Is the Backward Reaction at Equilibrium Non-Spontaneous or Reversible?

[himanshu121]

okay we say that reaction is spontaneous if $$\Delta G \le 0$$

consider a rxn at equilibrium

A + B $$\rightleftharpoons$$ C + D

for forward rxn be spontaneous i.e $$\Delta G \le 0$$

then by rule backward rxn will be non spontaneous

then WHY at equilibrium backward rxn taking place

 

[Monique]

Good question, maybe when the $$\Delta G$$ is small and there is a small energy barrier for the reverse?

 

[Bystander]

Originally posted by himanshu121
okay we say that reaction is spontaneous if $$\Delta G \le 0$$

Actually, "we" say no such thing; &Delta;G > 0 means that the reaction cannot occur in the direction written; &Delta;G = 0 means that no net reaction takes place in either direction; &Delta;G < 0 means that the reaction MAY occur in the direction written. The word "spontaneous" is used only in connection with discussion of whether the reaction is "reversible" or "irreversible." Follow so far?

consider a rxn at equilibrium

At equilibrium, &Delta;G = 0. The equilibrium state for a REVERSIBLY reacting system is defined as that state at which the free energy change for the forward and for the reverse reactions is zero.

Recall the definition of the equilibrium constant ---

&Delta;G⁰ = - RTlnK , where K = &Pi;a_i,prod/(&Pi;a_j,rctnt) --- in words, "The STANDARD state free energy change for the reaction equals the product of the gas constant, absolute temperature, and natural log of the equilibrium constant.

If you are interested in calculating free energies for the initial and final states of the system, the expression of interest is

&Delta;G_a = &Delta;G⁰_a + RTlna_a --- in words, the Gibbs free energy of reactant "a" at the temperature of interest, and at the activity a of interest (initial or final state) is calculated from the STANDARD state free energy of "a" and it's activity a.

A + B $$\rightleftharpoons$$ C + D

for forward rxn be spontaneous i.e $$\Delta G \le 0$$

then by rule backward rxn will be non spontaneous

then WHY at equilibrium backward rxn taking place

A "spontaneous" reaction is an "irreversible" reaction, such as the oxidation of a sugar molecule; compare this to a reversible reaction such as the formation/hydrolysis of an ester.

 

[Meninger]

At equilibrium the backward reaction takes place however at the same rate as the forward reaction.

A fundamental principle of reaction mechanism is based on the collision theory...chemical kinetics upon which the gibbs equation is based on. Think of it this way, even if there were the same amount of A B C D molecules a chemical reaction would take place more often between A and B than C and D. So you have to think about both the tendency of A and B, C and D to react as well as the number of molecules...the concentrations involved.

Remember the direction the reaction takes place is never completely one way. In other words we can only imply from delta G the overall direction of the reaction; the rate of the forward reaction in comparison to the reverse.

 

[himanshu121]

let me refine que for bystander , what happens if instead of spontaneous it is reversible. isn't $$\Delta G$$ is defined for reversible rxn , forward & backward rxn.

i would like to know how $$\Delta G$$ is related to collosion theory for conclusion
thanks for your views

 

[Meninger]

You can find the collision theory equation...I believe it is called the Arrhenius equation...under the rates of reactions chapter of a standard chemistry textbook. Collision theory is to define what happens between the before states and after states of a chemical reaction. For one a chemical reaction does not simply occur upon a collision. If three molecules in a trimolecular reaction collide and if the kinetic energy is great enough the energy will be used to form a activation complex. This complex may rearrange itself to form the product. This energy diagram can be likened onto a cyclist having to cycle up a valley to reach the other side. If the energy of products (the other side) is greater than the energy of the reactants than the reaction is endothermic. And thus the latter concept relates to enthalpy in the free energy equation delta G = delta H - T delta S.
One would have to factor in delta S which is related to the concentration partitioning of the solution.

 

[Bystander]

Originally posted by himanshu121
let me refine que for bystander , what happens if instead of spontaneous it is reversible. isn't $$\Delta G$$ is defined for reversible rxn , forward & backward rxn.

Certainly: &Delta;G for A+B reacting to form C+D is (assuming reactants and products to be in their standard states) G_C+G_D-G_A-G_B --- in words, the free energy change associated with reacting one mole of A with one mole of B to form one mole of C and one mole of D IN THEIR STANDARD STATES; the free energy change for the reverse is the negative of this quantity; at equilibrium the free energy change for reacting A and B to produce C and D (or the reverse) is zero --- the reactants and products are NOT NECESSARILY in their standard states.

i would like to know how $$\Delta G$$ is related to collosion theory for conclusion
thanks for your views

The free energy change for a reaction has NO relation to the kinetics of the reaction.