Chemical Kinetics and Equilibrium Principles

Chapter 12: Chemical Kinetics

1. Reaction Rate & Stoichiometry

For aA+bB→cC+dD:

Rate = −(1/a)∗Δ[A]/Δt = −(1/b)∗Δ[B]/Δt = (1/c)∗Δ[C]/Δt = (1/d)∗Δ[D]/Δt

• Rate is always positive. Reactants are negative, products positive.
• Ex: 2NO2​→2O2​. If Rate of disappearance of NO2​=0.010: Rate = −(1/2)∗(0.005 M/s. Rate of formation of O2​=0.005. Rate of formation of NO=20.005 M/s=0.010.

2. Rate Laws & Reaction Order

Rate = k[A]m[B]n

• k: Rate constant (temp-dependent)
• m,n: Reaction orders (exp. Determined)
• Overall Order = m+n
• Method of Initial Rates: Compare two experiments where only one reactant's conc. Changes. (Rate/Rate) = ([A]2​/[A]1​)
  • If Rate x2 when [A] x2 ⟹m=1 (1st order)
  • If Rate x4 when [A] x2 ⟹m=2 (2nd order)
  • If Rate x1 when [A] x2 ⟹m=0 (0 order)
• Steps: 1. Find orders (m,n). 2. Calc k using any exp.
• Ex: Rate = k[A][B]1. If Exp 1: [A]=0.100.10, Rate = 1.0×1. 1.0×1k(0.10)(0.10)1⟹k1.0 Ms−1.

3. Integrated Rate Laws & Half-Life (t1/2​)

• When to Use: Find conc. At time t, time for conc. To drop, determine order from plots.
• t1/2​ Concept: Time for reactant conc. To halve.
  • 1st Order: t1/2​ is independent of initial conc. (constant).
• Ex (1st Order): k=0.00504. [A]0​=0.500. Find [A] after 190 s. ln[A]=−ln(0.500 M) ln[A]=−0.6931=− [A]t​=e−1.65070.192 M

4. Collision Theory & Arrhenius Equation

• Collision Theory: For reaction: 1. Collisions must occur. 2. Sufficient Ea​. 3. Correct orientation.
• Activation Energy (Ea​): Min energy needed for reaction. Higher Ea​, slower reaction.
• Arrhenius Equation: Relates k,T,E. R=8.314K).
  • One-Point (Linear Plot): lnk=−lnA
    • Plot lnk vs. 1/T. Slope = −E/R.
  • Two-Point Form: ln(k2​/k1​)=−1/T)
• Ex: k1​=2.110 at 310 K. k2​=7.110 at 320 K. ln(7.1×110) = −(Ea​/8.314 J/mol⋅K1/310 K) 1.218=−Ea​≈1.010 J/mol=100

Chapter 13: Chemical Equilibrium

1. What is Equilibrium?

• Rate$_{\text{forward}}$ = Rate$_{\text{reverse}}$
• Concentrations of reactants/products are constant (not necessarily equal).
• Dynamic process: Reactions continue at equal rates.

2. Equilibrium Constant Expressions (Kc​, Kp​)

For aA(g)+bB(g)⇌cC(g)+dD(g)

• Kc​ (Concentrations): Kc​=([
  • Include only (g) and (aq) species. Exclude pure (s) and (l).
• Kp​ (Partial Pressures): Kp​=(
  • Include only (g) species.
• Relationship: Kp​=Kc​
  • Δn = (moles gaseous products) - (moles gaseous reactants)
  • R=0.08206atm/(mol⋅K)

3. Meaning of K Values

• K≫1 (large): Products favored at equilibrium.
• K≪1 (small): Reactants favored at equilibrium.
• K≈1: Significant amounts of both present.

4. Reaction Quotient (Q) & Predicting Direction

• Qc​=([ (uses initial or current concentrations)
• Comparing Q to K:
  • Q<K: Reaction shifts RIGHT (towards products) to reach equilibrium.
  • Q>K: Reaction shifts LEFT (towards reactants) to reach equilibrium.
  • Q=K: System is at equilibrium.

5. Calculating Equilibrium Concentrations (RICE Tables)

• When to Use: Given initial concentrations and K, find equilibrium concentrations.
• Steps:
  1. Reaction: Balanced Equation
  2. Initial: List initial conc.
  3. Change: Define 'x' based on stoichiometry and direction (from Q vs K).
  4. Equilibrium: Express final conc. In terms of 'x'.
  5. Substitute E-line into K expression and solve for 'x'.
  6. Calculate final concentrations.
• Algebraic Shortcuts:
  • "x is small" Approximation: If K is very small (<10 to 10) AND Initial Conc./K>100 (or 400).
    • Assume (A−xA or (A+xA.
    • Verify: If 'x' is less than 5% of the initial concentration it was subtracted from/added to, approximation is valid.
  • Perfect Square: If expression is (A−xx)=K, take K​ of both sides.
• Ex: CHCOOH(aq)⇌CH3​O(aq), Kc​=1.810. Initial [CH3​COOH]=1.0.

Kc​=(x∗x)/(1.0−x)=1.8×10−5

Check for approx.: 1.0/(1.8×10−5)≈55000≫400, so 'x' is small.

1.8×10−5≈x2/1.0⟹x2=1.8×10−5⟹x=0.00424

Verify: (0.00424/1.0)×100%=0.424%<5%, valid.

Eq. Conc: [CH3​COOH]≈1.0 M, [CH3​COO−]=[H3​O+]=0.00424 M.