Question

Consider the reaction data. A ⟶ products A⟶products T ( K ) T (K) k ( s − 1 ) k (s−1) 225 225 0.391 0.391 525 525 0.700 0.700 What two points should be plotted to graphically determine the activation energy of this reaction? To avoid rounding errors, use at least three significant figures in all values.

Answer 1

Answer:

The activation energy for the reaction is, 1.90682  KJ/mol.

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at 225 K = $0.391 s^{-1}$

$K_2$ = rate constant at 525 K = $0.700 s^{-1}$

$Ea$ = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

$T_1=225 K, T_2=525 K$

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

$\log (\frac{0.700 s^{-1}}{0.391 s^{-1}})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{225 K}-\frac{1}{525 K}]$

$Ea=1,906.82 J/mole=1.90682 KJ/mol$

Therefore, the activation energy for the reaction is, 1.90682  KJ/mol.