University College - Chemistry Chem 112A Lecture Notes - Lecture 12: Collision Frequency, Equilibrium Constant, Intermediate 1

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If reaction 1 is much slower than reaction 2, then reaction 2 is rate determining. B + i -> c rate2 = k_2[i][b] Overall: a + b -> c i = intermediate: step 2 much faster than step 1. I used as fast as it is formed. Step 1 rate determining: rate = rate 1 = k_1[a: step 2 slow compared to step -1. Meaning: k_1[a} = k_-1[i] k_1/k_-1 = [i]/[a] = k_eq(1,-1) P = steric factor (fraction of correctly oriented collisions. All reactions must overcome an activation energy (e_a). Forms the intermediate which is able to produce the final product. Arrhenius form of the constant: k = ae^(-e_a/(rt)) From k = ae^(-e_a/(rt)) ln(k) = ln(a) - e_a/(rt) If a plot of ln(k(t)) vs 1/t is made. Van t hoff: ln(k(t_2)/k(t_1)) = -deltah/r (1/t_2 - 1/t_1) ; k = equilibrium constant. lnk = ln(a_f/a_b) - deltah_r/(rt) A = arrhenius constant (takes into account speed and orientation, collision frequency factor)

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