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Solutions for The Study of Chemical Reactions
Cl Cl
H CH2CH3
Cl Cl
Cl Cl
H Cl
Cl CH2CH3
Cl CH2CH3
CH3CH2CH2CH3
hν9 (a) initiation
propagation
termination
(1) 
(2) Cl • +
(3)
 Cl • + • Cl
 Cl • + • CH2CH3
 CH3CH2 • + • CH2CH3
2 Cl •
+ • CH2CH3
+ • CH2CH3 + Cl •
(b) step (1) break Cl–Cl
step (2) break H–CH2CH3
 make H–Cl 
 step (2)
∆H° = + 242 kJ/mole (+ 58 kcal/mole)
∆H° = + 410 kJ/mole (+ 98 kcal/mole)
∆H° = – 431 kJ/mole (– 103 kcal/mole)
∆H° = −21 kJ/mole (−5 kcal/mole)
∆H° = +242 kJ/mole (+ 58 kcal/mole)
∆H° = – 339 kJ/mole (– 81 kcal/mole)
∆H° = −97 kJ/mole (−23 kcal/mole)
step (3) break Cl–Cl
 make Cl–CH2CH3 
 step (3)
(c) ∆H° for the reaction is the sum of the ∆H° values of the individual propagation steps:
− 21 kJ/mole + − 97 kJ/mole = − 118 kJ/mole
(− 5 kcal/mole + − 23 kcal/mole = − 28 kcal/mole)
hν
step (3) break Br–Br
 make Br–CH3 
 step (3)
∆H° = + 192 kJ/mole (+ 46 kcal/mole)
∆H° = – 293 kJ/mole (– 70 kcal/mole)
∆H° = −101 kJ/mole (−24 kcal/mole)
∆H° = + 192 kJ/mole (+ 46 kcal/mole)
∆H° = + 435 kJ/mole (+ 104 kcal/mole)
∆H° = – 368 kJ/mole (– 88 kcal/mole)
∆H° = +67 kJ/mole (+16 kcal/mole)
step (1) break Br–Br
step (2) break H–CH3
 make H–Br 
 step (2)
+ Br •+ • CH3
+ • CH3
2 Br •(1) 
(2) Br • +
(3)
initiation
propagation
10 (a)
Br CH3Br Br
H BrH CH3
Br Br
+ 67 kJ/mole + − 101 kJ/mole = − 34 kJ/mole
(+ 16 kcal/mole + − 24 kcal/mole = − 8 kcal/mole)
∆H° for the reaction is the sum of the ∆H° values of the individual propagation steps:(b)
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(a) first order: the exponent of [ (CH3)3CCl ] in the rate law = 1
(b) zeroth order: [ CH3OH ] does not appear in the rate law (its exponent is zero)
(c) first order: the sum of the exponents in the rate law = 1 + 0 = 1
12
(a) first order: the exponent of [cyclohexene] in the rate law = 1
(b) second order: the exponent of [ Br2 ] in the rate law = 2
(c) third order: the sum of the exponents in the rate law = 1 + 2 = 3
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