Question

The activation energy (E*) for 2N2O ---> 2N2 + O2 is 250 KJ. If the k for this reaction is 0.380/M at 1001oK, what will k be at 298oK? What will the half-life be at both temperatures?

Answer 1

Step-by-step explanation:

GIven that:

The activation energy = 250 kJ

k₁ = 0.380 /M

k₂ = ???

Initial temperature
`T_1 =` 1001 K

Final temperature
`T_2 =` 298 K

Applying the equation of Arrhenius theory.

`In (k_2)/(k_1 )= (Ea)/(R)( (1)/(T_1 )- (1)/(T_2))`

where ;

R gas constant = 8.314 J/K/mol

`In (k_2)/(0.380 )= (250 * 10^3)/(8,314)( (1)/(1001 )- (1)/(298))`

`In (k_2)/(0.380 )= -70.8655`

`(k_2)/(0.380 )= e^(-70.8655)`

`(k_2)/(0.380 )= 1.67303256 * 10^(-31)`

`{k_2}= 1.67303256 * 10^(-31) * {0.380 }`

`{k_2}= 6.3575 * 10^(-32)` /M .sec

Half life:

At 1001 K.

`t_(1/2) = (In_2)/(k_1)`

`t_(1/2) = (0.693)/(0.38)`

`t_(1/2) =` 1.82368 secc

At 298 K:

`t_(1/2) = (0.693)/(6.3575 * 10^(-32))`

`t_(1/2) =1.09 * 10^(31) \ sec`