | Microscopic Explanation |
|---|---|
| Concentration | More particles per unit volume $\rightarrow$ collision frequency $\uparrow$ |
| Surface Area | More particles exposed $\rightarrow$ collision frequency $\uparrow$ |
| Temperature | Major: More particles have $E \ge E_a$. Minor: Particles move faster (frequency $\uparrow$). |
| Catalyst | Provides an alternative pathway with Lower $E_a$. |
2.3 How Far? (Equilibrium)
1. Dynamic Equilibrium
- Rate of Forward reaction = Rate of Backward reaction.
- Concentrations of reactants and products remain constant (not necessarily equal).
- Occurs in a Closed System.
2. The Equilibrium Law ($K_c$)
3. Le Châtelier's Principle
When a system at equilibrium is subjected to a change, it will shift to minimize that change.
| Change | Shift Direction | Value of $K_c$ |
|---|---|---|
| Add Reactant | Right (Make Product) | No Change |
| Increase Pressure | To side with fewer gas moles | No Change |
| Add Catalyst | None (Rates increase equally) | No Change |
| Increase T (Exo) | Left (Absorb Heat) | Decreases |
| Increase T (Endo) | Right (Absorb Heat) | Increases |
Do not confuse Rate with Yield.
Increasing T in an Exothermic reaction increases Rate (faster) but decreases Yield (less product).
Advanced Theory
The following section is for HL Students ONLY.
Rate Laws • Mechanisms • Arrhenius • Gibbs Energy
Rate Expressions
1. The Rate Law
The rate law is derived experimentally, NOT from the balanced equation.
$mol^{1-\text{order}} dm^{3(\text{order}-1)} s^{-1}$
Example (Order 3): $mol^{-2} dm^6 s^{-1}$
2. Mechanisms & RDS
- RDS (Rate Determining Step): The Slowest Step. The reaction bottleneck.
- The Rate Law depends only on species involved in the RDS (and any prior steps).
- Complex Case: If Step 1 is a fast equilibrium before the RDS, substitute intermediate concentration using $K_{eq}$.
The Arrhenius Equation
Quantifies exactly how the Rate Constant ($k$) changes with Temperature.
- Y-axis: $\ln k$
- X-axis: $1/T$ (Kelvin)
- Gradient ($m$): $-E_a/R$
Gibbs Energy & Equilibrium
1. Reaction Quotient ($Q$)
Ratio of concentrations at any time, not just equilibrium.
Shift Right $\rightarrow$
Equilibrium
$\leftarrow$ Shift Left
2. Relating $\Delta G$ and $K$
This equation links Spontaneity to Equilibrium Position.
- If $\Delta G$ is negative, $\ln K$ is positive $\rightarrow K > 1$ (Products Favored).
- If $\Delta G$ is positive, $\ln K$ is negative $\rightarrow K < 1$ (Reactants Favored).
The Examiner's Vault
Strictly assessed on Reactivity 2 content.
Reaction: N2 + 3H2 → 2NH3.
Mixed: 2.0 mol N2 and 3.0 mol H2. What is the max NH3 produced?
Check limits: N2 (2/1=2), H2 (3/3=1). H2 is limiting.
Ratio H2:NH3 is 3:2. So 3.0 mol H2 → 2.0 mol NH3.
N2(g) + 3H2(g) ↔ 2NH3(g) $\Delta H = -92 kJ$.
Which change increases the value of $K_c$?
Only Temperature changes $K_c$. Reaction is Exothermic. Lowering T shifts Right, increasing products ($K_c$).
Rate doubles when [A] doubles. Quadruples when [B] doubles. What are the units of $k$?
Order A=1, B=2. Overall Order = 3.
Units: $mol^{1-3} dm^{3(3-1)} s^{-1} = mol^{-2} dm^6 s^{-1}$.
Sketch curves for T1 and higher temp T2. Explain rate increase.
- Axes labeled: Y=# Particles, X=Kinetic Energy. [1]
- T2 curve: Peak is lower and shifted to the right. [1]
- Explanation: Significantly larger area under curve past the $E_a$ line. More successful collisions. [2]
Step 1: NO + Cl2 ↔ NOCl2 (Fast Eq)
Step 2: NOCl2 + NO → 2NOCl (Slow)
Deduce Rate Law.
- Rate determined by Slow Step: Rate $= k_2 [NOCl_2][NO]$. [1]
- Substitute intermediate from Step 1: $[NOCl_2] = K_{eq} [NO][Cl_2]$.
- Final: Rate $= k [NO]^2 [Cl_2]$. [1]