Question

he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate constant of this reaction is at , what will the rate constant be at ?

Answer 1

The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

Answer: The rate constant at 324°C is $61.29M^{-1}s^{-1}$

Explanation:

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

$\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_{244^oC}$ = equilibrium constant at 244°C = $6.7M^{-1}s^{-1}$

$K_{324^oC}$ = equilibrium constant at 324°C = ?

$E_a$ = Activation energy = 71.0 kJ/mol = 71000 J/mol   (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = $244^oC=[273+244]K=517K$

$T_2$ = final temperature = $324^oC=[273+324]K=597K$

Putting values in above equation, we get:

$\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}$

Hence, the rate constant at 324°C is $61.29M^{-1}s^{-1}$