Following data is given for the reaction: CaCO3 (s – Class 11 Chemistry

NCERT Exemplar · Q37

Q. Following data is given for the reaction: CaCO3 (s) \to CaO (s) + CO2

(g)
Δf H° [CaO(s)] = – 635.1 kJ mol^-1
Δf H° [CO2(g)] = – 393.5 kJ mol^-1
Δf H° [CaCO3(s)] = – 1206.9 kJ mol^-1
Predict the effect of temperature on the equilibrium constant of the above reaction.

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Concept understanding — Phase Equilibrium

Phase Equilibrium: From Intuition to Precision

Imagine a glass of ice water on a warm day. The ice cubes are melting, but the water around them stays cold. If you wait long enough, the ice stops melting — not because the room got colder, but because the system has reached a balance. The ice and water coexist without any net change. That balance is phase equilibrium.

The Intuition: A Tug-of-War Between Two Tendencies

Every substance wants to exist in the phase (solid, liquid, or gas) that has the lowest energy under the given conditions. But there's a catch: nature also loves disorder (entropy). A solid is low-energy but highly ordered; a gas is high-energy but very disordered. At a specific temperature and pressure, these two opposing drives — minimising energy and maximising disorder — exactly cancel each other out. The result? Two phases can coexist peacefully, with molecules constantly jumping between them but no net change in the amount of each phase.

Think of it as a tug-of-war. On one side, molecules in the solid phase are tightly bound and low in energy. On the other side, molecules in the liquid phase have more freedom and higher entropy. At the melting point, the rope doesn't move — the rates of melting and freezing are equal.

The Precise Statement

Phase equilibrium is the condition in which two or more phases of a substance coexist in thermodynamic equilibrium, with no net transfer of mass between phases. For a pure substance, this occurs at a unique temperature for a given pressure (or vice versa), defined by the equality of the chemical potential (or Gibbs free energy per mole) in all phases.

In simpler terms: at equilibrium, the driving force for a molecule to move from phase A to phase B is exactly the same as the driving force to move from B to A. The system is dynamic — molecules are still crossing the boundary — but macroscopically static.

The Key Condition: Equal Chemical Potential

The rigorous condition for phase equilibrium in a pure substance is:

\mu_{\text{solid}} = \mu_{\text{liquid}} = \mu_{\text{gas}}

where $\mu$ (mu) is the chemical potential — the Gibbs free energy per mole. When chemical potentials are equal, no phase has a "preference" to grow at the expense of another.

Important

For a pure substance, phase equilibrium occurs only along specific lines on a phase diagram (the melting curve, boiling curve, and sublimation curve). At the triple point, all three phases coexist.

A Concrete Example: Water at 0°C

At 1 atm pressure and 0°C, ice and liquid water are in equilibrium. Here's what's happening at the molecular level:

Water molecules at the surface of the ice are vibrating. Some gain enough energy to break free and enter the liquid.
Simultaneously, water molecules in the liquid near the ice surface lose energy and attach to the ice crystal.
The rates of these two processes are equal. So the amount of ice and the amount of liquid water remain constant — even though individual molecules are constantly switching sides. …

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The decomposition of calcium carbonate is endothermic ( $\Delta_r H^\circ = +178.3 \text{ kJ mol}^{-1}$ ), so by van't Hoff's equation the equilibrium constant increases with temperature; heating favours the forward reaction (decomposition).

Why temperature affects equilibrium: the van't Hoff equation

Every equilibrium constant $K$ is tied to the standard Gibbs energy change through $\Delta_r G^\circ = -RT \ln K$ . Because $\Delta_r G^\circ = \Delta_r H^\circ - T \Delta_r S^\circ$ , the temperature dependence of $K$ is governed by the sign and magnitude of $\Delta_r H^\circ$ . The van't Hoff equation captures this:

\frac{d \ln K}{dT} = \frac{\Delta_r H^\circ}{RT^2}

When a reaction is endothermic ( $\Delta_r H^\circ > 0$ ), the derivative is positive: $K$ grows as $T$ rises. When exothermic ( $\Delta_r H^\circ < 0$ ), $K$ shrinks with heating. So the first task is to compute $\Delta_r H^\circ$ for the decomposition.

Step-by-step solution

1. Write the reaction enthalpy in terms of formation enthalpies.

For any reaction, the standard enthalpy change is

\Delta_r H^\circ = \sum \Delta_f H^\circ (\text{products}) - \sum \Delta_f H^\circ (\text{reactants})

Here the reaction is

\ce{CaCO3 (s) -> CaO (s) + CO2 (g)}

\Delta_r H^\circ = \bigl[\Delta_f H^\circ (\ce{CaO}) + \Delta_f H^\circ (\ce{CO2})\bigr] - \Delta_f H^\circ (\ce{CaCO3}).

2. Substitute the given data.

\begin{aligned} \Delta_r H^\circ &= \bigl[(-635.1) + (-393.5)\bigr] - (-1206.9) \\ &= -1028.6 + 1206.9 \\ &= +178.3 \text{ kJ mol}^{-1}. \end{aligned}

The positive sign tells us the decomposition is endothermic: energy must be supplied to break $\ce{CaCO3}$ into $\ce{CaO}$ and $\ce{CO2}$ .

\Delta_r H^\circ = +178.3 \text{ kJ mol}^{-1} \quad (\text{endothermic})

3. Apply the van't Hoff equation.

Because $\Delta_r H^\circ > 0$ , the slope $\frac{d \ln K}{dT} = \frac{\Delta_r H^\circ}{RT^2}$ is positive at every temperature. In other words, $\ln K$ increases with $T$ , which means $K$ itself increases.

4. Interpret the physical consequence.

A larger $K$ means the equilibrium position shifts toward products. At low temperature calcium carbonate is stable; as you heat it the equilibrium constant climbs and eventually decomposition becomes appreciable. This is exactly what happens in a lime kiln: $\ce{CaCO3}$ decomposes to quicklime ( $\ce{CaO}$ ) only at high temperature. …

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Q.Following data is given for the reaction: CaCO3 (s) → CaO (s) + CO2

Why temperature affects equilibrium: the van't Hoff equation

Step-by-step solution

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