Reaction equilibria

From chemical potential to $K$. The equilibrium condition for a reaction is that the weighted sum of chemical potentials of products equals that of reactants, each weighted by its stoichiometric coefficient $\nu_i$ (negative for reactants, positive for products). Substituting the general form $\mu_i=\mu_i^\circ+RT\ln a_i$ — where $a_i$ is the species activity, a dimensionless "effective concentration" referenced to a standard state ($a_i=p_i/p^\circ$ for an ideal gas, $a_i\approx c_i/c^\circ$ for a dilute solute) — and rearranging gives $\Delta_rG^\circ=-RT\ln K$ with

where: $a_i$ is the activity of species $i$ and $\nu_i$ its stoichiometric coefficient. Because every $\mu_i^\circ$ on the right of $\Delta_rG^\circ=-RT\ln K$ is a standard chemical potential — fixed once $T$ is fixed, independent of how much of each species is actually present — $K$ itself is a function of $T$ alone. This is the fact CBSE states but CY11001 proves: changing pressure, volume or concentration shifts the equilibrium composition to keep $K$ satisfied, but never changes $K$ itself.

The van't Hoff equation. Differentiate $\ln K=-\Delta_rG^\circ/(RT)$ with respect to $T$ and use the Gibbs-Helmholtz relation $\left(\partial(\Delta G^\circ/T)/\partial T\right)_P=-\Delta H^\circ/T^2$:

where: $\Delta_rH^\circ$ is the standard reaction enthalpy (assumed roughly constant over the temperature range considered) and $R=8.314\ \text{J mol}^{-1}\text{K}^{-1}$. Integrating between two temperatures gives the working form used in calculations:

This is structurally identical to the integrated Clausius-Clapeyron equation from phase equilibrium, because both descend from the same $d(\Delta G^\circ/T)/dT=-\Delta H^\circ/T^2$ mathematics.

Le Chatelier's rule, now a sign. For an exothermic reaction ($\Delta_rH^\circ<0$), the van't Hoff equation gives $d\ln K/dT<0$, so $K$ falls as $T$ rises — heating disfavours products. For an endothermic reaction ($\Delta_rH^\circ>0$), $K$ rises with $T$. The qualitative CBSE rule is exactly this sign, now derived rather than asserted.

For $\text{N}_2(g)+3\text{H}_2(g)\rightleftharpoons 2\text{NH}_3(g)$, $K_p=1.45\times10^{-5}$ at 500 K and $\Delta_rH^\circ=-92.4\ \text{kJ mol}^{-1}$ (exothermic, roughly constant over the range). Estimate $K_p$ at 600 K, and use the answer to find the degree of dissociation regime it implies is favoured.

Step 1. Apply the integrated van't Hoff equation with $T_1=500$ K, $K_1=1.45\times10^{-5}$, $T_2=600$ K. $\ln\dfrac{K_2}{K_1}=-\dfrac{\Delta_rH^\circ}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)=-\dfrac{(-92400)}{8.314}\left(\dfrac{1}{600}-\dfrac{1}{500}\right)$.

Step 2. Evaluate the bracket. $\frac{1}{600}=1.6667\times10^{-3}$, $\frac{1}{500}=2.0000\times10^{-3}$, so the difference is $-3.333\times10^{-4}\ \text{K}^{-1}$.

Step 3. Combine. $\ln\dfrac{K_2}{K_1}=-\dfrac{-92400}{8.314}\times(-3.333\times10^{-4})=11114.3\times(-3.333\times10^{-4})=-3.705$, so $\dfrac{K_2}{K_1}=e^{-3.705}=0.02461$ and $K_2=(1.45\times10^{-5})(0.02461)=3.57\times10^{-7}$.

Step 4. Interpret via the dissociation link. Since $K_p$ falls by roughly two orders of magnitude on heating, the equilibrium extent of the reverse reaction (ammonia decomposing back to $\text{N}_2$ and $\text{H}_2$) is strongly favoured at 600 K compared with 500 K — consistent with the general degree-of-dissociation relation $K_p=\dfrac{\alpha^2}{1-\alpha^2}\cdot\dfrac{P}{p^\circ}$ derived for a dissociating gas-phase equilibrium: a smaller $K_p$ at fixed $P$ forces a smaller equilibrium $\alpha$ for the forward process, i.e. less product. Answer: $K_p(600\ \text{K})\approx3.57\times10^{-7}$, roughly $1/40$ of its value at 500 K, obtained purely from $\Delta_rH^\circ$ — the quantitative version of "heating an exothermic reaction disfavours products."

Competitive stretch

Suppose instead you were given $K_p$ at both 500 K and 600 K but not $\Delta_rH^\circ$ — could you recover it? Yes: the same integrated van't Hoff equation is linear in the unknown $\Delta_rH^\circ$ once $K_1$, $K_2$, $T_1$ and $T_2$ are known, so it can be solved by rearrangement rather than integration. Using the two values already found, $K_1=1.45\times10^{-5}$ at $T_1=500$ K and $K_2=3.57\times10^{-7}$ at $T_2=600$ K, rearrange $\ln(K_2/K_1)=-\dfrac{\Delta_rH^\circ}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)$ for $\Delta_rH^\circ$: $\Delta_rH^\circ=-R\ln(K_2/K_1)\Big/\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)=-(8.314)(-3.705)/(-3.333\times10^{-4})=-92{,}400\ \text{J mol}^{-1}$, recovering the same $-92.4\ \text{kJ mol}^{-1}$ used to construct the problem — the check that the van't Hoff equation is a genuinely invertible relation between $\Delta_rH^\circ$ and the pair of equilibrium constants, exactly the way two points on a straight line fix its slope.

• Believing $K$ itself changes when reactant is added or pressure is changed — $K$ is a function of $T$ only; what changes is $Q$ relative to $K$, or (for pressure and volume changes) the equilibrium composition/extent that still satisfies the fixed $K_p$, never $K_p$ itself.
• Applying the integrated van't Hoff equation across a temperature range where $\Delta_rH^\circ$ is not roughly constant, such as one spanning a phase change of a reactant — the constant-$\Delta H$ assumption can fail over wide ranges, exactly as in Clausius-Clapeyron.
• Reversing the sign rule under exam pressure: an increase in $T$ increases $K$ for endothermic reactions ($\Delta_rH^\circ>0$) and decreases $K$ for exothermic reactions — students commonly remember it backwards.
• Forgetting that $K$ built from activities is dimensionless by construction (each activity already divides by $p^\circ$ or $c^\circ$); plugging a $K_p$ quoted with explicit pressure units directly into $\Delta_rG^\circ=-RT\ln K$ produces a dimensionally meaningless logarithm.

Ask the tutor: If you want Le Chatelier's pressure and concentration rules derived thermodynamically the same way the temperature rule was derived here, ask the AI tutor to work through the $Q$-versus-$K$ argument for those cases too.