Rate depends only on [A] (order = 1 w.r.t. A). Halving [A] decreases rate by factor of 2. [B] has no effect
Using ln(k₂/k₁) = (Eₐ/R)(T₂-T₁)/(T₁T₂); ln(2) = (Eₐ/8.314)(10/93000); Eₐ ≈ 50.4 kJ/mol
The pre-exponential factor accounts for collision frequency, proper orientation (steric factor), and the Maxwell-Boltzmann energy distribution of molecules
Q₁₀ = rate at (T+10)/rate at T. For 300K to 320K (two 10K intervals), rate = r × 2.5² = 6.25r
Km is a characteristic constant for an enzyme-substrate pair, representing substrate concentration when v = Vmax/2
[A]ₜ = [A]₀e^(-kt) = 100 × e^(-0.0045 × 100) = 100 × e^(-0.45) ≈ 41.2%
Z ∝ σ × [A] × [B] × √(T/M), where all these factors contribute to collision frequency
If Rate = k[A]^m[B]^n, then 12 = 2^m × 3^n. Testing: 2¹ × 3² = 2 × 9 = 18 (no); 2² × 3¹ = 12 (yes). So m=1, n=2, giving Rate = k[A][B]²
For first-order reaction, [A]ₜ = [A]₀(1/2)^(t/t₁/₂). For [A]ₜ = 1/4[A]₀, we need (1/2)^(t/30) = 1/4, so t/30 = 2, giving t = 60 minutes
Using ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂): ln(5.0/1.7) = (Ea/8.314)(1/320 - 1/330), solving gives Ea ≈ 50 kJ/mol