Question

The activation energy, Ea, for the reaction 2 N2O5 (g) LaTeX: \longrightarrow ⟶ 4 NO2 (g) + O2 (g) is 22 kJ/mol. What is the rate constant at 84.8°C if the rate constant is 1.868 sec-1 at 16.6°C? Enter to 3 decimal places. LaTeX: \ln\frac{k2}{k1}=\frac{Ea}{R}\left(\frac{1}{T1}\:-\frac{1}{T2}\right)

Answer 1

The rate constant
`k_(2)` at 84.8°C is
`k_(2)=6.423sec^(-1)`

Step-by-step explanation:

Taking the Arrhenius equation we have:

`ln(k_(2))/(k_(1))=(E_(a))/(R)((1)/(T_(1))-(1)/(T_(2)))`

Where
`k_(2)` is the rate constant at a temperature 2,
`k_(1)` is the rate constant at a temperature 1;
`T_(1)` is the temperature 1,
`T_(2)` is the temperature 2, R is the gas constant and
`E_(a)` is the activation energy.

Now, we need to solve the equation for
`k_(2)`, so we have:

`ln(k_(2))/(k_(1))=(E_(a))/(R)((1)/(T_(1))-(1)/(T_(2)))`

`ln({k_(2))-ln(k_(1))=(E_(a))/(R)((1)/(T_(1))-(1)/(T_(2)))`

`ln(k_(2))=E_(a)((1)/(T_(1))-(1)/(T_(2)))+ln(k_(1))`

Then we need to make sure that we are working with the same units, so:

`R=8.314(J)/(mol.K)`

`T_(1)=16.6^(o)C+273.15=289.75K`

`T_(2)=84.4^(o)C+273.15=357.95K`

And now we can replace the values into the equation:

`ln(k_(2))=(22000(J)/(mol))/(8.314(J)/(mol.K))((1)/(289.75K)-(1)/(357.95K))+ln(1.868sec^(-1))`

`ln(k_(2))=2646.139K(0.003451K^(-1)-0.002794K^(-1))+0.6249`

`ln(k_(2))=2.363sec^(-1)`

To solve the ln we have to apply e in both sides of the equation, so we have:

`e^{ln(k_(2))}=e^(2.363)sec^(-1)`

`k_(2)=6.423sec^(-1)`

Answer 2

10.37 s-1

Step-by-step 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