Review Session-7

# Review Session-7 - REVIEW SESSION‐ 7 ...

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Unformatted text preview: REVIEW SESSION‐ 7  Chemical Kine6cs  14.8) You study the eﬀect of temperature on the rate of reac8ons and graph  the ln(k) vs 1/T. How do the two graphs compare (a) if Ea of second reac8on  higher than Ea of ﬁrst reac8on, but both reac8ons have same frequency factor  (b) if Frequency factor of second reac8on > ﬁrst reac8on, but have same Ea  14.8) You study the eﬀect of temperature on the rate of reac8ons and graph  the ln(k) vs 1/T. How do the two graphs compare (a) if Ea of second reac8on  higher than Ea of ﬁrst reac8on, but both reac8ons have same frequency factor  (b) if Frequency factor of second reac8on > ﬁrst reac8on, but have same Ea  Ea1  ln(k)  Ea2  ln(k)  Ea1  Ea2  1/T  Ea2 > Ea1  Same frequency factor  1/T  Ea2 = Ea1 (same slope)  Diﬀerent frequency factor  14.51) The gas‐phase reac8on Cl+HBrHCl+Br has an overall enthalpy change  of ‐66kJ. The Ea of this reac8on is 7kJ. (a) Sketch the energy proﬁle for the  reac8on, label Ea and ΔE (b) What is the Ea for the reverse reac8on?  14.51) The gas‐phase reac8on Cl+HBrHCl+Br has an overall enthalpy change  of ‐66kJ. The Ea of this reac8on is 7kJ. (a) Sketch the energy proﬁle for the  reac8on, label Ea and ΔE (b) What is the Ea for the reverse reac8on?  Ea =7kJ  For the reverse reac8on: Ea= (66+7) = 73kJ  ΔE= ‐66kJ  14.52) For the elementary process N2O5 NO2+NO3, the Ea and overall ΔE are  154kJ/mol and 136 kJ/mol. (a) Sketch the energy proﬁle for the reac8on, label  Ea and ΔE (b) What is the Ea for the reverse reac8on?  14.52) For the elementary process N2O5 NO2+NO3, the Ea and overall ΔE are  154kJ/mol and 136 kJ/mol. (a) Sketch the energy proﬁle for the reac8on, label  Ea and ΔE (b) What is the Ea for the reverse reac8on?  For the reverse reac8on: Ea= (154‐136) = 18kJ  Ea =154kJ/mol  ΔE= 136kJ/mol  14.41)The reac8on SO2Cl2  SO2+Cl2 is ﬁrst order in SO2Cl2. Using the  kine8c data, determine the magnitude of ﬁrst order rate constant:       Time (s)       Pressure (SO2Cl2) (atm)       0          1.0       2500        0.947       5000        0.895       7500        0.848           10000        0.803     14.41)The reac8on SO2Cl2  SO2+Cl2 is ﬁrst order in SO2Cl2. Using the  kine8c data, determine the magnitude of ﬁrst order rate constant:       Time (s)       Pressure (SO2Cl2) (atm)     lnP       0          1.0              0         2500        0.947            ‐0.0545       5000        0.895            ‐0.111       7500        0.848            ‐0.165           10000        0.803            ‐0.219     14.41)The reac8on SO2Cl2  SO2+Cl2 is ﬁrst order in SO2Cl2. Using the  kine8c data, determine the magnitude of ﬁrst order rate constant:       Time (s)       Pressure (SO2Cl2) (atm)     lnP       0          1.0              0         2500        0.947            ‐0.0545       5000        0.895            ‐0.111       7500        0.848            ‐0.165           10000        0.803            ‐0.219     ‐k = Slope = 2.19X10‐5s‐1  lnP  T  14.55)A certain ﬁrst order reac8on has a rate constant of 2.75X10‐2 s‐1 at 20oC.  What is the value of k at 60oC of (a) Ea= 75.5kJ/mol?  14.55)A certain ﬁrst order reac8on has a rate constant of 2.75X10‐2 s‐1 at 20oC.  What is the value of k at 60oC of (a) Ea= 75.5kJ/mol?  ln(k1/k2) = ‐(Ea/R) (1/T1‐1/T2)  (a) Ea= 75.5kJ = 75500 J/mol         R = 8.314 J/mol   k1 = 2.75X10‐2s‐1   T1 = 273+20 = 293K   T2 = 273+60 = 333K    k2 = ?   k2 = 1.14s‐1   14.65)(a) Based on the following reac8on proﬁle, how many intermediates are  formed in the reac8on AD (b) How many transi8on states are there? (c)  Which step is fastest? (d) is the reac8on endothermic or exothermic?  C  B  A  D  14.65)(a) Based on the following reac8on proﬁle, how many intermediates are  formed in the reac8on AD (b) How many transi8on states are there? (c)  Which step is fastest? (d) is the reac8on endothermic or exothermic?   2  intermediates  (B and C)  3 transi8on states  CD is fastest step  C  B  A  D  Endothermic reac8on  Two step mechanism with a Fast Ini6al Step  •  A proposed mechanism is  Step 1:  NO + Br2  NOBr2     (fast)  Step 2:  NOBr2 + NO → 2 NOBr     (slow)  Step 1 includes the forward and reverse reac8ons.  Two step mechanism with a Fast Ini6al Step  •  The rate of the overall reac8on depends upon  the rate of the slow step.  •  The rate law for that step would be  Rate = k2 [NOBr2] [NO]  •  But how can we ﬁnd [NOBr2]?  Two step mechanism with a Fast Ini6al Step  •  NOBr2 can react two ways:  –  With NO to form NOBr  –  By decomposi8on to reform NO and Br2  •  The reactants and products of the ﬁrst step  are in equilibrium with each other.  •  Therefore,  Ratef = Rater  Two step mechanism with a Fast Ini6al Step  •  Because Ratef = Rater ,  k1 [NO] [Br2] = k−1 [NOBr2]  •  Solving for [NOBr2] gives us  k1  [NO] [Br2] = [NOBr2]  k−1  Two step mechanism with a Fast Ini6al Step   Subs8tu8ng this expression for [NOBr2] in  the rate law for the rate‐determining step  gives  Rate =  k2k1  k−1  [NO] [Br2] [NO]  = k [NO]2 [Br2]  Consider the following three‐step mechanism:   Ce4+  + Mn2+  →  Ce3+ + Mn3+   Ce4+  + Mn3+  →  Ce3+ + Mn4+   Mn4+  + Tl+  →  Mn2+  + Tl3+   (1) Write balance equa8on for overall reac8on?  (2) Intermediate(s) ?  (3) Iden8fy catalyst(s):   Consider the following three‐step mechanism:   Ce4+  + Mn2+  →  Ce3+ + Mn3+   Ce4+  + Mn3+  →  Ce3+ + Mn4+   Mn4+  + Tl+  →  Mn2+  + Tl3+   (1) Write balance equa8on for overall reac8on?  2 Ce4+ + Tl+   2 Ce3+ +Tl3+   (2) Intermediate(s) ?  (3) Iden8fy catalyst(s):   Consider the following three‐step mechanism:   Ce4+  + Mn2+  →  Ce3+ + Mn3+   Ce4+  +Mn3+  →  Ce3+ + Mn4+   Mn4+  + Tl+  →  Mn2+  + Tl3+   (1) Write balance equa8on for overall reac8on?  2 Ce4+ + Tl+   2 Ce3+ +Tl3+   (2) Intermediate(s) ?  Mn3+ and Mn4+  (3) Iden8fy catalyst(s):   Consider the following three‐step mechanism:   Ce4+  + Mn2+  →  Ce3+ + Mn3+   Ce4+  + Mn3+  →  Ce3+ + Mn4+   Mn4+  + Tl+  →  Mn2+  + Tl3+   (1) Write balance equa8on for overall reac8on?  2 Ce4+ + Tl+   2 Ce3+ +Tl3+   (2) Intermediate(s) ?  Mn3+ and Mn4+  (3) Iden8fy catalyst(s):   Mn2+  1) Assuming complete dissocia8on of the solute, how many grams of  KNO3  must be added to 275  of water to produce a solu8on that freezes at 14.5 ? The  freezing point for pure water is 0.0  and Kf = 1.86 (Mastering Chemistry)  1) Assuming complete dissocia8on of the solute, how many grams of  KNO3  must be added to 275  of water to produce a solu8on that freezes at 14.5 ? The  freezing point for pure water is 0.0  and Kf = 1.86 (Mastering Chemistry)  ΔT = 14.5 , Kf = 1.86 , i=2 (not given any info)   ΔT = Kf . m . i  m = ΔT/Kf . I  m = 14.5/(1.86x2) = 3.9molals  Molality = number of moles/kg of solvent  3.9 = n/ (0.275kg)   (assump8on)  n= 1.07 moles of KNO3 => 108g  ...
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## This note was uploaded on 09/13/2011 for the course CHEM 102 taught by Professor Todd during the Spring '08 term at UNC.

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