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What is the Relation between Equilibrium Constant, Reaction Quotient and Gibbs Energy?

  • Last Updated : 19 Jan, 2022

A scientist was observing a reaction and at a certain point and found the concentration of reactant is equal to the concentration of product and after some time and observed color of reactant is changing, the scientist found concentration of products is greater than the concentration of reactants, for example, it can be observed, the following reaction N2O4 ⇢ 2NO2. Here initially the reaction was in colorless form and after some time when one observes the reaction was in reddish-brown in color and wants to know what is actually happening and to know this scientists invented a new concept called equilibrium constant and here are some terms we need to know about the equilibrium constant. The rate of reaction is the speed at which a chemical reaction takes place. 

Equilibrium

It is the condition of a reversible chemical reaction in which there is no net change in the amount of reactant and product. At equilibrium rate of forwarding reaction = rate of backward reaction. Equilibrium outside equilibrium is recognized by the consistency of certain measurable properties like Intensity of color, density, concentration, and Pressure. Equilibrium is dynamic in nature but not static.

Equilibrium Constant

It is the ratio between the product of the concentration of products raised to their powers and the product of the concentration of reactants raised to their powers at equilibrium.                                                                                                 

Derivation of Equilibrium Constant

According to the law of mass action, the rate of any chemical reaction at a given temperature is directly proportional to the product of the molar concentration of reactants rising to a number of moles in their products. [Note: Law of mass action is applicable to reactions that are at chemical equilibrium only For physical changes the law of mass action cannot be applied]. Suppose if the reaction is, aA + bB ⇢ cC + dD. Now let’s apply the law of mass action to forward reaction rf is proportional to [A]a[B]b (Here f indicates the rate for forward reaction).   

rf = kf[A]a[B]b 

(Here kf is the rate constant for forwarding reaction). Now let’s apply the law of mass action to backward reaction, rb is proportional to [C]c[D]d 

(Here b indicates the rate for  backward reaction)    

rb = kb[C]c[D]d      

(Here kb is rate constant for backward reaction)                   

At equilibrium: Rate of forward reaction = rate for backward reaction                               

kf[A]a[B]b = kb[C]c[D]d      

So, kf/kb = [C]c[D]d/[A]a[B]b                        

bsp;         

[Note: kf/kb = kc] and kc is equilibrium constant in terms of concentration                           

kc = (Product of concentration of product raised to powers)/(product of the concentration of reactant raised to powers).

Illustration: Find equilibrium constant for the following reaction 2X ⇢ Y + Z

Answer: kc = (product of concentration of product raised to powers)/(product of concentration of reactent raised to powers)             Kc = (Concentration of Y)(Concentration of Z)/(Concentration of X)2                                                                                                            

Illustration: If the order of the reaction is X, A ⇢ B + C. Find rate of reaction.                                 

Answer: Rate = K[A]x

Reaction Co-efficient (Q)

It is useful to predict the direction of reaction at a given time at beginning of reaction Q value is 0 and as time proceeds Q value increases. K is calculated using concentrations at equilibrium whereas Q is calculated using concentrations that may or may not be at equilibrium.  

Relation between K and Q

  • Case 1: If Q = Kc then that reaction is at equilibrium       
  • Case 2: If Q is greater than Kc then that reaction takes place in the backward direction          
  • Case 3: If Q is less than Kc then that reaction takes place in the forward direction           

Example: IF Kc for the reaction is 2 × 10-3 at a given time and if reaction proceeds as 2X ⇢ Y + Z and if concentrations of x, y, z = 2 × 10-3. Now find in which dire

n the reaction proceeds.    

Answer: Q = [Concentration of Y][Concentration of z]/[Concentration of x]2 = (2 × 10-3 × 2 × 10-3)/(2 × 10-3) = 1.                                As Q is more than Kc so the reaction proceeds in backward direction                            

Entropy

Represents the unavailability of a system’s thermal energy for conversion into mechanical work, often interpreted as the degree of disorderliness or randomness of the system.                                                                                                                        

Gibb’s free energy

It is that part of the energy that can be used to calculate the maximum reversible potential that may be performed by a thermodynamic system at a constant temperature and constant pressure total energy = Gibb’s free energy + random energy. 

H = G + TS                                                                                             

So, G = H – TS                                                                                    

Example: Consider the following equation,                                                                    

4Fe + 3O2 ⇢ 2Fe203 if S = -550 J/K and H = -1650KJ at 298k. Find ∧G for reaction                  

Answer: G = H – TS = -1650 – 298 × 550/1000 = -1486KJ  

  1. If G = 0, then the reaction is said to be at equilibrium                                                        
  2. If G is less than 0 then the reaction is said to be a spontaneous reaction                  
  3. If G is more than 0 then the reaction is said to be a nonspontaneous reaction

Relationship between equilibrium constant and standard free energy change                    

G = G° + 2.303 RTlog10Q                                                

bsp;         

Here, G is Gibb’s free energy at standard state whereas G° is Gibb’s free energy at the nonstandard state and we know that at equilibrium G = 0 and Q = K                                

So at equilibrium: G° = -2.303 RTlog10K                                                             

Note: 

  1. If is less than 0 then the reaction is said to be a spontaneous reaction            
  2. If G° is more than 0 then the reaction is said to be a nonspontaneous reaction    

Illustration: If K = 102 for a reaction X ⇢ Y + Z which is at equilibrium at 300k find G° for reaction.

Answer: G° = -2.303 RTlog10K = -2.303 × 8 × 300 × 2 × log10 = 11054.4

  • If k is greater than 1 (if the product of the concentration of products raised to powers is more than a product of the concentration of reactants raised to powers) then will be less than 0 then the reaction is said to be a spontaneous reaction. 
  • If k is less than 1 (If the product of the concentration of products raised to powers is less than a product of the concentration of reactants raised to powers) then G° will be more than 0 then the reaction is said to be a nonspontaneous reaction.

Sample Problems

Question 1: If G° for the reaction is -4.606 Kcal at equilibrium then find the equilibrium constant for the following reaction X ⇢ Y + Z at 227°C.

Solution:

G° = -2.303 RTlog10

-4.606 = -2.303RTlog10K                                                                                         &nbs

nbsp;  

K = 102

Question 2: If in a reaction, the reaction quotient Q, is greater than K in a gas phase, then what happens to the reaction?

Answer:

If Q = Kc then that reaction is at equilibrium. If Q is less than Kc then that reaction takes place in forward direction. If Q is greater than Kc it implies that concentration of products is more than concentration of reactents then the reaction takes place in backward direction till attains equilibrium.   

Question 3: For a chemical reaction that has reached dynamic equilibrium at a certain temperature, which condition is incorrect?  

Answer:

At equilibrium reaction never halts by the time of equilibrium only some part of reaction would be completed whereas at equilibrium, there will equal rate of forward reaction and reverse reaction, concentrarion of reactents and products is constant    

Question 4: In the reaction, A + 2B ⇢ 2C, if 2 moles of A, 3.0 moles of B, and 2.0 moles of C are placed in a 2L flask and equilibrium moles of C is 0.5 mol. The equilibrium constant (Kc) for the reaction is?

Solution:

 A + 2B ⇢ 2C                                                                           

Initial moles, 2, 3, 2                                                                  

sp;          

At equilibrium, (2 – x) (3 – 2x) 2 – 2x                                                                       

Given: Number of moles of C = 0.5                                                                          

So, 2 – 2x = 0.5      

This implies, x = 0.75                                          

Equilibrium moles: A = (2 – 0.75) = 1.25 mol                                                                     

B = (3 – 2 × 0.75) = 1.5 mol                                                                                      

C = 0.5mol                                                                                                  

Equilibrium Concentration: [A] = 1.25/2M; [B] = 1.5/2 M; [C] = 0.5/2 M                      

K = [(0.5/2)2 /((1.5/2)2× (1.25/2)) = 0.148                                                                                                                            

Question 5: K₁ and K₂ are equilibrium constants for reactions (i) and (ii).                              

p;       

(i) N₂(g) + O2(g) = 2NO(g) 

(ii)NO(g) = N₂(g) + 1/202(g)

Answer:

N₂ + O₂ = 2NO K1                                                                                             

NO = 1/2N₂ + 1/20₂ K₂                                                                                

The reaction is halved and reversed K2 = √1/K1  

Question 6: A change in the free energy of a system at constant temperature and pressure will be: Gsystem = Hsystem – TSsystem. At constant temperature and pressure, if a reaction has positive values of ∧H and ∧S.From this you can deduce that the reaction.  

Answer:                               

Gsystem = Hsystem – TSsystem                                                           

As both H and S are positive, and if TS is greater than Hsystem then only G will be negative And if G will be negative then the reaction will be spontaneous and this to occur TS must be high and for this to be high T must be high.      

Question 7: For an equilibrium reaction, if the value of standard Gibb’s free energy. Gº is zero, then the value of the equilibrium constant, K will equal to?  

Solution:                                                

G° = -2.303 RTlog10K                                                                                              

0 = log10K                                                                                                             

K is equal to  1


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