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 [tex]61.29M^{-1}s^{-1}[/tex]
Explanation:
To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:
[tex]\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]K_{244^oC}[/tex] = equilibrium constant at 244°C = [tex]6.7M^{-1}s^{-1}[/tex]
[tex]K_{324^oC}[/tex] = equilibrium constant at 324°C = ?
[tex]E_a[/tex] = Activation energy = 71.0 kJ/mol = 71000 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
[tex]T_1[/tex] = initial temperature = [tex]244^oC=[273+244]K=517K[/tex]
[tex]T_2[/tex] = final temperature = [tex]324^oC=[273+324]K=597K[/tex]
Putting values in above equation, we get:
[tex]\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}[/tex]
Hence, the rate constant at 324°C is [tex]61.29M^{-1}s^{-1}[/tex]
