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15. 𝑘 𝑒𝑞,2 =
𝑘1
𝑘−1
⇒ [ 𝐴 – 𝐵 ] 𝐸𝑄
𝑘 𝑒𝑞,1 =
𝑘2
𝑘−2
⇒ [ 𝐵– 𝐶 ] 𝐸𝑄
[ 𝐴 – 𝐵 ] 𝐸𝑄
[ 𝐵 – 𝐶 ] 𝐸𝑄
=
𝑘 𝑒𝑞,2
𝑘 𝑒𝑞,1
=
𝑘1
𝑘−1
𝑘2
𝑘−2
Para determinar o valor da razão de concentrações no equilíbrio é necessário
conhecer as k (constantes de velocidade cinéticas) das reações inversas.
20.33
[A] = [A]0 e –
(k
1
+k
2)t
ln [A]t = ln [A]0 + ln e –
(k
1
+k
2)t
ln [A]t = ln [A]0 – (k1 + k2)t
y = b + at
INTERCESSÃO: ln [A]0
COEFICIENTE ANGULAR: – (k1 + k2)
20.41
ln k = ln A - Ea∕R (1∕T) A = -54,274
ln k = -54,274 + 42651,862 (1∕T) R2
= 0,596
-Ea∕R = 42651,862K-1
Ea = 42651,862K-1
. R
Ea = 42651,862K-1
. 8,314J. k-1
.mol-1
Ea = -354607,581 J∕mol
Ea = -3,546 x 105
J∕mol
16. 20.43) 𝑙𝑛 (
𝐾1
𝐾2
) = (−
𝐸𝑎
𝑅
) (
1
𝑇1
−
1
𝑇2
)
Onde, T1= 295K, T2= 305K e K2= 2K1
𝑙𝑛 (
𝐾1
2𝐾1
) = (−
𝐸𝑎
𝑅
) (
1
295
−
1
305
)
𝑙𝑛 (
1
2
) = (−
𝐸𝑎
𝑅
) (1,1114𝑥10−4
𝐾−1)
−0,693
1,1114𝑥10−4 𝐾−1
= (−
𝐸𝑎
𝑅
)
- Ea = - 6236,703R
Ea = 6236,703R JK/Kmol
Ea = 6236,703R J/mol
20.44
Dados: T = 22ºC = 295K; k = 1011
s-1
k = A e-Ea∕RT
e-Ea∕RT
= k∕A
ln e -Ea∕RT
= ln k∕A
-Ea∕RT = ln k∕A
ln k – ln k = -Ea∕RT
ln A = -Ea∕RT +ln k
ln A = -249 x 103
J.mol-1
∕8,314J. k-1
.mol-1
x 295K
ln A = +101,524 + 25,328
ln A = 126,892
17. A = e126,892
= 1,23 x 1055
20.45
Dados: k = 9,9 x 10-12
cm3
∕s; T = 1153K; Ea = 1,9kJ.mol-1
k = A e-Ea∕RT
A = k∕e-Ea∕RT
A = 9,9 x 10-12
cm3
.s-1
∕exp [1,9kJ.mol-1
∕8,314J. k-1
.mol-1
.1153K]
A = 1,21 x 10-11
(cm3
)s-1
20.56
Cl2 + CH4 CHCl + HCl
a)
𝑣 = 𝑘[𝐶𝑙2][𝐶𝐻4]
b) Fazendo os experimentos de determinação de velocidade de reação por
método das velocidades iniciais para determinar a ordem da reação para
cada um dos reagentes. Se a ordem experimental concordar com a
predição teórica, há um indício de que o mecanismo proposto está
correto.