In the previous section, we saw how change in concentration affects equilibrium. In this section, we will see three more factors which affect equilibrium
Factor 2: Pressure
• This can be written in 8 steps:
1. Changes in pressure affects equilibrium of reactions in which at least one reactant or product is a gas
2. Suppose that, the system is inside a cylinder fitted with a piston
• Let P1 and V1 be the pressure and volume at equilibrium
• If we push down the piston to make the volume $\mathbf\small{\rm{\frac{V_1}{2}}}$, what will be the change in pressure?
3. We know that PV = a constant
• So P1V1 = P2V2
⇒ $\mathbf\small{\rm{p_1V_1=p_2 \times \frac{V_1}{2}}}$
⇒ P2 = 2P1
• That means, pressure becomes double the initial value
4. We know that:
• P1 and P2 are the total pressures
♦ Total pressure of the system is the sum of the partial pressures
♦ Partial pressure of a gas depends on it’s number of moles
• Now, total pressure has increased to a new value P2
♦ So partial pressures will also increase
5. According to Le Chatelier’s principle, the system will try to reduce the effect of this increased pressure
• That means, the pressure must decrease from P2
• For that, the reaction will proceed in that direction which will result in a lesser number of moles
6. An example:
• In the reaction CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g), there are four gaseous moles on the reactant side, but only two gaseous moles in the product side
• So, when the pressure is increased, the reaction moves in the forward direction
7. Another example:
• In the reaction C(s) + CO2(g) ⇌ 2CO(g), there is only one gaseous mole on the reactant side, but two gaseous moles in the product side
• So, when the pressure is increased, the reaction moves in the backward direction
8. Note that, with regard to 'change in pressure':
• We need not consider solids and liquids because, a change in pressure will not affect their volumes
• If volumes are not affected, concentrations are also not affected
Effect of adding an inert gas
• This can be written in 2 steps:
1. Consider a system in equilibrium
• If an inert gas is added into the system, it will not take part in the reaction
• So the concentrations of reactants and products will not change
2. Thus we can write:
In a system at equilibrium, if we add an inert gas with out changing the pressure, the equilibrium will not be affected
Factor 3: Temperature
• Recall the two factors that we have already seen:
(i) When concentrations are changed, the Kc does not change. The concentrations re-adjust to bring back Kc to the original value
(ii) When pressure is changed, the Kc(or Kp) does not change. The concentrations re-adjust to bring back Kc(or Kp) to the original value
• But the third factor is different. When temperature changes, Kc(or Kp) attains a new value
• This can be explained using an example. It can be written in 5 steps:
1. Consider the reaction: H2(g) + I2(g) ⇌ 2HI(g); ΔH = -10.4 kJ mol-1
• It is clear that, the forward reaction is exothermic. 10.4 kJ mol-1 will be released
2. At 500 K, the Kp of this reaction is: 160
• If the temperature is increased to 700 K, the Kp becomes 54
◼ We see that, when temperature is increased, the equilibrium constant decreases
the following steps from (3) to (5) will give the reason:
3. According to Le Chatelier’s principle, the reaction will move in that direction which reduces the effect
• In our present case, temperature is increased. So the system tries to reduce the temperature
4. For that, the reaction will move in that direction which absorbs heat
♦ The forward reaction in (1) is exothermic which releases heat
♦ The backward reaction is endothermic which absorbs heat
• So the backward reaction is preferred
• So more reactants will be produced and more products will be consumed
5. That means, numerator (in the expression $\mathbf\small{\rm{K_p=\frac{(p_C)^c (p_D)^d}{(p_A)^a (p_B)^b}}}$) decreases
• Thus the equilibrium constant decreases
Factor 4: Catalyst
This can be written in 4 steps:
1. A catalyst is added to speed up the reaction
2. But it speeds up both forward and backward reactions
• Thus the equilibrium is attained in a smaller time
3. Since the rates of both forward and backward reactions are increased, the concentrations of reactants and products remain the same
4. So the value of equilibrium constant does not change
Now we will see some solved examples:
Solved example 7.36
Does the number of moles of reaction products increase, decrease or remain
same when each of the following equilibria is subjected to a decrease in pressure
by increasing the volume?
(a) PCl5 (g) ⇌ PCl3(g) + Cl2(g)
(b) CaO(s) + CO2(g) ⇌ CaCO3(s)
(c) 3Fe(s) + 4H2O(g) ⇌ Fe3O4(s) + 4H2(g)
Solution:
Part (a):
• Total number of gaseous moles in the reactant side = 1
• Total number of gaseous moles in the product side = 2
◼ When pressure decreases, the system will try to increase the number of gaseous moles
♦ So the reaction will proceed in the forward direction
♦ Thus the number of moles of reaction products will increase
Part (b):
• Total number of gaseous moles in the reactant side = 2
• Total number of gaseous moles in the product side = 1
◼ When pressure decreases, the system will try to increase the number of gaseous moles
♦ So the reaction will proceed in the backward direction
♦ Thus the number of moles of reaction products will decrease
Part (c):
• Total number of gaseous moles in the reactant side = 4
• Total number of gaseous moles in the product side = 4
◼ When pressure decreases, the system will try to increase the number of gaseous moles. But in this case, the number of gaseous moles is same for both reactants and products
♦ So the decrease in pressure has no effect in the direction
♦ Thus the number of moles of reaction products will remain the same
Solved example 7.37
Which of the following reactions will get affected by increasing the pressure?
Also, mention whether change will cause the reaction to go into forward or
backward direction.
(i) COCl2(g) ⇌ CO(g) + Cl2(g)
(ii) CH4(g) + 2S2(g) ⇌ CS2(g) + 2H2S(g)
(iii) CO2(g) + C(s) ⇌ 2CO(g)
(iv) 2H2(g) + CO(g) ⇌ CH3OH(g)
(v) CaCO3(s) ⇌ CaO(s) + CO2(g)
(vi) 4NH3(g) + 5O2(g) ⇌ 4NO(g) + 6H2O(g)
Solution:
Part (i):
• Total number of gaseous moles in the reactant side = 1
• Total number of gaseous moles in the product side = 2
◼ When pressure increases, the system will try to decrease the number of gaseous moles
♦ So the reaction will proceed in the backward direction
Part (ii):
• Total number of gaseous moles in the reactant side = 3
• Total number of gaseous moles in the product side = 3
◼ When pressure increases, the system will try to decrease the number of gaseous moles. But in this case, the number of gaseous moles is same for both reactants and products
♦ So the increase in pressure has no effect in the direction
Part (iii):
• Total number of gaseous moles in the reactant side = 1
• Total number of gaseous moles in the product side = 2
◼ When pressure increases, the system will try to decrease the number of gaseous moles
♦ So the reaction will proceed in the backward direction
Part (iv):
• Total number of gaseous moles in the reactant side = 3
• Total number of gaseous moles in the product side = 1
◼ When pressure increases, the system will try to decrease the number of gaseous moles
♦ So the reaction will proceed in the forward direction
Part (v):
• Total number of gaseous moles in the reactant side = 0
• Total number of gaseous moles in the product side = 1
◼ When pressure increases, the system will try to decrease the number of gaseous moles
♦ So the reaction will proceed in the backward direction
Part (vi):
• Total number of gaseous moles in the reactant side = 9
• Total number of gaseous moles in the product side = 10
◼ When pressure increases, the system will try to decrease the number of gaseous moles
♦ So the reaction will proceed in the backward direction
Solved example 7.38
The equilibrium constant for the following reaction is 1.6 ×105 at 1024K
H2 (g) + Br2 (g) ⇌ 2HBr(g)
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a
sealed container at 1024K.
Solution:
1. At the beginning of the reaction, only HBr is present
• So we have to write the given equation in reverse order:
2HBr(g) ⇌ H2 (g) + Br2 (g)
• The equilibrium constant of this reaction will be: $\mathbf\small{\rm{\frac{1}{1.6 \times 10^5}}}$
2. At equilibrium, let x be the pressure of H2 and Br2
• Then the pressure of HBr will be (10.0 - 2x)
3. Thus the equilibrium constant will be given by: $\mathbf\small{\rm{\frac{x^2}{(10.0-2x)^2}}}$
• So we can write: $\mathbf\small{\rm{\frac{x^2}{(10.0-2x)^2}=\frac{1}{1.6 \times 10^5}}}$
4. Taking reciprocals, we get: $\mathbf\small{\rm{\frac{(10.0-2x)^2}{x^2}=\frac{1.6 \times 10^5}{1}}}$
⇒ $\mathbf\small{\rm{\frac{(10.0-2x)}{x}=\sqrt{\frac{16 \times 10^4}{1}}=400}}$
• Thus we get x = 0.0248
5. So we can write:
• At equilibrium,
♦ pH2 = pBr2 = x = 0.0248 bar
♦ pHBr = (10 - 2x) = (10 - 0.0496) = 9.95 bar
Link to a few more solved examples are given below:
In the next section, we will see the ionic equilibrium in solution
Previous
Contents
Next
Copyright©2021 Higher secondary chemistry.blogspot.com
No comments:
Post a Comment