Question

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

Answer 1

Explanation:

The given data is as follows.

   $K_{1} = 2.57 M^{-1}s^{-1}$ ,     $T_{1}$ = 701 K

   $K_{2}$ = ?

   $T_{2}$ = 525 K

  $E_{a} = 150 kJ/ mol \times \frac{1000 J}{1 kJ}$

             = 150000 J / mol

Now, we will calculate rate constant as follows.

      $ln[\frac{K_{2}}{K_{1}}] = (\frac{E_{a}}{R}) \times [(\frac{1}{T_{1}}) - (\frac{1}{T_{2}})]$

where,       R = 8.314 J per mol K

Putting the given values into the above formula as follows.

      $ln[\frac{K_{2}}{K_{1}}] = (\frac{E_{a}}{R}) \times [(\frac{1}{T_{1}}) - (\frac{1}{T_{2}})]$

   $ln [\frac{K_{2}}{2.57}] = \frac{150000 J/mol}{8.314 J/K mol} \times [(\frac{1}{701}) - (\frac{1}{525})]$

       $ln [\frac{K_{2}}{2.57}]$ = -8.628

    $\frac{K_{2}}{2.57}$ = anti ln[-8.628]

    ${K_{2}}{2.57}$ = 0.000179

    $K_{2} = 0.000179 \times 2.57$

    $K_{2}$ = 0.00046

or,            = $4.6 \times 10^{-4} L/mol s$

Thus, we can conclude that the rate constant at 525 K is $4.6 \times 10^{-4} L/mol s$.