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Van’t Hoff Equation and the Temperature Dependence of Equilibrium Constants

van't hoff equation

Subject: General Chemistry Chapter: Chemical Equilibrium Topic: Relating Kc and Temperature Through the Vant Hoff Equation Answer included
van't hoff equation van’t Hoff equation integrated van’t Hoff equation equilibrium constant temperature dependence of K Kc Kp reaction enthalpy
Accepted answer Answer included

van't Hoff equation

The van't hoff equation links the equilibrium constant \(K\) to temperature through the standard reaction enthalpy \(\Delta H^\circ\), quantifying how equilibrium shifts when \(T\) changes.

Thermodynamic foundation

Equilibrium constants and standard Gibbs energy are connected by \[ \Delta G^\circ = -RT\ln K, \] while standard thermochemistry connects \(\Delta G^\circ\), \(\Delta H^\circ\), and \(\Delta S^\circ\) through \[ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ. \] Combining these relations yields a linear dependence of \(\ln K\) on \(1/T\) when \(\Delta H^\circ\) is approximately constant over the temperature range.

Linear form (constant \(\Delta H^\circ\), \(\Delta S^\circ\) over the interval): \[ \ln K = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T}\right) + \frac{\Delta S^\circ}{R}. \]

Two-temperature (integrated) form: \[ \ln\!\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right). \]

Differential form: \[ \frac{d(\ln K)}{dT}=\frac{\Delta H^\circ}{RT^2}. \]

Meaning of the sign of \(\Delta H^\circ\)

The sign of \(\Delta H^\circ\) controls the temperature trend. Endothermic reactions satisfy \(\Delta H^\circ > 0\), giving larger \(K\) at higher \(T\). Exothermic reactions satisfy \(\Delta H^\circ < 0\), giving smaller \(K\) at higher \(T\). This trend is a quantitative expression of Le Châtelier’s principle.

Standard forms and typical use

Form What it provides Common context in general chemistry Key assumptions
\(\ln K = -\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T}\right) + \dfrac{\Delta S^\circ}{R}\) Straight-line relation of \(\ln K\) vs \(1/T\); slope \(= -\Delta H^\circ/R\) Extracting \(\Delta H^\circ\) from multiple \(K\) values measured at different \(T\) \(\Delta H^\circ\) and \(\Delta S^\circ\) treated as constant over the range
\(\ln\!\left(\dfrac{K_2}{K_1}\right) = -\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)\) Estimating \(K_2\) at a new temperature from \(K_1\) and \(\Delta H^\circ\) Relating \(K_c\) or \(K_p\) to temperature changes for an equilibrium system \(\Delta H^\circ\) approximately constant; temperatures in kelvin
\(\dfrac{d(\ln K)}{dT}=\dfrac{\Delta H^\circ}{RT^2}\) Instantaneous sensitivity of \(K\) to \(T\) Conceptual interpretation of how strongly equilibrium responds near a given \(T\) Thermodynamic standard state; smooth \(K(T)\) behavior

Numerical example (two-temperature form)

A representative endothermic reaction has \(\Delta H^\circ = +50.0\ \text{kJ}\cdot\text{mol}^{-1}\). At \(T_1 = 298\ \text{K}\), the equilibrium constant is \(K_1 = 1.50\). The equilibrium constant at \(T_2 = 350\ \text{K}\) follows from the integrated van’t Hoff equation:

\[ \ln\!\left(\frac{K_2}{1.50}\right) = -\frac{50\,000\ \text{J}\cdot\text{mol}^{-1}}{8.314\ \text{J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}} \left(\frac{1}{350}-\frac{1}{298}\right). \] \[ \left(\frac{1}{350}-\frac{1}{298}\right)\ \text{K}^{-1} \approx -4.9856\times 10^{-4}\ \text{K}^{-1}, \quad \ln\!\left(\frac{K_2}{1.50}\right) \approx 2.998. \] \[ \frac{K_2}{1.50} \approx e^{2.998} \approx 20.05, \quad K_2 \approx 1.50 \times 20.05 \approx 30.1. \]

The increase in \(K\) with temperature is consistent with \(\Delta H^\circ > 0\).

Visualization: \(\ln K\) versus \(1/T\)

Van’t Hoff plot: ln K vs 1/T is linear when ΔH° is constant A chart plots ln K on the vertical axis and 1/T on the horizontal axis. Two labeled points correspond to (T = 298 K, K = 1.50) and (T = 350 K, K ≈ 30.1). A straight line connects them with negative slope. Text notes slope = -ΔH°/R. Van’t Hoff plot (constant ΔH°): ln K vs 1/T Slope = −ΔH°/R, intercept = ΔS°/R (standard-state form) 0.0027 0.0030 0.0033 0.0035 −0.8 0.8 2.4 4.0 1/T (K⁻¹) ln K T = 298 K K = 1.50 T = 350 K K ≈ 30.1 Interpretation • Leftward movement means higher temperature (smaller 1/T). • A negative slope corresponds to ΔH° > 0 (endothermic). • Magnitude of slope |slope| = |ΔH°|/R sets sensitivity to temperature. • Straight-line behavior indicates ΔH° approximately constant over the range.
A van’t Hoff plot uses \(1/T\) on the horizontal axis, so higher temperature corresponds to smaller \(1/T\) (movement to the left). Linear behavior of \(\ln K\) versus \(1/T\) indicates an approximately constant \(\Delta H^\circ\) over the interval, with slope \(-\Delta H^\circ/R\).

Common pitfalls

Temperature scale: kelvin is required in every form of the van’t Hoff equation; Celsius introduces large systematic error.

Logarithm base: natural logarithm is implied by \(\ln\); converting to \(\log_{10}\) changes constants by a factor of \(\ln 10\).

Units consistency: \(\Delta H^\circ\) in joules per mole matches \(R\) in \(\text{J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}\).

Meaning of \(K\): equilibrium constants are defined in terms of activities; concentration-based constants (\(K_c\)) and pressure-based constants (\(K_p\)) use standard conventions and approximations.

Temperature range: strong curvature in \(\ln K\) vs \(1/T\) signals non-constant \(\Delta H^\circ\) (heat-capacity effects), requiring a more detailed treatment.

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