IB DP Chemistry - R2.3.3 Magnitude of K and temperature dependence - Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – R2.3.3 Magnitude of K and temperature dependence – Study Notes – New Syllabus
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Reactivity 2.3.3 – Magnitude, Direction, and Temperature Dependence of the Equilibrium Constant
Reactivity 2.3.3 – Magnitude, Direction, and Temperature Dependence of the Equilibrium Constant
The equilibrium constant \( K \) provides a quantitative measure of the position of equilibrium for a chemical reaction. It tells us whether the reaction favors products or reactants, how far the reaction proceeds, and how equilibrium shifts with temperature changes.
1. Interpreting the Magnitude of \( K \)
The value of \( K \) (usually \( K_c \)) reflects the ratio of concentrations of products to reactants at equilibrium:
| Magnitude of \( K \) | Relative Amounts at Equilibrium | Extent of Reaction | Position of Equilibrium |
|---|---|---|---|
| \( K \gg 1 \) | Products ≫ Reactants | Reaction goes nearly to completion | Far to the right |
| \( K > 1 \) | More Products than Reactants | Reaction favors products | Right of center |
| \( K = 1 \) | Products ≈ Reactants | Significant amounts of both | Balanced |
| \( K < 1 \) | More Reactants than Products | Reaction favors reactants | Left of center |
| \( K \ll 1 \) | Reactants ≫ Products | Very little reaction occurs | Far to the left |
2. Relationship Between \( K \) Values for Forward and Reverse Reactions
At a fixed temperature, the equilibrium constant for a reverse reaction is the reciprocal of the constant for the forward reaction.
- If forward reaction: \( aA + bB \rightleftharpoons cC + dD \) has \( K_{\text{forward}} \),
- Then reverse reaction: \( cC + dD \rightleftharpoons aA + bB \) has
\( K_{\text{reverse}} = \dfrac{1}{K_{\text{forward}}} \)
This relationship arises because the roles of products and reactants are switched in the equilibrium expression.
3. Effect of Temperature on the Value of \( K \)
The value of \( K \) depends only on temperature. Changes in pressure, concentration, or use of a catalyst do not affect the value of \( K \).
The way \( K \) responds to temperature depends on whether the reaction is endothermic or exothermic:
| Reaction Type | Increase in Temperature | Decrease in Temperature |
|---|---|---|
| Endothermic (ΔH > 0) | \( K \) increases Equilibrium shifts right | \( K \) decreases Equilibrium shifts left |
| Exothermic (ΔH < 0) | \( K \) decreases Equilibrium shifts left | \( K \) increases Equilibrium shifts right |
Summary: Increasing temperature favors the endothermic direction, while decreasing temperature favors the exothermic direction. The equilibrium constant adjusts accordingly to reflect the new position.
Note: A catalyst does not change the value of \( K \). It only helps the system reach equilibrium faster.
Example
At 400 K, the following equilibrium has a constant of \( K = 4.0 \times 10^3 \):
\( \text{A}(g) + \text{B}(g) \rightleftharpoons \text{C}(g) + \text{D}(g) \)
Predict the extent of reaction and calculate the equilibrium constant for the reverse reaction.
▶️Answer/Explanation
The large value of \( K = 4.0 \times 10^3 \) indicates the reaction favors products heavily — it lies far to the right and goes nearly to completion.
The equilibrium constant for the reverse reaction is:
\( K_{\text{reverse}} = \dfrac{1}{4.0 \times 10^3} = 2.5 \times 10^{-4} \)
This small value shows that the reverse reaction hardly proceeds — it favors reactants.
Example
The synthesis of NO2 from NO and O2 is represented by the exothermic reaction:
\( 2\text{NO}(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}_2(g) \quad \Delta H = -114 \, \text{kJ/mol} \)
At 298 K, \( K_c = 1.2 \times 10^6 \), and at 700 K, \( K_c = 1.0 \times 10^{-3} \).
Based on this information, answer the following:
- Which direction is favored at each temperature?
- What does this imply about the extent of reaction at low and high temperatures?
- How does the temperature affect the position of equilibrium and the value of \( K \)?
▶️Answer/Explanation
At 298 K, \( K_c = 1.2 \times 10^6 \), which is much greater than 1. This means the forward reaction is strongly favored — equilibrium lies far to the right, and the reaction goes nearly to completion, forming mostly NO2.
At 700 K, \( K_c = 1.0 \times 10^{-3} \), which is much less than 1. The reverse reaction is now favored — equilibrium lies to the left, and very little NO2 is formed.
Because the reaction is exothermic (\( \Delta H = -114 \, \text{kJ/mol} \)), increasing temperature adds heat, which shifts equilibrium to the left (reactants), reducing \( K \). Decreasing temperature favors the forward (exothermic) reaction and increases \( K \).
This example illustrates how:
- A large \( K \) means near-complete reaction (low T),
- A small \( K \) means limited reaction (high T),
- And for exothermic reactions, increasing T decreases \( K \) and product formation.
Example
The equilibrium reaction below occurs at 800 K:
\( \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \)
At this temperature, \( K_c = 2.5 \times 10^{-2} \).
Predict:
- Which direction the equilibrium lies at 800 K,
- The effect of increasing temperature on the position of equilibrium and value of \( K \),
- Which species appear in the \( K_c \) expression.
▶️Answer/Explanation
Direction of equilibrium at 800 K:
\( K_c = 2.5 \times 10^{-2} \) is much less than 1, meaning the reactants (CaCO3) are favored. The equilibrium lies to the left.
Temperature effect:
The decomposition of CaCO3 is endothermic (heat is required to break it apart). Increasing temperature supplies heat, shifting the equilibrium right to favor products (CaO and CO2). Therefore, the value of \( K \) will increase with temperature.
Species in \( K_c \) expression:
Solids are excluded from the equilibrium expression. Only CO2(g) appears:
\( K_c = [\text{CO}_2] \)