Given:

• T₁ = 500 K, k₁ = 3.2 × 10⁻⁴ s⁻¹
• T₂ = 600 K, k₂ = 1.5 × 10⁻² s⁻¹
• R = 8.314 J/(mol·K)

(a) Calculate Ea

Use two-point form:

$\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

Calculate ln(k₂/k₁):

$\frac{k_2}{k_1} = \frac{1.5 \times 10^{-2}}{3.2 \times 10^{-4}} = 46.9$

$\ln(46.9) = 3.85$

Calculate 1/T₁ - 1/T₂:

$\frac{1}{500} - \frac{1}{600} = 0.00200 - 0.00167 = 0.000333 \text{ K}^{-1}$

Solve for Ea:

$3.85 = \frac{E_a}{8.314} \times 0.000333$

$E_a = \frac{3.85 \times 8.314}{0.000333} = 96,100 \text{ J/mol}$

$E_a = 96 \text{ kJ/mol}$

Answer (a): Ea = 96 kJ/mol

(b) Calculate A

Use Arrhenius equation at either temperature (use T₁ = 500 K):

$k_1 = Ae^{-E_a/RT_1}$

$A = k_1 e^{E_a/RT_1}$

Calculate Ea/(RT₁):

$\frac{E_a}{RT_1} = \frac{96100}{8.314 \times 500} = \frac{96100}{4157} = 23.1$

Calculate A:

$A = (3.2 \times 10^{-4}) \times e^{23.1}$

$e^{23.1} = 1.1 \times 10^{10}$

$A = 3.5 \times 10^6 \text{ s}^{-1}$

Answer (b): A = 3.5 × 10⁶ s⁻¹

Check with T₂: k₂ = (3.5 × 10⁶)e^(-96100/(8.314×600)) = (3.5 × 10⁶)e^(-19.25) = (3.5 × 10⁶)(4.3 × 10⁻⁹) = 1.5 × 10⁻² ✓

Given:

• Exothermic: ΔH = -50 kJ/mol
• Ea(forward) = 80 kJ/mol

Energy Diagram:

Energy
  ↑
  |    Transition State
  |         ↗ ↘
  |      ↗     ↘  Ea(reverse) = 130 kJ
  |   ↗           ↘
  |  ↗  Ea(fwd)     ↘
  | ↗   80 kJ        ↘
Reactants              Products
  |                    ↓ ΔH = -50 kJ
  |___________________
  |__________________|
  
  Reaction Progress →

Key points:

1. Reactants start at reference level
2. Transition state is 80 kJ above reactants
3. Products are 50 kJ below reactants (exothermic)

Calculate Ea(reverse):

Relationship for exothermic:

$E_{a,\text{reverse}} = E_{a,\text{forward}} - \Delta H$

But ΔH = -50 kJ (negative), so:

$E_{a,\text{reverse}} = 80 - (-50) = 80 + 50 = 130 \text{ kJ/mol}$

Or think of it as:

• Forward: reactants → transition state = +80 kJ
• Total drop: reactants → products = -50 kJ
• Reverse: products → transition state = +130 kJ

Answer: Ea(reverse) = 130 kJ/mol

General relationships:

Exothermic (ΔH < 0):

• Ea,reverse > Ea,forward
• Ea,reverse = Ea,forward + |ΔH|

Endothermic (ΔH > 0):

• Ea,forward > Ea,reverse
• Ea,forward = Ea,reverse + ΔH

Always: $E_{a,\text{forward}} - E_{a,\text{reverse}} = \Delta H$

If catalyst added:

• Both Ea,forward and Ea,reverse decrease by same amount
• ΔH unchanged (path doesn't affect thermodynamics)
• Transition state lower, but products/reactants same

Slope m = −9500 ⇒ Ea = −mR = 9500×8.314 = 7.90×10^4 J/mol = 79.0 kJ/mol. Intercept b = 18.2 ⇒ A = e^{18.2} ≈ 9.91×10^7 (units of s⁻¹ for first-order).

Compute k = A e^{−Ea/RT}. Use R = 8.314 J/(mol·K). Convert Ea to J/mol. For I: exponent = −85000/(8.314×298) = −34.3 ⇒ k₁ ≈ 2.0×10^12 × e^{−34.3} = 2.0×10^12 × 1.2×10^{−15} ≈ 2.4×10^{−3} s⁻¹. For II: exponent = −65000/(8.314×298) = −26.2 ⇒ k₂ ≈ 4.0×10^10 × e^{−26.2} = 4.0×10^10 × 4.3×10^{−12} ≈ 0.172 s⁻¹. Reaction II is faster by factor ≈ k₂/k₁ ≈ 0.172 / 2.4×10^{−3} ≈ 71.

Use two-point form with k₂/k₁ = 2 and T₁ = 300 K, T₂ = 300 + ΔT. ln(2) = (Ea/R)(1/T₁ − 1/T₂). Solve approximately with T₂ ≈ T₁ + ΔT and small ΔT: 1/T₁ − 1/T₂ ≈ ΔT/T₁^2. Thus ln(2) ≈ (Ea/R)(ΔT/T₁^2) ⇒ ΔT ≈ ln(2)·R·T₁^2 / Ea. Plug in: ΔT ≈ 0.693×8.314×(300)^2 / 95000 ≈ (0.693×8.314×90000)/95000 ≈ (5187)/95000 ≈ 0.0546×10^2 ≈ 5.46 °C. So roughly a 5–6 °C increase doubles the rate (depends on Ea and starting T).

Factor = k_cat/k_uncat = e^{−Ea_cat/RT} / e^{−Ea_uncat/RT} = e^{(Ea_uncat − Ea_cat)/(RT)}. ΔEa = 40 kJ/mol = 4.0×10^4 J/mol; RT = 8.314×350 ≈ 2910 J/mol. Factor ≈ e^{40000/2910} = e^{13.7} ≈ 9.0×10^5. So k increases by ~9×10^5.