Question

At $407 \, K$ the rate constant of a chemical reaction is $9.5\times 10^{- 5}s^{- 1}$ and at $420 \, K$ , the rate constant is $1.9\times 10^{- 4}s^{- 1}$ . The frequency factor of the reaction is $x\times 10^{5}s^{- 1}$ .The value of 'x' is.
Report your answer by rounding it up to nearest whole number.

Answer 1

The Arrhenius equation is,
$\log _{10} \frac{\frac{k_{2}}{k_{1}}}{k_{1}}=\frac{E_{a}}{2.303 \times R}\left[\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right]$
Given $k _{1}=9.5 \times 10^{-5} s ^{-1} ; k _{2}=1.9 \times 10^{-3} s ^{-1}$
$R =8.314\, J\,mol ^{-1} K ^{-1}$
$T_{1}=407\, K$ and $T_{2}=420\, K$
Substituting the values in Arrhenius equation.
$\log _{10} \frac{1.9 \times 10^{-4}}{9.5 \times 10^{-5}}=\frac{E_{a}}{2.303 \times 8.314}\left[\frac{420-407}{420 \times 407}\right]$
$E_{a}=75782.3\, J\, mol ^{-1}$
Applying now $\log k_{1}=\log A-\frac{E_{a}}{2.303 R T_{1}}$
$\log 9.5 \times 10^{-5}=\log A-\frac{75782.3}{2.303 \times 8.314 \times 407}$
$\log \frac{\text { A }}{9.5 \times 10^{-5}}=\frac{75782.3}{2.303 \times 8.314 \times 407}=9.7246$
So, $A=5.0 \times 10^{5} s ^{-1}$