1
$\begingroup$

If this is the case, can you give a brief explanation why is this always the case?

$\endgroup$
3
  • $\begingroup$ If you push a boulder up a hill, does the potential energy increase or decrease? $\endgroup$ – Bob Mar 13 '17 at 0:12
  • $\begingroup$ That actually depends on what you mean by "energy." If you are talking about free energy, then that would always be the case. $\endgroup$ – Zhe Mar 13 '17 at 1:05
  • $\begingroup$ The reaction can still occur to some extent but the equilibrium condition favours the reactants, whereas if the reaction is exothermic the equilibrium lies on the product side. $\endgroup$ – porphyrin Mar 13 '17 at 9:01
0
$\begingroup$

This question brings up the broader point of what it means for a reaction to be favored. Your question describes thermodynamic favorability and that is just dependent on the relative energy of the products and the reactants. An analogy to this is to think of energy levels as literal heights subjected to gravity. If you have a ball on the ground (reactants) and you want to put it on a shelf (products), you have to put in energy. But, once it is in the product state, gravity will want to drive the ball back to the reactant state because it is lower in energy.

Another type of favorability is kinetic favorability; this relates to the rate at which the forward and reverse reactions can occur. There are cases where a thermodynamically favorable reaction doesn't occur (or rather occurs at an incredibly slow rate) due it being kinetically unfavorable. An example of this is diamond and graphite; it would be thermodynamically favorable for diamond to convert to graphite, however the reaction to do so is so kinetically unfavorable that you are unlikely to see it occur.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.