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Chemistry · Ch 8 — Physical and Chemical Equilibrium

Van't Hoff Equation

8.9

Van't Hoff Equation

The Van't Hoff equation gives the precise, quantitative relationship between temperature and the equilibrium constant K. It follows directly from two relations already familiar from thermodynamics: the standard free energy change of a reaction is related to its equilibrium constant by

ΔG0=−RTln⁡K(1)\Delta G^0 = -RT\ln K \qquad (1)

and ΔG0\Delta G^0 is also related to the standard enthalpy and entropy changes by

ΔG0=ΔH0−TΔS0(2)\Delta G^0 = \Delta H^0 - T\Delta S^0 \qquad (2)

Substituting (1) into (2) gives −RTln⁡K=ΔH0−TΔS0-RT\ln K = \Delta H^0 - T\Delta S^0, and dividing through by −RT-RT and rearranging gives

ln⁡K=−ΔH0RT+ΔS0R(3)\ln K = -\frac{\Delta H^0}{RT} + \frac{\Delta S^0}{R} \qquad (3)

Differentiating equation (3) with respect to temperature (treating ΔH0\Delta H^0 and ΔS0\Delta S^0 as constant over a small temperature range) gives the differential form of the Van't Hoff equation:

d(ln⁡K)dT=ΔH0RT2(4)\frac{d(\ln K)}{dT} = \frac{\Delta H^0}{RT^2} \qquad (4)

Integrating equation (4) between two temperatures T1T_1 and T2T_2, with their respective equilibrium constants K1K_1 and K2K_2, gives the integrated form of the Van't Hoff equation:

ln⁡K2K1=ΔH0R(1T1−1T2)or, in base-10 form,log⁡K2K1=ΔH02.303R(T2−T1T1T2)(5)\ln\frac{K_2}{K_1} = \frac{\Delta H^0}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \qquad \text{or, in base-10 form,} \qquad \log\frac{K_2}{K_1} = \frac{\Delta H^0}{2.303R}\left(\frac{T_2-T_1}{T_1T_2}\right) \qquad (5)

Worked Problem. For an equilibrium reaction, KP=0.0260K_P = 0.0260 at 25∘25^\circC, with ΔH=32.4\Delta H = 32.4 kJ mol−1^{-1}. Calculate KPK_P at 37∘37^\circC. Here T1=25+273=298T_1 = 25+273 = 298 K, T2=37+273=310T_2 = 37+273 = 310 K, ΔH=32400\Delta H = 32400 J mol−1^{-1}, R=8.314R = 8.314 J K−1^{-1}mol−1^{-1}. Then

log⁡KP2KP1=324002.303×8.314×310−298298×310=3240019.147×1292380=0.2198\log\frac{K_{P2}}{K_{P1}} = \frac{32400}{2.303\times8.314}\times\frac{310-298}{298\times310} = \frac{32400}{19.147}\times\frac{12}{92380} = 0.2198

so KP2KP1=antilog(0.2198)=1.6588\dfrac{K_{P2}}{K_{P1}} = \text{antilog}(0.2198) = 1.6588, and KP2=0.026×1.6588=0.0431K_{P2} = 0.026\times1.6588 = 0.0431. …

Misc 8.9-worked-problemProblem: Kp at a new temperature from ΔH

Worked out. For an equilibrium reaction, KP=0.0260K_P = 0.0260 at 25∘25^\circC, ΔH=32.4\Delta H = 32.4 kJ mol−1^{-1}. Calculate KPK_P at 37∘37^\circC. With T1=298T_1=298 K, T2=310T_2=310 K, R=8.314R=8.314 J K−1^{-1}mol−1^{-1}: log⁡KP2KP1=324002.303×8.314×310−298298×310=0.2198\log\dfrac{K_{P2}}{K_{P1}} = \dfrac{32400}{2.303\times8.314}\times\dfrac{310-298}{298\times310} = 0.2198, so KP2KP1=antilog(0.2198)=1.6588\dfrac{K_{P2}}{K_{P1}} = \text{antilog}(0.2198) = 1.6588, giving $K_{P2} = 0.026 \times 1.6588 = …

Misc 8.9-real-worldReal-world box: oxygen exchange between maternal and fetal blood

Worked out. In a pregnant woman, both maternal and fetal haemoglobin reversibly bind oxygen in the placenta, where the mother's and fetus's blood vessels lie close together: Hb(mother)+O2⇌HbO2(mother)Hb(\text{mother}) + O_2 \rightleftharpoons HbO_2(\text{mother}) and Hb(fetus)+O2⇌HbO2(fetus)Hb(\text{fetus}) + O_2 \rightleftharpoons HbO_2(\text{fetus}). The equilibrium constant for fetal haemoglobin's oxygenation is higher (due to its greater affinity for oxygen), so oxygen is effectively transferred from the mother's blood to the fetal haemoglobin in the pl …