2
$\begingroup$

I was trying to answer a question from Zumdahl and Zumdahl's Chemistry textbook which asks me to show that the rate constant K is related to the forward and reverse reaction constants: $$K=\frac{K_\mathrm{f}}{K_\mathrm{r}}.$$

In answering the question it is given that: $$K_\mathrm{r}=Ae^{\frac{-(E_\mathrm{a}-\Delta G)}{RT}}$$

And something is lacking in my understanding here. As I understand it the activation energy of the reverse reaction is equal to $E_\mathrm{a}$ of the forward reaction plus the free energy. That takes you back to the transition state - but the transition state is defined as the point at which products always become reactants, so how are reactants being formed from the products?

$\endgroup$
  • 1
    $\begingroup$ Are $K_\mathrm{f/r}$ rate constants? $\endgroup$ – Martin - マーチン Jan 14 '16 at 7:56
  • $\begingroup$ You can find your answer here, chemed.chem.purdue.edu/genchem/topicreview/bp/ch22/react.html (take care with your second equation, above). $\endgroup$ – user1945827 Jan 14 '16 at 9:56
  • $\begingroup$ In the reverse reaction, the products and reactants and switched over from the forward reaction. Products are just what is made from a chemical reaction, not necessarily your desired or intended 'product'. $\endgroup$ – Spontification Jan 14 '16 at 14:20

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.