A reversible reaction has both the forward and the backward reaction occurring at the same time. I struggle to understand how this is logically possible. One reaction pathway is always endothermic, while the other reaction pathway is exothermic. Hence under the same conditions, how is it theoretically possible for both reaction pathways to occur at the same time?
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$\begingroup$ Being exothermic is neither required neither sufficient condition for the thermodynamic tendency to undergo the reaction. $\endgroup$– PoutnikJun 3 at 7:00
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4 Answers
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Once there is a product molecule there is a chance that it can be converted back into reactant. Of course at the start of the reaction there is v. little product so v. little reactant can be re-formed, but each individual reactant molecule still has the same chance of forming a product or a (different) chance of a product molecule reforming a reactant. At equilibrium the rate of forwards and backwards rates are equal.
The reason an endothermic reaction (positive enthalpy change) occurs can be due to entropy (making the free energy negative, $A\to 2B, \Delta S^\text{o} \gt 0$ ) but there is also a Boltzmann distribution of energies in all molecules so some amount of product will be formed even if energetically 'uphill' wrt. reactant. The activation energies will primarily determine the rate constants and so equilibrium constant and so amounts of reactant and product.
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$\begingroup$ "Once there is a product molecule there is a chance that it can be converted back into reactant" this is the basis of the principle of microscopic reversibility $\endgroup$– PAEPJun 9 at 12:10
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This is a consequence of being taught incomplete truths as absolutes and blindly believing it. At first Chemical reactions are ABSOLUTE; They happen! A bit later there are complications, possible side reactions, CO as well as CO2 is formed in combustion, ice below zero Celsius does not melt![unless pressure is applied]. The final straw! Reactions can go in reverse!! They can reach equilibrium and both forward and reverse reactions are happening simultaneously!! It is time to learn that everything we have been taught or think we know is but a snippet of the truth.
Perhaps a better way is to start with simple equilibria with a loose understanding of the kinetic theory of matter and the First and Second Laws of Thermodynamics: Molecules are in motion, Energy is conserved, Entropy generally increases and all are important. Molecules must be moving fast enough, Energy must be transferred and entropy must increase [somewhere]. At equilibrium ΔG = 0 = ΔH - TΔS. ΔH and ΔS have the same sign. In one direction the reaction is energy driven in the other it is entropy driven. When ΔH and ΔS are of different sign or one is zero the reaction cannot be reversed. Think of a thermonuclear bomb, expansion of a [ideal] gas into a vacuum, racemization of an enantiomer, respiration, photosynthesis.
A reversible reaction can reach equilibrium from either direction. The mechanism is the same for each direction. Since rates are a function of the motion of molecules and concentrations and the activation energies, what is changing at constant temperature is the concentrations of reactants in each direction. Eventually concentrations are reached where the rates in both directions are the same, equilibrium. The position is determined by the two activation energies that are incorporated into the rate constants.
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Typically when we think about thermodynamics, we're not describing an individual molecule but rather the whole system. So thinking about this just in terms of endothermic and exothermic reactions is not sufficient. It's a matter of kinetics, occasionally two molecules will collide with enough energy and in the correct orientation, such that some of the energy of the collision will be used to create the less energetically favorable molecule.
If you're trying to explain this with thermodynamics, in spontaneous reactions the energy required to overcome the activation energy of the reverse reaction comes from the reaction energy of the forward reaction. Whereas, in non-spontaneous reactions the energy required comes from the surroundings.
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1$\begingroup$ You have the idea; at equilibrium that is about what happens locally. However, without a constant source of energy to maintain T, Temperature will change altering the equilibrium conditions. This is effected by blackbody radiation. In the lab or reactor a constant temperature bath or oven is used to maintain temperature. At equilibrium the reaction rates are equal and DeltaG = zero. Ponder that for each reaction. How is it possible for an endo and exothermic reaction to have the same rate? $\endgroup$– jimchmstJun 9 at 5:41
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I don't think that both reactions happen at the same time. It goes one way and then the other. One could associate that with a prey-predator model, where the preys are the reactants and the predator the products. If there too much preys the predators will thrive (right to left reaction) and vice-versa. The reaction react to the way that is the more favorable. Two quantities that can help you study such reactions are the Equilibrium constant along side the reaction quotient. These quantities can tell you which of the two pathways is the more favorable given the current state of your system.
There will be some inaccuracies since I am not a chemist.
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3$\begingroup$ Every chemical equilibrium, that looks statical, has dynamic nature on molecular level, with opposite reactions going both ways at equal rates. $\endgroup$– PoutnikJun 3 at 6:58
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1$\begingroup$ No, as soon as a product molecule is formed there is a chance that it well reform reactant. $\endgroup$ Jun 3 at 8:34
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$\begingroup$ Not really a chance, the rate is k2[products] where Keq= k1/k2 $\endgroup$– jimchmstJun 4 at 4:18