The rate constants of a reaction at $500 k and 700 k are 0.02 s^{= 1}$ and $0.07 s^{- 1}$ respectively. Calculate the values of $E_{a}$ and $A$ .

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Solution

Given,
$k_{1} = 0.02, T_{1} = 500 K k_{2} = 0.07, T_{2} = 700 K$

Calculation of $E_{a}$

We know that,

$log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

Put given values in above formula, we get,

$log \frac{0.07}{0.02}$

$= (\frac{E_{a}}{2.303 \times 8.314 J K^{- 1} m o l^{- 1}}) [\frac{700 - 500}{700 \times 500}]$

$0.544 = E_{a} \times 5.714 \times 10^{- 4} / 19.15$

$\Rightarrow E_{a} = 0.544 \times \frac{19.15}{5.714} \times 10^{+ 4} = 18230.8 J$

$= 18.2 k J$

Calculation of $A$ Using Arrhenius equation

Since $k = A e^{- E a / R T}$

$0.02 = A e^{- 18230.8 / 8.314 \times 500}$

$\Rightarrow A = 0.02 / 0.012 = 1.67$

Final answer: $E_{a} = 18230.8 A = - 1.67$

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The rate constants of a reaction at 500k and 700k are 0.02s=1 and 0.07 s−1 respectively. Calculate the values of Ea and A.

The rate constants of a reaction at $500 k and 700 k are 0.02 s^{= 1}$ and $0.07 s^{- 1}$ respectively. Calculate the values of $E_{a}$ and $A$ .