The reaction quotient (Q) is a measure used to determine the direction in which a chemical reaction will proceed to reach equilibrium. It is calculated using the same expression as the equilibrium constant (K), but with the current concentrations or partial pressures of the reactants and products at any given moment, rather than at equilibrium.
Syllabus reference R2.3.5
Reactivity 2.3.5 - The reaction quotient, Q, is calculated using the equilibrium expression with non-equilibrium concentrations of reactants and products. (HL)
- Calculate the reaction quotient Q from the concentrations of reactants and products at a particular time, and determine the direction in which the reaction will proceed to reach equilibrium.
The reaction quotient
Mathematically, for a general reaction aA + bB ⇌ cC + dD, the reaction quotient is given by Q = [C]^c[D]^d / [A]^a[B]^b. By comparing Q to K, one can predict the direction of the reaction: if Q < K, the reaction will proceed forward to form more products; if Q > K, the reaction will proceed in reverse to form more reactants; and if Q = K, the system is already at equilibrium.
Example:
In the Haber process the equilibrium reaction is written as:
N2(g) + 3H2(g) ⇋ 2NH3(g)
When the system is at equilibrium at a given set of temperature and pressure conditions, the value of kc may be determined.
kc = [NH3]2/[N2][H2]3
Before the equilibrium is established, or at ANY OTHER GIVEN TIME, the value of the reaction quotient equals:
Q = [NH3]2/[N2][H2]3
When the reaction quotient is equal to kc, the system has arrived at equilibrium and no further change can occurs in the concentrations of reactants or products, unless the conditions are changed.
When the conditions are changed, the value of Qc no longer equals kc and the system moves to re-establish the equilibrium.
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Worked examples
Q235-01 Consider the endothermic
reaction below:
5CO(g) + I2O5(g) ⇋ 5CO2(g) + I2(g)
According to Le Chatelier's principle, which change would result in an increase in the amount of CO2?
- Increasing the temperature
- Decreasing the temperature
- Increasing the pressure
- Decreasing the pressure
Answer
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As it is an endothermic reaction in the forward direction an increase
in temperature pushes the equilibrium to the side of the products. There
is no change in the number of moles and so the equilibrium is unaffected
by pressure change.
The correct response is A.
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Q235-02 The compounds N
2O
4
and NO
2 produce an equilibrium mixture according to the equation below:
N2O4 (g) ⇋ 2NO2(g) ........ ∆H = +57.2 kJmol-1
An increase in the equilibrium concentration of NO2 can be produced
by increasing which of the following factors:
- I. Pressure
- II. Temperature
Answer
|
Increasing the temperature pushes the reaction in the direction of
ENDOthermic change, ie to the right hand side.
Increasing the pressure pushes the reaction in the direction of the
fewer number of moles of gas, ie to the left hand side in this case.
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Q235-03 For the following system
at equilibrium, which change will shift the position of equilibrium to the right?
CH3COOH(l) + C2H5OH(l) ⇋
CH3COOC2H5(l) + H2O(l) .....
∆H < 0
- Adding a catalyst
- Increasing the pressure
- Removing water
- Increasing the temperature
Answer
|
Remove from one side of an equilibrium and you pull the reaction towards
that same side. If you remove one of the reacting components of the
REVERSE reaction, then the rate of the reverse reaction becomes less
than that of the forward reaction. The system then reacts faster in
the forward direction, making more of the right hand side.
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Q235-04 Which change will shift
the position of equilibrium to the right in this reaction?
N2(g) + 3H2(g) ⇋
2NH3(g) ... ∆H = - 92 kJ
- Increasing the temperature.
- Decreasing the pressure.
- Adding a catalyst.
- Removing the ammonia from the equilibrium mixture.
Answer
|
The sign of ∆H is negative therefore it is exothermic in the
forward direction. Increasing temperature favours the direction of ENDOthermic
change ie to the left hand side in this case. II does not go
to the right hand side.
Decreasing the pressure forces the equilibrium to move towards the
side of greater moles of gas to re-establish the equilibrium concentrations.
Catalysts do NOT affect equilibria.
Removing ammonia disturbs the equilibrium concentrations and the reaction
responds by making more ammonia to re-establish the equilibrium concentration
proportions - this is the correct response.
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Q235-05
2H2O(l) ⇋
H3O+(aq) + OH-(aq)
The equilibrium constant for the reaction above is 1.0 x 10-14
at 25ºC and 2.1 x 10-14 at 35ºC. What can be concluded
from this information?
- [H3O+] decreases as the temperature is raised
- [H3O+] is greater than [OH-]
at 35ºC
- Water is a stronger electrolyte at 25ºC
- The ionization of water is endothermic.
Answer
|
Kc has a larger value at the higher temperature. Thus the temperature
increases and the proportion of products increases. Temperature increase
favours the direction of endothermic change therefore this reaction
is endothermic in the forward direction.
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Q235-06 Which change increases
the amount of NH
4+ in the below reaction?
NH3(g) + H2O(l) ⇋
NH4+(aq) + OH-(aq)....
∆H > 0
- Decreasing the temperature
- Decreasing the pressure
- Removing water
- Adding an acid
Answer
|
The sign of ∆H is positive (it is greater than 0) therefore it
is endothermic in the forward direction. Increasing temperature favours
the direction of ENDOthermic change ie to the right hand side in this
case. Decreasing the temperature will move the equilibrium position
to the left hand side.
Decreasing the pressure forces the equilibrium to move towards the
side of greater moles of gas to re-establish the equilibrium concentrations,
in this case the left (notice the state symbols)
Removing water disturbs the equilibrium concentrations and the reaction
responds by making more water to re-establish the equilibrium concentration
proportions. This pulls the reaction to the left to restore the value
of Kc.
Adding acid reacts with the OH- ions from the right hand side disturbing
the equilibrium proportions that are then re-established by the reaction
moving to the right hand side making more NH4+.
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Q235-07 Consider the equilibrium
reaction:
4NH3(g) + 3O2(g) ⇋ 2N2(g) + 6H2O(g) ΔH = -1268 kJ
Which change will cause the reaction to shift to the right?
- Increase the temperature
- Decrease the volume of the container.
- Add a catalyst to speed up the reaction.
- Remove the gaseous water by allowing it to react and be absorbed by KOH.
Answer
|
The reaction is exothermic in the forward direction, thus increasing
the temperature drives the equilibrium in the reverse direction.
Decreasing the volume increases the pressure. The equilibrium has 7
moles of gas on the left and 8 moles of gas on the right, hence increasing
the pressure drives the reaction to the left hand side.
Adding a catalyst has no effect on the position of equilibrium.
Removing the gasous water disturbs the equilibrium and the system must
respond by making more water, i.e. by moving
to the right to re-establish the equilibrium proportions.
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Q235-08 Which changes will shift
the position of equilibrium to the right in the following reaction?
2CO2(g) ⇋
2CO(g) + O2(g)
- I. adding a catalyst
- II. decreasing the oxygen concentration
- III. increasing the volume of the cylinder
- I and II only
- I and III only
- II and III only
- I, II and III
Answer
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I - catalysts have no effect on equilibrium.
II - decreasing the oxygen concentration pulls the reaction to the
right to re-establish the equilibrium proportions.
II - increasing the volume decreases the pressure. There are two moles
of gas on the left and three moles of gas on the right, hence a decrease
in pressure moves the reaction towards the side of the most moles (the
right) to re-establish the equilibrium proportions.
Corrct response II and III only
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Q235-09 The reaction below is
an important step in the production of sulfuric acid. An increase in which of
the following, will increase the ratio of SO
3(g) to SO
2(g)
at equilibrium?
2SO2(g) + O2(g) ⇋
2SO3(g) ΔH = -197 kJ
- Pressure only
- Temperature
- Both pressure and temperature
- Neither pressure nor temperature
Answer
|
Increase pressure moves the reaction in the directin of fewer moles,
in this case to the right. This increases the ratio of SO3(g)
to SO2(g) at equilibrium. Increased temperature moves the
position of equilibrium towards the side of endothermic change. in this
case the left. This reduces the ratio of SO3(g) to SO2(g)
at equilibrium.
Correct response A
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Q235-10 The equation for a reaction
used in the manufacture of nitric acid is:
4NH3(g) + 5O2(g) ⇋
4NO(g) + 6H2O(g) ΔH =
-900 kJ
Which changes occur when the temperature of the reaction is increased?
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Position of equilibrium
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Value of Kc
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A.
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shifts to the left
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increases
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B.
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shifts to the left
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decreases
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C.
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shifts to the right
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increases
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D.
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shifts to the right
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decreases
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Answer
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When the temperature is increased the position of equilibrium moves
towards the direction of endothermic change, in this case it shifts
to the left. The value of Kc decreases as there is less product (right
hand side) and more reactants (left hand side) at equilibrium.
Correct response B
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