Topic 10/11 Equilibrium 1/2 Flashcards
Define the term ‘dynamic equilibrium’.
The rate of the forward reaction is equal to the rate of the reverse reaction.
(Hence, the concentrations of reactants and products do not change.)
Give an essential condition for an equilibrium mixture.
● Equilibrium occurs in a closed system where reactants and products cannot escape.
OR
● Macroscopic properties do not change with time.
State Le Chatelier’s principle.
If a system at equilibrium is altered, the position of equilibrium moves in the direction that reduces the effect of the initial change.
What effect would increasing the temperature have on the position of equilibrium?
CH 4(g) + H 2 O (g) ⇌ CO (g) + 3H 2(g)
[ΔH°= +210 kJ mol-1]
The equilibrium position shifts to
the right.
(This is because the forward reaction is endothermic
shown by the +ve ∆H value.)
What effect would increasing the pressure have on the position of equilibrium?
CH 4(g) + H 2 O (g) ⇌ CO (g) + 3H 2(g)
[ΔH°= +210 kJ mol-1]
The equilibrium position shifts to the left.
(This is because the rhs of the equation has more moles of gas than the LHS. The side with fewer moles is favoured.)
Suggest and explain why an industrial chemist may use a high pressure for the production of hydrogen from:
CH 4(g) + H 2 O (g) ⇌ CO (g) + 3H 2(g)
[ΔH°= +210 kJ mol-1]
- The high pressure increases the collision frequency, increasing the rate of reaction.
- This is a compromise pressure between an economically viable rate of reaction and a slightly lower yield of hydrogen from the equilibrium reaction
What effect does a catalyst have on the position of
equilibrium?
No effect.
(This is because a catalyst affects rate of forward and reverse reactions equally rsulting in no overall effect.)
What condition affects the value of K c ?
Temperature only
For the reaction below, deduce an expression for K c .
2[A] + 3[B] + [C] ⇌ [D]+ 4[E]
’ [D][E] ^ 4
K c = ———————-
[A] ^ 2 [B] ^ 3 [C]
Deduce units for the value of K c when:
‘ [D][E] ^ 4
K c = ———————-
[A] ^ 2 [B] ^ 3 [C]
mol ^ -1 dm ^ 3
What type of system is K c relevant for?
Homogeneous systems in equilibrium.
What does K c being greater of lesser than 1 suggest for the position of equilibrium?
Greater than 1 = shifted to the right
Lesser than 1 = shifted to the left
What effect does decreasing the temperature in an endothermic reaction have on K c ?
K c decreases
(The endothermic reaction isn’t favoured so equilibrium shifts
to the left.)
What effect does increasing the temperature in an endothermic reaction have on K c ?
K c increases
(The endothermic reaction is favoured so equilibrium shifts to
the right.)
What effect does decreasing the temperature in an exothermic reaction have on K c ?
K c increases
(The exothermic reaction is favoured so equilibrium shifts to
the right.)
What effect does increasing the temperature in an exothermic reaction have on K c ?
K c decreases
(The exothermic reaction isn’t favoured so equilibrium shifts
to the left.)
What is partial pressure?
Each gas’s contribution to the total pressure
How would you calculate the partial pressure of a gas?
Partial pressure p = mole fraction x total pressure
A reaction is represented by aA (g) + bB (g) ⇌ cC (g) + dD (g), what
Kp for the system?
’ pC^c pD^d
Kp = ———————–
pA^a pB^b
How do you calculate the units for Kp?
Write out the units for the partial pressures/ concentrations in the same arrangement as the Kp/Kc equation and cancel out/multiply together.
For Kp usually in Pa, kPa, atm etc. DO NOT CHANGE UNITS
For Kc units for concentration is mol dm -3
What is the effect of increasing temperature on Kp for an endothermic reaction?
Equilibrium shifts to the right, so partial pressures of products increase, so Kp increases
What is the effect of increasing temperature on Kp/Kc for an endothermic reaction?
Equilibrium shifts to the left, so partial pressures/concentration of reactants increase, so Kp decreases
What is the effect of increasing the overall pressure on Kp for this reaction?
Pressure does not affect Kp as, if moles of gas are not the
same on each side), either top or bottom of Kp expression will
have a total pressure term that does not cancel.
What will be the kinetic effect of increasing the temperature and pressure for any reaction?
Increasing both will increase the rate of reaction as:
Temperature - many more particles have energy greater than or equal to the activation energy → more successful collisions per second
Pressure - more particles in the same volume → more successful collisions per second.
What is the effect of changing concentration or pressure or by the addition of a catalyst on the value of equilibrium constant?
There is no effect
