Question

For the reaction, where M is an inert gas atom (such as He or Ar),

O + NO + M --> NO₂ + M

the rate constant k exibits thte following temperature behavior:

at 300K, k = 6 x 10⁹ L²mol^-2s^-1

at 1000K, k = 3 x 10¹⁰ L²mol^-2s^-1

Determine the activation energy and the prexponential factor using the Arrhenius equation.

Answer 1

Given,  O + NO + M --> NO2 + M

at T = 300K , k= 6*109 L2mol-2s-1

at 1000K, k = 3 x 1010 L2mol-2s-1

Arehenius law states that ,

k = k₀ e ^-E/RT

Where, E = activation energy (J)

R = ideal gas constat (L-atm/mol.K)

T= temperature (K)

k = rate constant

k0 = proportionality constant

k = k₀ e ^-E/RT

Apply log on both sides,

==> ln k = ln k0 + -E/RT ln e ==> ln k = ln k0 + -E/RT

at temperature T1 , k = k1 then

==> ln k1 = ln k0 + -E/RT1 --------------> (1)

at temperature T2, k = k2 then

==> ln k2 = ln k0 + -E/RT2 -----------------> (2)

now, (1) - (2) gives,

ln k1- lnk2 = ln k0 + -E/RT1 - ( ln k0 + -E/RT2 )

ln ( k1/k2) =  -E/RT1 + E/RT2 = -E /R [ ( 1/T2 ) - (1/T1 ) ]

Therefore, ln ( k1/k2) = -E /R [ ( 1/T2 ) - (1/T1 ) ]

Now, substitute, T1 =300K, and k1 = 6 x 10⁹ L2mol-2s-1

and T2 = 1000K, and k2 = 3 x 10¹⁰ L2mol-2s-1

therefore, ln (  6 x 10⁹ / 3 x 10¹⁰) = -E /0.082[ ( 1/1000 ) - (1/300 ) ] ==> E = -56.5 L-atm

  E = -56.5 L-atm = -56.5 * 101.33 joules (J) = -5.731 *10³ J

Therefore, Activation energy , E = -5.731 *10³ J