Question

薯條

想請問一下 圈起來的部分 兩個除不盡的分數相減的有效位數要怎麼取

134 ky/mol
Two-Point Equation Relating k and T
The Arrhenius equation can be expressed in a different form by following the pro
dure used with the Clausius-Clapeyron equation in Chapter 9. At two different
reperatures, 7 and 71
In k2 = In A
E
RT 2
ins=1
Ink, = In A-
E
RT
Subtracting the second equation from the first eliminates In A, and we obtain
In k
Ink,
E
R
1
T
(11.7a
k
In
E 1
RT
– 8.31 J/mol K, the activation energy is expressed in joules per mole.
(11.75
Taking R
EXAMPLE
11.8 GRADED
Consider the first-order decomposition of A. The following is known about it:
KR
the rate constant doubles when the temperature increases from 15°C to 25°C.
the rate constant for the decomposition at 40°C is 0:0/25 s"
What is the activation energy for the decomposition
ANALYSIS
Information given:
k at 15°C(k); k at 25°C k, = 2k17
a
Information implied:
R value
Asked for:
E
STRATEGY
Take 15°C (288 K) as T where the rate constant is ki.
Take 25°C (298 K) as T2 where the rate constant is k2 = 2k1.
Substitute into Equation 11.7b.
ka E 1 1 1
In
R T T2
0.000111
SOLUTION
1
1
(0.693)(8.31) =
288
298
2
1
2k1
In-
ki
E
1
8.31 J/mol·K 288 298
E = 4.9 X 104 J/mol = 49 kJ/mol
4,949 k]
conti

化學 有效位數