Question

A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, assuming they are all at the same temperature and that each starts with the same initial concentration.
E, 50 kJ/mol E,-350 kJ/mol 50 kJ/mol
A = 1.5 × 10-7 s-i A = 1.9 × 10-7 s-i A = 1.5 × 10-7 s-1
Fraction of molecules
The exponential term in the Arrhenius equation is equal to the fraction of molecules, f, with kinetic energy greater than or equal to the activation energy: f=e?Ea/(R?T). Most scientific calculators have an exfunction as the second function of the LN button.
B. A certain reaction with an activation energy of 165 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the higher temperature to f at the lower temperature?

Answer 1

Answer:

A) $E_{a} = 350KJ/mol$, $E_{a} = 50KJ/mol$, $E_{a} = 50KJ/mol$

     A = 1.5×$10^{-7}s^{-1}$, A = 1.9×$10^{-7} s^{-1}$, A=1.5×$10^{-7} s^{-1}$

B) 4.469

Explanation:

From Arrhenius equation

      $K=Ae^{\frac{E_{a} }{RT} }$

where; K = Rate of constant

            A = Pre exponetial factor

            $E_{a}$ = Activation Energy

             R = Universal constant

             T = Temperature in Kelvin

Given parameters:

$E_{a} =165KJ/mol$

$T_{1}=505K$

$T_{2}=525K$

$R=8.314JK^{-1}mol^{-1}$

taking logarithm on both sides of the equation we have;

$InK=InA-\frac{E_{a} }{RT}$

since we have the rate of two different temperature the equation can be derived as:

$In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )$

$In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1} }.(\frac{1}{505} -\frac{1}{525} )$

$In(\frac{K_{2} }{K_{1} } )$= 19846.04×7.544×$10^{-5}$ = 1.497

$\frac{K_{2} }{K_{1} }$ =$e^{1.497}$ = 4.469