Question

Name Wkst #16 CHEM& 162 1. Consider the following proposed reaction mechanism. 12 = 21 1 + O₂ → 10₂ + 0 0+ 03 - 202 1 +10, -12 + O2 A. What is the overall reaction? B. What are the intermediates and catalysts? Intermediates: Catalysts: C. If the first step is in equilibrium and the second step is slow. what is the overall rate law? Remember, intermediates cannot appear in the final rate law! Rate = D. On the back, sketch an accurate reaction-energy diagram. Label the axes, the EA, all transition states, and intermediates for each step, then label the overall the AH , reactants and products for this exothermic reaction. 2. A certain reaction is found to have a rate constant of 452 M-Is- at 20°C. When the rate is measured at a second temperature, it is found to be 1695 M-15 - What is the second temperature? The activation energy for this reaction is 201 kJ/mol Useful information: In-96) R = 0.08206 Latm/mol. K = 8.314/mol

Answer 1

81. . I +102 I K 21 It og hy 20, +0 0 +03 K3 202 ka , I +02 If we sum of the all left hand side and right hand sides of each steps then we will get the overall sobed scheme - e It K, 21 2 I 102 to e 202 1 + 1 + O2 + 0 + 0 + I₂ + O2 I2 + 203 X 302 + I2 min + 102 Ka Overall rate 203 302 constant The overall soon is; 20 12 302 (I, as catalyst) Intermediates I, 102, 0 Calatyst o Iz 0 at © Now, the rate of the formation of product, peate the dłow] - Kq [I] [I Oz Nw, applying Steady-state - Apporoximation ubich tells us the rate of formation of intermediates

are zeno. ise. d11] at =0 d [102] at d[o] dt Now, d[1] at = k, [12] - Kn [1] - K2 [7][02] - K4 [1] [102] 2 m, [1] - at K2 [03 3 ma [202] = have; K, [12] K-1 [1] kq K, LI>) K-1 tk 2 [02] + kq { [Og Now, d[102] = ky [1] [03] - K4 [13] [102] = 0 K₂ [1][03) Kq [I] Kq Pulting [102] valme in en 6 we KI [12] KitK₂ [03] + kq x K₂ [03] + 2K2103] Now, putting the values of (I) and [202] in equa Ô we KI[12] K2 [03] Kq x K-1 + 2k₂ [03] K, Kz [ I ] [03 K-1 + 2K₂ [03 by the given condition ist step is in eqlim and show irl. K₂ << 1 So, k-1 + 2 K₂ [03] ř Kay have: K4 d[02] 1 dt 2nd step is very

②. k, = 452M7 st T, - 20°C = (20273) K = 293k T K₂ = 1695 Mist Tz = ? EA = 201 kJ. molt R = 0.08206 L-atmakt mot mot 8,314 J, ki mot 201 x 103 J, molt

EA (T, -T2) T, Т. EA R KI (72 - ) en e R Now, en K2 R ln K L 8.314 J. Kt. mot x ln 452 1695 T - Tz m Kz EA T, Tz 201103J. molt T, - Tz = -5.47810-57 mm TiTz

-S:47 xo-Sk 293 К — Т. 245K x T2. от 243K - Т. - - 0.016 Т, - СР, 243 Ka 0.98 Т, от, т. = 217.11 K = (297.77- 2 13)*4 - 24. 172