Respuesta :
Answer:
The rate constant [tex]k_{2}[/tex] at 84.8°C is [tex]k_{2}=6.423sec^{-1}[/tex]
Explanation:
Taking the Arrhenius equation we have:
[tex]ln\frac{k_{2}}{k_{1}}=\frac{E_{a}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})[/tex]
Where [tex]k_{2}[/tex] is the rate constant at a temperature 2, [tex]k_{1}[/tex] is the rate constant at a temperature 1; [tex]T_{1}[/tex] is the temperature 1, [tex]T_{2}[/tex] is the temperature 2, R is the gas constant and [tex]E_{a}[/tex] is the activation energy.
Now, we need to solve the equation for [tex]k_{2}[/tex], so we have:
[tex]ln\frac{k_{2}}{k_{1}}=\frac{E_{a}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})[/tex]
[tex]ln({k_{2})-ln(k_{1})=\frac{E_{a}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})[/tex]
[tex]ln(k_{2})=E_{a}(\frac{1}{T_{1}}-\frac{1}{T_{2}})+ln(k_{1})[/tex]
Then we need to make sure that we are working with the same units, so:
[tex]R=8.314\frac{J}{mol.K}[/tex]
[tex]T_{1}=16.6^{o}C+273.15=289.75K[/tex]
[tex]T_{2}=84.4^{o}C+273.15=357.95K[/tex]
And now we can replace the values into the equation:
[tex]ln(k_{2})=\frac{22000\frac{J}{mol}}{8.314\frac{J}{mol.K}}(\frac{1}{289.75K}-\frac{1}{357.95K})+ln(1.868sec^{-1})[/tex]
[tex]ln(k_{2})=2646.139K(0.003451K^{-1}-0.002794K^{-1})+0.6249[/tex]
[tex]ln(k_{2})=2.363sec^{-1}[/tex]
To solve the ln we have to apply e in both sides of the equation, so we have:
[tex]e^{ln(k_{2})}=e^{2.363}sec^{-1}[/tex]
[tex]k_{2}=6.423sec^{-1}[/tex]
Answer:
10.37 s-1
Explanation:
From
k= A e-^Ea/RT
Given
Ea=22KJmol-1
T=16.6+273= 289.6K
k= 1.868 sec-1
R= 8.314JK-1mol-1
A???
Hence
A= k/e^-Ea/RT
A= 1.868/e-(22000/8.314×289.6)
A= 1.7 ×10^4
Substitute into to find k at 84.8°C
k= 1.7×10^4× e-(22000/8.314×357.8)
k=10.37 s-1
