Questions: The van't Hoff Equation: Temperature Dependence of Equilibrium

Question 1

A reaction is exothermic with ΔH° = –50 kJ/mol. A chemist increases the temperature of the system at equilibrium. According to the van't Hoff equation, what happens to K?

• A. K increases, because higher temperature always drives reactions forward
• B. K decreases, because higher temperature favors the endothermic (reverse) direction, shifting equilibrium toward reactants
• C. K stays constant, because K is a thermodynamic quantity independent of temperature
• D. K increases, because exothermic reactions release more heat at higher temperatures

Answer: (B)

For an exothermic reaction (ΔH° < 0), d(ln K)/dT < 0 — K decreases as temperature rises. Higher temperature favors the endothermic (reverse) direction. This is the quantitative foundation of Le Chatelier's principle: adding heat shifts an exothermic equilibrium toward reactants. Option A reflects the common misconception that temperature always drives forward reactions; in fact, temperature drives equilibrium toward the endothermic direction.

Question 2

Using the integrated van't Hoff equation, a researcher finds K₂/K₁ > 1 when T₂ > T₁. What can they conclude about the reaction?

• A. The reaction is exothermic, because K increased with temperature
• B. The reaction is endothermic, because K increased with temperature
• C. The reaction has ΔH° = 0, because K always increases with temperature
• D. The reaction is spontaneous, because K > 1 indicates products are favored

Answer: (B)

From the integrated form ln(K₂/K₁) = –(ΔH°/R)(1/T₂ – 1/T₁): if T₂ > T₁, then (1/T₂ – 1/T₁) is negative. For ln(K₂/K₁) to be positive (K increasing), we need ΔH° > 0 — the reaction must be endothermic. Endothermic reactions have K that increases with temperature because higher temperature provides energy to drive the thermodynamically uphill forward reaction.

Question 3

A van't Hoff plot of ln K vs 1/T is curved rather than linear. This indicates that the reaction's enthalpy of reaction is zero.

• T. True
• F. False

Answer: False

Curvature in a van't Hoff plot indicates that ΔH° is not constant over the temperature range — not that it is zero. Curvature arises when the heat capacities of products and reactants differ significantly (large ΔCp), so ΔH° itself changes with temperature. A zero ΔH° would produce a horizontal line (slope = 0), not a curved one. Significant curvature signals that the Kirchhoff equation correction is needed.

Question 4

The van't Hoff equation and the Arrhenius equation look similar mathematically because they describe the same physical phenomenon.

• T. True
• F. False

Answer: False

They look similar (both have RT in the denominator) but describe fundamentally different quantities. Van't Hoff governs K, the equilibrium constant — a thermodynamic quantity telling you where equilibrium lies. Arrhenius governs k, the rate constant — a kinetic quantity telling you how fast a reaction proceeds. A reaction can have large K (thermodynamically favorable) but tiny k (kinetically slow), or vice versa. These are completely independent.

Question 5

Why does the van't Hoff equation contain ΔH° rather than ΔG°, even though ΔG° directly determines K?

Model answer: ΔG° = ΔH° – TΔS°, so ΔG° already contains temperature explicitly. When you differentiate ΔG°/T with respect to T (via the Gibbs-Helmholtz equation), the TΔS° term cancels, leaving only ΔH°/T². This is why ΔH° — not ΔG° — controls how K changes with temperature. ΔG° tells you the equilibrium position at one temperature; ΔH° tells you how that position shifts as temperature changes.

This is a mathematically elegant result: the temperature sensitivity of K depends only on the reaction enthalpy, not the full Gibbs energy. It means you can predict K at any temperature using only calorimetric data (ΔH°), without needing the entropy contribution at every temperature.