5
$\begingroup$

Recently my teacher finished teaching the chapter Chemical kinetics in which I have gone through Arrhenius equation.Can we estimate the change in entropy using it? I have done my calculation, considering the reversible reaction, $\ce{A <=> B}$ which has forward and backward rate constants $\mathrm{k_f}$ and $\mathrm{k_b}$ with pre-exponential factors $\mathrm{A_f}$ , $\mathrm{A_b}$ with activation energies of $\mathrm{E_f}$ and $\mathrm{E_b}$ respectively.

According to Arrhenius Equation,

$\mathrm{k_f= A_f.e^{-E_f/RT}}$ and

$\mathrm{k_b=A_b.e^{-E_b/RT}}$

$\mathrm{ k_f/k_b = {A_f/A_b}.e^{{-E_f+E_b}/RT}}$

and $\mathrm{k_f/k_b}$ = $\mathrm{{A_f/A_b}.e^{-\Delta H^0/RT}} $

$\mathrm{k_f/k_b}$=$\mathrm{k_{eq}}$=$\mathrm{e^{-\Delta G^0/RT}}$

but, $\mathrm{\Delta G^0 = \Delta H^0 - T \Delta S^0}$

which gives $\mathrm{\Delta S^0 = Rln(A_f/A_b)}$

Will the final expression give an approximate value of change in entropy of reaction?

$\endgroup$
1
  • $\begingroup$ The thermodynamic quantities are the enthalpy and entropy of activation and are different to 'normal' $\Delta H^\text{o}$ etc. but come from the transition state approach to reactions where reactants and products are assumed to be in equilibrium. We write down directly $K=(k_BT/h)\exp(-\Delta G^{o*}/RT)$ $\endgroup$ – porphyrin Feb 25 at 8:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.