K c is the equilibrium constant expressed in terms of molar concentrations at a given temperature. For the reaction aA + bB ⇌ cC + dD: K c = [C] c [D] d / [A] a [B] b

Which species are excluded from a K c expression and why?

Solids are excluded from K c expressions. The concentration of a solid is constant and does not change, so it is incorporated into the value of K c rather than written separately.

Write the K c expression for: N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g) K c = ..........

K c = [NH 3 ] 2 / ([N 2 ][H 2 ] 3 ) Products raised to their stoichiometric coefficients appear in the numerator; reactants in the denominator.

How do you determine the units of K c ?

Substitute mol dm -3 for each concentration term and cancel. The overall units depend on the difference between the total powers in the numerator and denominator. K c may be unitless if all powers cancel.

What does a large value of K c indicate about the equilibrium position?

A large K c value indicates the equilibrium position lies towards the products (right-hand side). The concentration of products is much greater than the concentration of reactants at equilibrium.

In a K c calculation, given moles at equilibrium in a known volume, what must you do before substituting into the expression?

Convert moles to concentrations using: concentration = moles / volume (in dm 3 ) All concentrations must be in mol dm -3 before substituting into the K c expression.

True or False? Adding a reactant to an equilibrium mixture changes the value of K c .

False. K c only changes with temperature . Adding a reactant shifts the equilibrium position but the system adjusts to restore the same value of K c .

For the reaction: Ag + (aq) + Fe 2+ (aq) ⇌ Ag (s) + Fe 3+ (aq) K c = ..........

K c = [Fe 3+ (aq)] / ([Fe 2+ (aq)][Ag + (aq)]) Ag (s) is a solid and is excluded from the expression.

An equilibrium mixture contains 0.235 mol CH 3 COOH and 0.035 mol C 2 H 5 OH and 0.182 mol each of CH 3 COOC 2 H 5 and H 2 O in 500 cm 3 . Calculate K c .

[CH 3 COOH] = 0.470 mol dm -3 | [C 2 H 5 OH] = 0.070 mol dm -3 | [CH 3 COOC 2 H 5 ] = 0.364 mol dm -3 | [H 2 O] = 0.364 mol dm -3 K c = (0.364 × 0.364) / (0.070 × 0.470) = 4.03 (no units)

What does it mean if a reactant is zero order ?

K p is the equilibrium constant expressed in terms of the partial pressures of gaseous reactants and products. For aA (g) + bB (g) ⇌ cC (g) + dD (g): K p = (pC) c (pD) d / (pA) a (pB) b

Which species are excluded from K p expressions?

Solids and liquids are excluded from K p expressions in heterogeneous reactions. Only gaseous species contribute to partial pressures and are included in the expression.

For CaCO 3 (s) ⇌ CaO (s) + CO 2 (g) K p = ..........

K p = pCO 2 CaCO 3 and CaO are both solids and are excluded from the K p expression. Only the gaseous CO 2 contributes.

What is dynamic equilibrium ?

partial pressure = mole fraction × total pressure Mole fraction = moles of that gas / total moles of all gases.

Calculate K p for 2SO 2 (g) + O 2 (g) ⇌ 2SO 3 (g) given: pSO 2 = 1.0 × 10 6 Pa | pO 2 = 7.0 × 10 6 Pa | pSO 3 = 8.0 × 10 6 Pa

K p = (pSO 3 ) 2 / [(pSO 2 ) 2 × pO 2 ] K p = (8.0 × 10 6 ) 2 / [(1.0 × 10 6 ) 2 × (7.0 × 10 6 )] K p = 9.1 × 10 -6 Pa -1

What does Le Chatelier's principle state?

Substitute Pa for each partial pressure term and cancel. The units depend on the net difference in powers of pressure between numerator and denominator. K p may be unitless if all powers cancel.

True or False? The pH at the half equivalence point of a weak acid–strong base titration equals the pK a of the weak acid.

False. Neither K p nor K c changes with pressure. Pressure changes shift the position of equilibrium but the system adjusts to maintain the same value of K p or K c .

Write the K p expression for: N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g) K p = ..........

K p = (pNH 3 ) 2 / (pN 2 × (pH 2 ) 3 ) The units are Pa -2 as there are 2 powers in the numerator and 4 in the denominator.

What is the equilibrium constant, K c ?

Temperature is the only factor that changes the value of K c or K p . Changes in concentration, pressure and the addition of a catalyst all shift the position of equilibrium but leave the value of K unchanged.

What does a large value of K c indicate about the equilibrium position?

K c decreases . Increasing temperature shifts the equilibrium to the left (towards reactants), decreasing the ratio of [products] to [reactants] and therefore reducing K c .

How does a catalyst affect the value of K c ?

K c increases . Increasing temperature shifts the equilibrium to the right (towards products), increasing the ratio of [products] to [reactants] and therefore raising K c .

True or False? A more positive standard electrode potential means a species is a stronger reducing agent.

False. A catalyst speeds up both forward and reverse reactions equally, allowing equilibrium to be reached faster but having no effect on the position of equilibrium or the value of K p .

Does increasing pressure change the value of K p ?

No. Increasing pressure does not change K p . The equilibrium position shifts to reduce the effect of the increased pressure, restoring the same value of K p at the new position.

For the equilibrium AB (aq) + CD (aq) ⇌ AC (aq) + BD (aq), ΔH = +180 kJ mol -1 . Which change would increase K c ?

Increasing temperature would increase K c . The forward reaction is endothermic, so increasing temperature shifts the equilibrium towards products, raising K c .

A catalyst allows an equilibrium to be reached .......... but does not change ..........

A catalyst allows an equilibrium to be reached faster but does not change the position of the equilibrium or the value of K .\nIt lowers activation energy equally for both forward and reverse reactions.

For the endothermic reaction 2HI (g) ⇌ H 2 (g) + I 2 (g), what happens to [HI] when temperature increases?

[HI] decreases . Increasing temperature shifts the endothermic equilibrium to the right , consuming HI and producing more H 2 and I 2 .

For the reaction 2SO 2 (g) + O 2 (g) ⇌ 2SO 3 (g), ΔH is negative. What happens to K c when temperature increases?

K c decreases . Increasing temperature shifts the exothermic equilibrium to the left, decreasing [SO 3 ] and increasing [SO 2 ] and [O 2 ], which reduces the value of K c .

For an endothermic reaction, increasing temperature causes K c to ..........

For an endothermic reaction, increasing temperature causes K c to increase . The equilibrium shifts towards products, increasing [products] and decreasing [reactants], raising the ratio and therefore K c .

True or False? A reaction with a negative ΔG is spontaneous.

True. Temperature is the only factor that permanently changes K p (and K c ). Other factors such as pressure or concentration shift the equilibrium position but not the value of K.

For the equilibrium 2A (g) + B (g) ⇌ 2C (g), ΔH = +6.5 kJ mol -1 . What change would increase K p ?

Increasing temperature would increase K p . The reaction is endothermic, so higher temperature shifts the equilibrium towards products (C), increasing the ratio of products to reactants and raising K p .

K c is highest at low temperatures . For exothermic reactions, lower temperature shifts the equilibrium towards products, giving a greater ratio of [products] to [reactants] and a higher value of K c .

True or False? Entropy always increases in a chemical reaction.

False. For an exothermic reaction, increasing temperature shifts the equilibrium to the left (towards reactants), reducing the ratio of [products] to [reactants] and decreasing K c .

What is the relationship between the total entropy change and the equilibrium constant ?

ΔS total = R ln K Where R is the gas constant and K is the equilibrium constant (K c or K p ). Rearranging: K = e (ΔS total /R)

The equilibrium constant can be calculated using: K = ..........

K = e (ΔS total /R) Where ΔS total is the total entropy change and R is the gas constant (8.314 J K -1 mol -1 ).

What is the partial pressure of a gas in a mixture?

At equilibrium, the total entropy change for both the forward and reverse reactions is equal.\nThe system has reached the maximum entropy position. Any movement towards reactants or products would decrease total entropy.

True or False? The enthalpy of hydration of an ion becomes more exothermic as the ion's charge density increases.

True. For a reaction to reach equilibrium, ΔS total must be positive in both directions from the extremes towards the equilibrium position. This means both directions are spontaneous.

What is the mole fraction of a gas?

A very large K indicates the equilibrium position lies far towards the products (right-hand side).\nConversely, a very small K indicates the equilibrium position lies far towards the reactants (left-hand side).

The entropy change of the surroundings during a reaction is given by which expression?

ΔS surroundings = −ΔH / T Where ΔH is the enthalpy change of the reaction and T is the absolute temperature in kelvin.

True or False? Dissolving an ionic solid is always an exothermic process.

True. At equilibrium, moving towards either pure reactants or pure products would decrease total entropy (making ΔS total negative). Only the equilibrium mixture gives maximum total entropy.

Is the decomposition of CaCO 3 spontaneous at 293 K given ΔH = +177.9 kJ mol -1 and ΔS sys = +160.4 J K -1 mol -1 ?

ΔS surr = −177900 / 293 = −607.2 J K -1 mol -1 ΔS total = +160.4 + (−607.2) = −446.8 J K -1 mol -1 ΔS total is negative , so the reaction is not spontaneous at 293 K.

International A Level (IAL)ChemistryEdexcelFlashcards4: Rates, Equilibria & Further Organic ChemistryChemical Equilibria

Chemical Equilibria (Edexcel International A Level (IAL) Chemistry): Flashcards

Exam code: YCH11

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FrontEquilibrium Constant, Kc
Define K_c

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Define K_c
K_c is the equilibrium constant expressed in terms of molar concentrations at a given temperature.
For the reaction aA + bB ⇌ cC + dD:
K_c = [C]^c[D]^d / [A]^a[B]^b
Which species are excluded from a K_c expression and why?
Solids are excluded from K_c expressions.
The concentration of a solid is constant and does not change, so it is incorporated into the value of K_c rather than written separately.
Write the K_c expression for:
N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)
K_c = ..........
K_c = [NH₃]² / ([N₂][H₂]³)
Products raised to their stoichiometric coefficients appear in the numerator; reactants in the denominator.
How do you determine the units of K_c?
Substitute mol dm^-3 for each concentration term and cancel.
The overall units depend on the difference between the total powers in the numerator and denominator. K_c may be unitless if all powers cancel.
What does a large value of K_c indicate about the equilibrium position?
A large K_c value indicates the equilibrium position lies towards the products (right-hand side).
The concentration of products is much greater than the concentration of reactants at equilibrium.
In a K_c calculation, given moles at equilibrium in a known volume, what must you do before substituting into the expression?
Convert moles to concentrations using:
concentration = moles / volume (in dm³)
All concentrations must be in mol dm^-3 before substituting into the K_c expression.
True or False?
Adding a reactant to an equilibrium mixture changes the value of K_c.
False.
K_c only changes with temperature. Adding a reactant shifts the equilibrium position but the system adjusts to restore the same value of K_c.
For the reaction:
Ag⁺ (aq) + Fe²⁺ (aq) ⇌ Ag (s) + Fe³⁺ (aq)
K_c = ..........
K_c = [Fe³⁺(aq)] / ([Fe²⁺(aq)][Ag⁺(aq)])
Ag (s) is a solid and is excluded from the expression.
An equilibrium mixture contains 0.235 mol CH₃COOH and 0.035 mol C₂H₅OH and 0.182 mol each of CH₃COOC₂H₅ and H₂O in 500 cm³. Calculate K_c.
[CH₃COOH] = 0.470 mol dm^-3 | [C₂H₅OH] = 0.070 mol dm^-3 | [CH₃COOC₂H₅] = 0.364 mol dm^-3 | [H₂O] = 0.364 mol dm^-3
K_c = (0.364 × 0.364) / (0.070 × 0.470) = 4.03 (no units)
What does it mean if a reactant is zero order?
K_p is the equilibrium constant expressed in terms of the partial pressures of gaseous reactants and products.
For aA (g) + bB (g) ⇌ cC (g) + dD (g):
K_p = (pC)^c(pD)^d / (pA)^a(pB)^b
Which species are excluded from K_p expressions?
Solids and liquids are excluded from K_p expressions in heterogeneous reactions.
Only gaseous species contribute to partial pressures and are included in the expression.
For CaCO₃ (s) ⇌ CaO (s) + CO₂ (g)
K_p = ..........
K_p = pCO₂
CaCO₃ and CaO are both solids and are excluded from the K_p expression. Only the gaseous CO₂ contributes.
What is dynamic equilibrium?
partial pressure = mole fraction × total pressure
Mole fraction = moles of that gas / total moles of all gases.
Calculate K_p for 2SO₂ (g) + O₂ (g) ⇌ 2SO₃ (g) given:
pSO₂ = 1.0 × 10⁶ Pa | pO₂ = 7.0 × 10⁶ Pa | pSO₃ = 8.0 × 10⁶ Pa
K_p = (pSO₃)² / [(pSO₂)² × pO₂]
K_p = (8.0 × 10⁶)² / [(1.0 × 10⁶)² × (7.0 × 10⁶)]
K_p = 9.1 × 10^-6 Pa^-1
What does Le Chatelier's principle state?
Substitute Pa for each partial pressure term and cancel.
The units depend on the net difference in powers of pressure between numerator and denominator. K_p may be unitless if all powers cancel.
True or False?
The pH at the half equivalence point of a weak acid–strong base titration equals the pK_a of the weak acid.
False.
Neither K_p nor K_c changes with pressure. Pressure changes shift the position of equilibrium but the system adjusts to maintain the same value of K_p or K_c.
Write the K_p expression for:
N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)
K_p = ..........
K_p = (pNH₃)² / (pN₂ × (pH₂)³)
The units are Pa^-2 as there are 2 powers in the numerator and 4 in the denominator.
What is the equilibrium constant, K_c?
Temperature is the only factor that changes the value of K_c or K_p.
Changes in concentration, pressure and the addition of a catalyst all shift the position of equilibrium but leave the value of K unchanged.
What does a large value of K_c indicate about the equilibrium position?
K_c decreases.
Increasing temperature shifts the equilibrium to the left (towards reactants), decreasing the ratio of [products] to [reactants] and therefore reducing K_c.
How does a catalyst affect the value of K_c?
K_c increases.
Increasing temperature shifts the equilibrium to the right (towards products), increasing the ratio of [products] to [reactants] and therefore raising K_c.
True or False?
A more positive standard electrode potential means a species is a stronger reducing agent.
False.
A catalyst speeds up both forward and reverse reactions equally, allowing equilibrium to be reached faster but having no effect on the position of equilibrium or the value of K_p.
Does increasing pressure change the value of K_p?
No. Increasing pressure does not change K_p.
The equilibrium position shifts to reduce the effect of the increased pressure, restoring the same value of K_p at the new position.
For the equilibrium AB (aq) + CD (aq) ⇌ AC (aq) + BD (aq), ΔH = +180 kJ mol^-1. Which change would increase K_c?
Increasing temperature would increase K_c.
The forward reaction is endothermic, so increasing temperature shifts the equilibrium towards products, raising K_c.
A catalyst allows an equilibrium to be reached .......... but does not change ..........
A catalyst allows an equilibrium to be reached faster but does not change the position of the equilibrium or the value of K.\nIt lowers activation energy equally for both forward and reverse reactions.
For the endothermic reaction 2HI (g) ⇌ H₂ (g) + I₂ (g), what happens to [HI] when temperature increases?
[HI] decreases.
Increasing temperature shifts the endothermic equilibrium to the right, consuming HI and producing more H₂ and I₂.
For the reaction 2SO₂ (g) + O₂ (g) ⇌ 2SO₃ (g), ΔH is negative. What happens to K_c when temperature increases?
K_c decreases.
Increasing temperature shifts the exothermic equilibrium to the left, decreasing [SO₃] and increasing [SO₂] and [O₂], which reduces the value of K_c.
For an endothermic reaction, increasing temperature causes K_c to ..........
For an endothermic reaction, increasing temperature causes K_c to increase.
The equilibrium shifts towards products, increasing [products] and decreasing [reactants], raising the ratio and therefore K_c.
True or False?
A reaction with a negative ΔG is spontaneous.
True.
Temperature is the only factor that permanently changes K_p (and K_c). Other factors such as pressure or concentration shift the equilibrium position but not the value of K.
For the equilibrium 2A (g) + B (g) ⇌ 2C (g), ΔH = +6.5 kJ mol^-1. What change would increase K_p?
Increasing temperature would increase K_p.
The reaction is endothermic, so higher temperature shifts the equilibrium towards products (C), increasing the ratio of products to reactants and raising K_p.
What is K_p?
K_c is highest at low temperatures.
For exothermic reactions, lower temperature shifts the equilibrium towards products, giving a greater ratio of [products] to [reactants] and a higher value of K_c.
True or False?
Entropy always increases in a chemical reaction.
False.
For an exothermic reaction, increasing temperature shifts the equilibrium to the left (towards reactants), reducing the ratio of [products] to [reactants] and decreasing K_c.
What is the relationship between the total entropy change and the equilibrium constant?
ΔS_total = R ln K
Where R is the gas constant and K is the equilibrium constant (K_c or K_p).
Rearranging: K = e^(ΔS_^total^/R)
The equilibrium constant can be calculated using:
K = ..........
K = e^(ΔS_^total^/R)
Where ΔS_total is the total entropy change and R is the gas constant (8.314 J K^-1 mol^-1).
What is the partial pressure of a gas in a mixture?
At equilibrium, the total entropy change for both the forward and reverse reactions is equal.\nThe system has reached the maximum entropy position. Any movement towards reactants or products would decrease total entropy.
True or False?
The enthalpy of hydration of an ion becomes more exothermic as the ion's charge density increases.
True.
For a reaction to reach equilibrium, ΔS_total must be positive in both directions from the extremes towards the equilibrium position. This means both directions are spontaneous.
What is the mole fraction of a gas?
A very large K indicates the equilibrium position lies far towards the products (right-hand side).\nConversely, a very small K indicates the equilibrium position lies far towards the reactants (left-hand side).
The entropy change of the surroundings during a reaction is given by which expression?
ΔS_surroundings = −ΔH / T
Where ΔH is the enthalpy change of the reaction and T is the absolute temperature in kelvin.
True or False?
Dissolving an ionic solid is always an exothermic process.
True.
At equilibrium, moving towards either pure reactants or pure products would decrease total entropy (making ΔS_total negative). Only the equilibrium mixture gives maximum total entropy.
Is the decomposition of CaCO₃ spontaneous at 293 K given ΔH = +177.9 kJ mol^-1 and ΔS_sys = +160.4 J K^-1 mol^-1?
ΔS_surr = −177900 / 293 = −607.2 J K^-1 mol^-1
ΔS_total = +160.4 + (−607.2) = −446.8 J K^-1 mol^-1
ΔS_total is negative, so the reaction is not spontaneous at 293 K.

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