In the previous section, we saw the basic features of equilibrium. We also saw an activity which demonstrates the dynamic nature of equilibrium. In this section, we will see the equilibrium in chemical reactions in greater detail
• We have seen some basics about equilibrium in chemical reactions in the first section of this chapter. We saw that:
♦ In the forward reaction, the reactants react together to give the products
♦ In the backward reaction, the products react together to give the reactants
♦ Both forward and backward reactions occur simultaneously
• In the initial stages of the reaction:
♦ The rate of forward reaction
♦ is greater than
♦ The rate of backward reaction
• At equilibrium:
♦ The rate of forward reaction
♦ is equal to
♦ The rate of backward reaction
We can learn some more details with the help of a graph. It can be written in 16 steps:
1. Consider the reaction: A + B ⇌ C + D
2. As this reaction proceeds, we record the following items at regular time intervals:
♦ Quantity of reactant A
♦ Quantity of reactant B
♦ Quantity of product C
♦ Quantity of product D
3. We are recording the quantities at regular time intervals
• So, if the time interval is 2 minutes, we take the quantities when the stop watch shows 0, 2, 4, 6 . . . , minutes
4. Now we plot a graph
♦ Time is plotted along the x-axis
♦ Quantity (or concentration) is plotted along the y-axis
5. The points are plotted according to the following 4 steps:
(i) We will be having a vertical line through the 0 min mark on the x-axis
(This vertical line is the y-axis)
• On that vertical line, we mark two points:
♦ Concentration of reactant A at 0 min
♦ Concentration of product C at 0 min
(Concentration of C at 0 min will be zero because, no products are formed at 0 min)
(ii) We will be having a vertical line through the 2 min mark on the x-axis
• On that vertical line, we mark two points:
♦ Concentration of reactant A at 2 min
♦ Concentration of product C at 2 min
(iii) We will be having a vertical line through the 4 min mark on the x-axis
• On that vertical line, we mark two points:
♦ Concentration of reactant A at 4 min
♦ Concentration of product C at 4 min
- - -
- - -
(iv) In this way, all the readings related to A and C are plotted
♦ Draw a smooth red curve through the points related to A
♦ Draw a smooth green curve through the points related to C
• The result graph is shown in fig.7.3 below:
 |
| Fig.7.3 |
6. We can plot a graph using the readings related to B and D also. The graph will be similar. There is no need to plot a separate graph for B and D
♦ So along the red curve, we write: A or B
♦ Also along the green curve, we write: C or D
• The following steps from (7) to (16) give us a detailed analysis of the curves
7. Consider the red curve
• In the initial stages, it is a falling curve. That means, in the initial stages, the concentration of reactant A goes on decreasing
• This is expected. We know that, more and more A is continuously being used up. As time passes by, there will be lesser and lesser A available
8. But in the later stages, the red curve becomes horizontal
• This is not expected. We expect the concentration of A to become lesser and lesser and finally become zero
• That is., we expect the red curve to fall continuously and finally meet the x-axis
• The horizontal nature of the curve indicates that, even if we wait for a very long time, there will be some quantity of A remaining
9. So a question arises:
Why is A not completely used up?
• The answer is that, the products C and D reacts together to give back small quantities of A and B. This is the backward reaction
10. Next, consider the green curve
• In the initial stages, it is a rising
curve. That means, in the initial stages, the concentration of product C
goes on increasing
• This is expected. We know that, more and more C is
continuously being formed. As time passes by, there will be more and
more C available
11. But in the later stages, the green curve becomes horizontal
• This is not expected. We want the concentration of C to become larger and larger and reach a maximum
•
The horizontal nature of the curve indicates that, even if we wait for a
very long time, there will be no further increase in the quantity of C
12. So a question arises:
Why is C not increasing further?
•
The answer is that, some of the newly formed C is being used up in the backward reaction
13. Consider the magenta vertical dashed line
• It is drawn just at the beginning of the horizontal portions
• So we have an important point:
The point of intersection of the magenta vertical dashed line and the x-axis
• This point gives us the time at which equilibrium is attained
14. Once equilibrium is attained,
♦ The rate of forward reaction
♦ will be equal to
♦ The rate of backward reaction
(i) That means:
♦ The number of reactant molecules used up in unit time
♦ will be equal to
♦ The number of reactant molecules formed in unit time
(ii) That also means:
♦ The number of product molecules formed in unit time
♦ will be equal to
♦ The number of product molecules used up in unit time
◼ From (i) it is clear that, once equilibrium is attained, the concentration of reactants remain unchanged
◼ From (ii) it is clear that, once equilibrium is attained, the concentration of products remain unchanged
15. Note that, at equilibrium,
♦ the concentration of products
♦ is greater than
♦ the concentration of reactants
• We know this because:
♦ The horizontal portion of green curve
♦ is at a higher level than
♦ The horizontal portion of red curve
◼ So the manufacturer need not worry too much. Even if products are being converted back into reactants, at equilibrium, he has a higher concentration of products
• Also, by making suitable changes to temperature and pressure, he can change the equilibrium in such a way that, more products are obtained
♦ We will see those details in later sections
16. It is important to note that, the same equilibrium can be reached by starting the reaction with products C and D, with no presence of A and B
• This is because, as soon as C and D are brought together, they will start to react with each other
Dynamic equilibrium in chemical reactions
• Consider the reaction: N2(g) + 3H2(g) ⇌ 2NH3(g)
• After some time, the reaction will attain equilibrium
• Even after attaining equilibrium, both forward and backward reactions will be still going on
• Haber proved this using isotopes. His experiment can be explained in 15 steps:
1. Take H2 and N2 in a container
• Maintain a high temperature and pressure inside the container. This is because, the forward reaction is favored under high temperature and pressure
2. Measure the quantities of N2, H2 and NH3 at regular intervals
3. Plot the Concentration vs Time graph. It will be as shown in fig.7.4(a) below:
 |
| Fig.7.4 |
◼ We see that:
• The red curve is high above in the graph
♦ That means, we are taking a greater quantity of H2
♦ This is because, we need 3 mol H2 for every one mol N2
◼ Also we see these:
• Initially the concentration of H2 decreases
♦ This is because, H2 is being used up in the forward reaction
• Initially the concentration of N2 also decreases
♦ This is because, N2 is also being used up in the forward reaction
• Initially the concentration of NH3 increases
♦ This is because, NH3 is being formed in the forward reaction
4. But after some time, all the three graphs become horizontal
• This indicates equilibrium
5. Next, we want N2 to react with D2 to form ND3
♦ That is., we want the reaction: N2(g) + 3D2(g) ⇌ 2ND3(g)
♦ [D2 (Deuterium) is the isotope of H2]
• For that, take D2 and N2 in a second container
• The quantities must be same as those taken in the first container
♦ Quantity of D2 must be same as that of H2 in the first container
♦ Quantity of N2 must be same as that of N2 in the first container
• Maintain a high temperature and
pressure inside the container. This is because, the forward reaction is
favored under high temperature and pressure
6. Measure the quantities of N2, H2 and NH3 at regular intervals
7. Plot the Concentration vs Time graph. It will be as shown in fig.7.4(b) above
◼ We see that:
• The graph is fig.b is similar to the graph in fig.a
• That means, at equilibrium,
♦ Quantity of D2 remaining will be same as that of H2 in the first container
♦ Quantity of N2 remaining will be same as that of N2 in the first container
♦ Quantity of ND3 formed will be same as that of NH3 in the first container
8. The position of the vertical magenta dashed line is also same in both the graphs (a) and (b)
• That means, the 'time at which equilibrium is attained' is same for both cases
9. Now we have two mixtures
• In the first container, we have a mixture of N2, H2 and NH3
♦ This mixture is at a temperature T and pressure P
♦ This mixture is at equilibrium
• In the second container, we have a mixture of N2, D2 and ND3
♦ This mixture is also at the same temperature T and pressure P
♦ This mixture is also at equilibrium
10. The two mixtures are mixed together in a third container
♦ The temperature and pressure in this third container are the same T and P
◼ We would expect the following composition in the third container:
• Quantity of H2:
♦ Same as that in container 1
• Quantity of N2:
♦ Double of that in container 1
• Quantity of D2:
♦ Same as that in container 2
• Quantity of NH3:
♦ Same as that in container 1
• Quantity of ND3:
♦ Same as that in container 2
11. If we analyze the composition of the mixture in the third container, by ordinary methods, we will see that, the results mentioned in (10) are true
12. But if we analyze using a mass spectrometer, we will see that:
• The mixture in the third container has the following three compounds also:
♦ NH2D, NHD2 and HD
13. We would not expect to find these three compounds because:
• In the first container, there was no D2
♦ So in the first container, there will not be any NH2D, NHD2 or HD
• In the second container, there was no H2
♦ So in the second container, there will not be any NH2D, NHD2 or HD
14. So clearly, the three compounds are formed after the inter mixing in the third container
• But in the third container, every thing is in equilibrium. There is no change in composition. Then how did the new compounds form?
• The answer can be written in 3 steps:
(i) Even at equilibrium, the forward and backward reactions are taking place simultaneously
(ii) So in the new forward reaction, N2 reacts with both H2 and D2 to give
NH2D and NHD2
(iii) Also, in the new backward reaction, both NH2D and NHD2 get decomposed
15. If the equilibrium was static, there would have been no mixing of isotopes in this way
◼ So the formation of NH2D, NHD2 and HD is clear evidence for the dynamic nature of equilibrium
• Let us analyze another reaction: H2(g) + I2(g) ⇌ 2HI(g)
• It is the formation of Hydrogen iodide from H2 and I2
• We can analyze this reaction with the help of the graph in fig.7.5(a) below:
 |
| Fig.7.5 |
• The analysis can be written in 9 steps:
1. In the initial stages, concentration of H2 and I2 (red curve) decreases
• This is because, they are being use up in the forward reaction
2. Also in the initial stages, concentration of HI (green curve) increases
• This is because, it is being formed in the forward reaction
3. We would expect the red curve to fall continuously and reach the x-axis
• But in reality, it stops falling and becomes horizontal
• This is because, H2 and I2 gets newly formed due to the backward reaction
4. We would expect the green curve to rise continuously and reach a maximum
• But in reality, it stops rising and becomes horizontal
• This is because, HI begins to get decomposed due to the backward reaction
5. As time passes by,
♦ the rate of forward reaction
♦ becomes equal to
♦ the rate of backward reaction
(i) That means:
♦ Number of molecules of H2 and I2 used up in forward reaction
♦ becomes equal to
♦ Number of molecules of H2 and I2 newly formed in backward reaction
(ii) That also means:
♦ Number of molecules of HI being decomposed in backward reaction
♦ becomes equal to
♦ Number of molecules of HI newly formed in forward reaction
◼ From (i) it is clear that, once equilibrium is attained, the concentration of reactants remain unchanged
◼ From (ii) it is clear that, once equilibrium is attained, the concentration of products remain unchanged
• This condition is indicated by the horizontal nature of the red and green curves
6. We know that, one mol H2 reacts with one mol I2 to give two mol HI
• So, if we take n mol H2 and n mol I2, we will get 2n mol HI
7. Suppose that, the curves in fig.7.5(a) is obtained when n mol each of H2 and I2 are mixed
• Theoretically, we must receive 2n mol HI
• But the reaction will not proceed up to the formation of 2n mol HI
• This is because, some of the newly formed HI will be decomposed back to H2 and I2
8. Suppose that, initially we take 2n mol HI
• Then this HI will decompose to give n mol H2 and n mol I2
• This is shown graphically in fig. 7.5(b) above
♦ For better comparison, time is plotted from right to left
• Theoretically, we must receive n mol H2 and n mol I2
• But the decomposition will not proceed to that extent
• This is because, some of the newly formed H2 and I2 will combine to give back HI
• So in this decomposition reaction also, an equilibrium will be attained
• This equilibrium is indicated by the horizontal portions in fig.b
9. Let us note an interesting point. It can be written in 3 steps:
(i) Number of moles:
♦ In step (7), we took n mol each of H2 and I2
♦ In step (8), we took 2n mol HI
• So the number of moles is the same in two cases
(ii) If number of moles is same, the number of atoms will also be the same. Then the curves will coincide
• That is:
♦ The green curve in fig.b
♦ will be at the same level as
♦ The green curve in fig.a
• Also:
♦ The red curve in fig.b
♦ will be at the same level as
♦ The red curve in fig.a
(iii) Since the two curves in each graph coincide, we can combine them. This is shown in fig.c
• By making suitable changes to temperature and pressure, the manufacturer can force the magenta dashed line to be more towards the right
• Then at equilibrium, there will be a greater concentration of HI
• In later sections, we will see how this can be achieved
•
In the next section, we will see equilibrium constant in chemical reactions
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