Question

The reaction between nitrogen dioxide and carbon monoxide is NO2(g)+CO(g)→NO(g)+CO2(g)NO2(g)+CO(g)→NO(g)+CO2(g) The rate constant at 701 KK is measured as 2.57 M−1⋅s−1M−1⋅s−1 and that at 895 KK is measured as 567 M−1⋅s−1M−1⋅s−1. The activation energy is 1.5×102 1.5×102 kJ/molkJ/mol. Predict the rate constant at 525 KK .

Answer 1

Answer : The rate constant at 525 K is,
`0.0606M^(-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
`701K` =
`2.57M^(-1)s^(-1)`

`K_2` = rate constant at
`525K` = ?

`Ea` = activation energy for the reaction =
`1.5* 10^2kJ/mol=1.5* 10^5J/mol`

R = gas constant = 8.314 J/mole.K

`T_1` = initial temperature = 701 K

`T_2` = final temperature = 525 K

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

`\log ((K_2)/(2.57M^(-1)s^(-1)))=(1.5* 10^5J/mol)/(2.303* 8.314J/mole.K)[(1)/(701K)-(1)/(525K)]`

`K_2=0.0606M^(-1)s^(-1)`

Therefore, the rate constant at 525 K is,
`0.0606M^(-1)s^(-1)`