In GP: b = ar and c = ar².
Given a + ar + ar² = 39 and a² + a²r² + a²r⁴ = 651.
From the first equation: a(1 + r + r²) = 39.
From the second: a²(1 + r² + r⁴) = 651.
Dividing these equations and solving yields r = 2, which satisfies both conditions when a = 3.
From Vieta's formulas: α + β = 4/3 and αβ = k/3.
Using α² + β² = (α+β)² - 2αβ, we get 10/9 = (4/3)² - 2(k/3) = 16/9 - 2k/3.
Solving: 10/9 = 16/9 - 2k/3 gives 2k/3 = 6/9 = 2/3, so k = 1/3.
Using ln(k₂/k₁) = (Eₐ/R)[(T₂-T₁)/(T₁T₂)]; ln(k₂/k₁) = (60000/8.314)[(300)/(600×300)] ≈ 6.2; k₂/k₁ ≈ 500
Adding all steps and canceling intermediates (Cl, H): Cl₂ + H₂ → 2HCl
k = A·exp(-Eₐ/RT) = 2 × 10¹³ × exp(-50000/8.314×300) = 2 × 10¹³ × exp(-20.03) ≈ 3.4 × 10⁻³ s⁻¹
Rate = k₂[I][B]. From equilibrium: [I] = K₁[A][B]. Therefore: Rate = k₂K₁[A][B]², which is second-order in B
Rate = k₂[AB][C]. From fast equilibrium: [AB] = K₁[A][B]. So Rate = k₂K₁[A][B][C], making it third-order overall
A = k/e^(-Ea/RT) = (2 × 10⁻⁵)/e^(-80000/2.478) ≈ 5.6 × 10¹³ s⁻¹
When [B]₀ >> [A]₀, the concentration of B doesn't change significantly during the reaction, so it can be incorporated into the rate constant, making the reaction appear first-order in A only.
In the Lindemann mechanism: Step 1 (fast equilibrium): A + A ⇌ A* + A, Step 2 (slow): A* → products. The slow step is rate-determining.