Respuesta :
Answer:
(a) [tex]F_{496/ K}=1.411\times 10^{-17}[/tex]
(b) [tex]F_{528/ K}=1.482\times 10^{-16}[/tex]
(c) [tex]\frac {F_{528/ K}}{F_{496/ K}}=10.5[/tex]
Explanation:
Using the expression,
[tex]{k}=Ae^{-\dfrac{E_{a}}{RT}}[/tex]
Where,
[tex]k\ is\ the\ rate\ constant\ at\ T[/tex]
[tex]E_a[/tex] is the activation energy
A is the pre-exponential factor
R is Gas constant having value = 8.314 J / K mol
The factor, [tex]e^{-\dfrac{E_{a}}{RT}}[/tex] is the fraction of molecules which have value equal to or grater than activation energy.
Thus,
[tex]F=e^{-\dfrac{E_{a}}{RT}}[/tex]
(a) Given, [tex]E_a[/tex] = 160.0 kJ/mol = 160000 J/mol (As 1 kJ = 1000 J)
T = 496 K
Thus, applying values as:
[tex]F=e^{-\dfrac{160000}{8.314\times 496}}[/tex]
[tex]F=\frac{1}{7.08749\times 10^{16}}[/tex]
[tex]F_{496/ K}=1.411\times 10^{-17}[/tex]
(b) Given, [tex]E_a[/tex] = 160.0 kJ/mol = 160000 J/mol (As 1 kJ = 1000 J)
T = 528 K
Thus, applying values as:
[tex]F=e^{-\dfrac{160000}{8.314\times 528}}[/tex]
[tex]F=\frac{1}{6.74917\times 10^{15}}[/tex]
[tex]F_{528/ K}=1.482\times 10^{-16}[/tex]
(c) Ratio :
[tex]\frac {F_{528/ K}}{F_{496/ K}}=\frac{1.482\times 10^{-16}}{1.411\times 10^{-17}}=10.5[/tex]
