2
$\begingroup$

In chemical kinetics , we learn that in a reversible reaction ,both the forward reaction and reverse reaction occur at the same time. At equilibrium , the rates are the same. For example:Production of ammonia NH3 .During the production of ammonia both the forward & reverse reaction is present .

$\ce{ 3H_2(g) + N_2(g) ⟶ 2NH_3(g)}$ forward reaction

$\ce{2NH_3(g) ⟶ 3H_2(g) + N_2(g)}$ reverse reaction

Now come to the Gibbs free energy.

Under standard condition(298K) , it can be calculated that the reaction is forwardly favored (forward reaction is spontaneous,and thus the decomposition of NH3 is non-spontaneous). I understand H2 and N2 will produce more NH3.My question is at this situation will NH3 decompose into H2 and N2 to some extent(though it is a non-spontaneous process)?.But chemical kinetics tells that both reaction(not necessarily same rate) have to be present to create equilibrium in future .

$\endgroup$
  • $\begingroup$ It does not matter in this case that the forward reaction is energetically favourable because the activation energy (energy barrier) between reactants and product is so large that the forward reaction rate is effectively zero at room temperature. i.e. Transition state energy is huge wrt. RT. The barrier is also extremely large for the reverse reaction, therefore reaction rate is also effectively zero at room temperature. Equilibrium exists but rates effectively zero. At high temperature and pressure, and most importantly with a catalyst, the reaction will occur, see Haber process. $\endgroup$ – porphyrin Dec 6 at 11:48
4
$\begingroup$

Under standard conditions (100kPa, 0°C), nitrogen and hydrogen do not react with each other in any measurable way. It follows logically that also ammonia cannot decompose (as it would to a small extent in an equillibrium reaction), because otherwise it would decompose completely with time, although it is stable.

In equillibrium, the forward and backward reaction rates must be identical and finite.

$\endgroup$
  • $\begingroup$ Here standard condition is 25 degree Celcius $\endgroup$ – thephysicist Sep 17 '17 at 19:22
  • $\begingroup$ @thephysicist No. What you mean is NTP. en.wikipedia.org/wiki/… $\endgroup$ – Karl Sep 17 '17 at 19:27
  • $\begingroup$ okay . I edited my question $\endgroup$ – thephysicist Sep 17 '17 at 19:33
  • $\begingroup$ @thephysicist 298 or 273 make absolutely no difference. $\endgroup$ – Karl Sep 17 '17 at 19:35
2
$\begingroup$

In equilibrium, as you say, forward and backward reactions have same velocity of conversion (that is equilibrium). Out of equilibrium, you could probably see that the non-spontaneous direction of the process has a very small rate.

Microscopically, a molecule of product can collide another molecule absorbing its energy, and then -because of the high energy- be converted in a reactant. Following your example, a very small quantity of $\ce{NH3}$ produced will be converted into a reactant. In short: it is improbable but occurs.

$\endgroup$
  • 2
    $\begingroup$ It's so improbable it will not occur. Not in a measureable way, not in any reasonable amount of time. $\endgroup$ – Karl Sep 17 '17 at 22:48
  • $\begingroup$ @Karl well, I don't think so. $k_{inv}$ will depend on activation energy of the inverse process.., I mean, it will be produced..at least that's what our theories implies.. :) $\endgroup$ – santimirandarp Sep 18 '17 at 7:04
  • $\begingroup$ You are misapplying the theories. To get to ammonia is a multistep reaction, and each of the preliminary steps has a terribly low propability at STP, and they have to happen in very close proximity etc. This is not like winning the lottery, it's like winning the lottery on your birthday, three years in a row. And even then you only get a single molecule. Perhaps. $\endgroup$ – Karl Sep 18 '17 at 18:15

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.