Rate = k₂[I][B]. From equilibrium: [I] = K₁[A][B]. Therefore: Rate = k₂K₁[A][B]², which is second-order in B
A straight line in [A] vs time plot indicates [A] = [A]₀ - kt, which is the integrated rate law for zero-order reactions
Q₁₀ = rate at (T+10)/rate at T. For 300K to 320K (two 10K intervals), rate = r × 2.5² = 6.25r
Rate = k₂[AB][C]. From fast equilibrium: [AB] = K₁[A][B]. So Rate = k₂K₁[A][B][C], making it third-order overall
Since the reaction is first-order in B and zero-order in A, Rate ∝ [B]. Increasing [B] by 4 times increases Rate by 4 times
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%
A catalyst works by providing an alternative mechanism with lower Ea, thus increasing reaction rate without being consumed
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