Question

For the second-order reaction NO( g) + O 3( g) → NO 2( g) + O 2( g), the rate constant has been measured to be 1.08 × 10 7 M –1 s –1 at 298 K and the activation energy has been measured to be 11.4 kJ/mol over the temperature range 195 K to 304 K. What is the rate constant at 207 K? ( R = 8.3145 J K –1 mol –1)

Answer 1

Answer : The rate constant at 207 K is,
`4.49* 10^6M^(-1)s^(-1)`

Explanation :

According to the Arrhenius equation,

`K=A* e^{(-Ea)/(RT)}`

or,

`\log ((K_2)/(K_1))=(Ea)/(2.303* R)[(1)/(T_1)-(1)/(T_2)]`

where,

`K_1` = rate constant at
`298K` =
`1.08* 10^7M^(-1)s^(-1)`

`K_2` = rate constant at
`207K` = ?

`Ea` = activation energy for the reaction =
`11.4kJ/mol=11400J/mol`

R = gas constant = 8.314 J/mole.K

`T_1` = initial temperature = 298 K

`T_2` = final temperature = 207 K

Now put all the given values in this formula, we get:

`\log ((K_2)/(1.08* 10^7M^(-1)s^(-1)))=(11400J/mol)/(2.303* 8.314J/mole.K)[(1)/(298K)-(1)/(207K)]`

`K_2=4.49* 106M^(-1)s^(-1)`

Therefore, the rate constant at 207 K is,
`4.49* 10^6M^(-1)s^(-1)`