In the previous section, we saw how equilibrium constant is related to Gibbs energy. In this section, we will see the factors affecting equilibrium
• We know that, equilibrium is attained when:
The concentrations of reactants and products remain constant
◼ Such a constant concentration is achieved when two conditions are satisfied:
Condition 1:
♦ The 'number of molecules of reactants' used up during a time interval t in the forward reaction
♦ is equal to
♦ The ,number of molecules of reactants' produced during the same time interval t in the backward reaction
Condition 2:
♦ The 'number of molecules of products' produced during a time interval t in the forward reaction
♦ is equal to
♦ The 'number of molecules of products' used up during the same time interval t in the backward reaction
• The ‘number of molecules in a time interval t’ is simply, the rate of a reaction
• So we can combine the two conditions as:
Equilibrium is achieved when the rate of forward reaction becomes equal to the rate of the backward reaction
• We have already seen the above facts in the previous sections
• In the present section, we are going to analyze the situations where the equilibrium gets affected
• What do we mean by saying ‘equilibrium gets affected’ ?
The answer can be written in just two steps:
(i) It simply means that, the existing equilibrium suffers a ‘change’
(ii) So a 'new equilibrium' will be reached
• The manufacturer will try to obtain the ‘new equilibrium’ in such a way that the two conditions are satisfied:
♦ The concentrations of the products is maximum
♦ The concentrations of the reactants is minimum
• By adjusting the ‘factors which affect equilibrium’ the manufacturer will be able to satisfy the two conditions
• So it is very important that we clearly understand those factors
• Consider any one factor. It may be concentration, temperature, pressure etc.,
• When we consider any one of those factors, a question immediately arises:
If we bring about a change in that factor, in which direction will the reaction proceed?
• For example:
♦ If the temperature is increased, will the reaction proceed in the forward direction, leading to ‘more products?
♦ Or, will the reaction proceed in the backward direction, leading to more reactants?
Le Chatelier’s principle helps us to find answers to such questions
The principle can be stated in 3 steps:
1. Consider a system in equilibrium
• Note down the factors like: concentrations, temperature, pressure etc.,
2. The system is in equilibrium because, those factors are remaining constant
• If a ‘change’ is brought about in any one of those factors, the equilibrium will change
3. The change will be in such a way that:
The effect of the ‘change’ gets reduced
• Based on Le Chatelier's principle, we can now analyze various factors
Factor 1: Concentration
This can be written in 2 steps:
1. Concentrations of various reactants and products is an important factor that keeps the system in equilibrium
• This is clear from the expression that we use to find Kc: $\mathbf\small{\rm{K_c=\frac{[C]^c [D]^d}{[A]^a [B]^b}}}$
2. Now we want to know this:
How will the equilibrium be affected, if we change the concentrations of one or more reactants or products ?
• The answer is given by Le Chatelier’s principle. It can be written in two steps:
(i) When a reactant or product is added into the system, we are increasing the concentration
♦ In such a situation, the reaction will proceed in that direction which consumes the excess concentration
(ii) When a reactant or product is removed from the system, we are decreasing the concentration
♦ In such a situation, the reaction will proceed in that direction which replenishes the lost concentration
Let us apply this information to a practical situation. It can be written in steps:
1. Consider the reaction: H2(g) + I2(g) ⇌ 2HI(g)
• Let it be at equilibrium
2. Suppose that some H2 is added to the system
• Then the equilibrium is disturbed
3. The system will try to reduce the effect of ‘excess H2’
• For that, the system will try to use up the excess H2
4. This can be achieved by moving in the forward direction
• That is., more H2 will react with I2 to produce more HI
• When this happens:
♦ The concentration of H2 decreases from the ‘new total value’
♦ The concentration of I2 decreases from the equilibrium value
♦ The concentration of HI increase from the equilibrium value
5. Consider the expression for Kc: $\mathbf\small{\rm{K_c=\frac{[HI]^2}{[H_2][I_2]}}}$
• When some H2 is added in step (2) above, [H2] increases
♦ So Kc decreases
6. In order to bring Kc back, [H2] must be reduced
• For that, some H2 combines with I2 to produce more HI
• Due to this combination, [I2] decreases and [HI] increases
7. But [H2] does not decrease all the way back to the original value
• This fact is shown pictorially in fig.7.6 below. It can be explained in 6 steps
 |
| Fig.7.6 |
(i) In the graph
♦ Time is plotted along the x-axis
♦ Concentration is plotted along the y-axis
(ii) Up to time t1 (first vertical dashed line), the system is in equilibrium
♦ The [H2] up to t1 can be denoted as [H2]AB
♦ This [H2]AB is constant. This is indicated by the horizontal nature of line AB
(iii) The excess H2 is added at t1
♦ Due to the addition of H2, the [H2] rises to point C
♦ The [H2] in this situation can be denoted as [H2]C
(iv) The reaction between excess H2 and already existing I2 begins at t1 itself
(v) So from t1 on wards, the decrease in [H2] begins
♦ That is., the concentration begins to decrease from [H2]C
✰ This is indicated by the yellow curve CD
♦ It decreases to a new value
♦ This new value is [H2]DE
(vi) We see that:
♦ [H2]DE is lesser than [H2]C
♦ [H2]DE is greater than [H2]AB
♦ That is., the [H2] never returns to the original value [H2]AB
8. In a similar way, we can analyze the concentrations of I2 and HI also
• [I2]PQ represents the initial concentration of I2
♦ Due to the reaction with H2, [I2]PQ decreases to a new value: [I2]RS
• [HI]UV represents the initial concentration of HI
♦ Due to the increased production, [HI]UV increases to a new value: [HI]WX
9. It is clear that:
♦ The final value [H2]DE is different from the initial value [H2]AB
♦ The final value [I2]RS is different from the initial value [I2]PQ
♦ The final value [HI]WX is different from the initial value [HI]UV
• The final values are all different from their respective initial values
• But yet, the differences occur in such a way as to restore the equilibrium constant back to the original Kc
• That is: $\mathbf\small{\rm{K_c=\frac{[HI]_{UV}^2}{[H_2]_{AB}[I_2]_{PQ}}=\frac{[HI]_{WX}^2}{[H_2]_{DE}[I_2]_{RS}}}}$
• The above example helps us to obtain a clear idea about how the equilibrium is affected when the concentration of one of the reactants/products is increased
• We can cause a decrease also. That is., we can remove one or more of the reactants/products from the system
• For example, if we remove some HI from the above system at equilibrium, the system will try to replenish the lost concentration by producing more HI
♦ That is., the system will move in the forward direction
♦ So more H2 and I2 will be consumed
• This method is commonly adopted by manufacturers
♦ They can thus convert more and more reactants into products
• Let us see two such examples:
◼ In the reaction N2(g) + 3H2(g) ⇌ 2NH3(g), the product NH3 is liquefied and removed
♦ This causes the system to convert more and more N2 and H2 into NH3
♦ NH3 (ammonia) is a commercially important substance
◼ In the reaction CaCO3(s) ⇌ CaO(s) + CO2(g), the product CO2 is continuously removed
♦ This causes the system to convert more and more CaCO3 into CaO and CO2
♦ CaO (calcium oxide) is a commercially important substance
Thus we have seen how 'change in concentration' affects equilibrium.
In the next section, we will see the effect of 'change in pressure'
Previous
Contents
Next
Copyright©2021 Higher secondary chemistry.blogspot.com
No comments:
Post a Comment