Equilibrium Processes: Features, Difference and Sample Questions

Equilibrium processes include:

1. Physical Equilibrium

Physical equilibrium happens when different phases of the same substance are converting back and forth at the same rate. Think about ice cubes in a drink at exactly 0°C - the ice is melting, but water is also freezing back onto the ice at the same speed, so the amount of ice stays constant. NCERT exercise covers a number of questions on Equilibrium processes.

This doesn't just happen with phase changes though. You see it with things dissolving too. If you keep adding salt to water until no more will dissolve, you've got a saturated solution. Even though it looks like nothing's happening, salt crystals are actually dissolving while dissolved salt is crystallizing out at the same rate.

Temperature and pressure really matter here. Warm up that ice-water mixture and suddenly more ice melts than freezes. Change the pressure on a gas-liquid system and you'll shift which phase is favored based on how dense each phase is.

2. Chemical Equilibrium

Chemical equilibrium is when a reversible reaction reaches a point where the forward and backward reactions happen at equal speeds. In that sulfur dioxide reaction you mentioned, SO₂ and O₂ are combining to make SO₃, but at the same time, SO₃ is breaking down back into SO₂ and O₂. When these happen at the same rate, you're at equilibrium.

The big difference from physical equilibrium is that you're actually breaking and making chemical bonds here. We can describe how far the reaction goes with an equilibrium constant - basically a number that tells you whether you'll have more reactants or products when things settle out.

The cool thing is that you can nudge these reactions around. Add more of one reactant and the reaction will make more products to balance things out. Heat up an exothermic reaction and it'll actually shift backward because the system tries to absorb that extra heat.

The following table explains the difference between chemical and physical equilibrium:

NEET and CUET questions test equilibrium types and calculations. Students must follow the below-given strategies:

• Identify the equilibrium type (physical or chemical) to apply relevant principles.
• Use equilibrium constants ( $K_c$ for chemical, $K_p$  for gases, or vapor pressure for physical) to quantify states.
• Set up ICE tables for chemical equilibria to find concentrations.
• For physical equilibria, relate equilibrium to saturation points.
• Practice problems with varied initial conditions for NEET and CUET entrance examinations.

The followins questions are asked in entrane exams like IISER and JEE Main:

Problem 1: For ${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\rightleftharpoons {\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$ in a closed vessel at 298 K , the vapor pressure is 23.8 mm Hg . What's the equilibrium partial pressure of ${\mathrm{H}}_2\mathrm{O}$ (g)? 

Solution: $p_{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)}=23.8\mathrm{m}\mathrm{m}\mathrm{H}\mathrm{g}$

Problem 2: For ${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{N}\mathrm{H}}_3\left(\mathrm{g}\right),K_c=0.5$ at 700 K . Initial: 1 mol ${\mathrm{N}}_2,2\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{H}}_2$  in 1 L . Find $\left[{\mathrm{N}\mathrm{H}}_3\right]$ at equilibrium. 

Solution: Initial: $\left[{\mathrm{N}}_2\right]=1\mathrm{M},\left[{\mathrm{H}}_2\right]=2\mathrm{M}$ . Let $\left[{\mathrm{N}\mathrm{H}}_3\right]=2\mathrm{x},\left[{\mathrm{N}}_2\right]=1-\mathrm{x},\left[{\mathrm{H}}_2\right]=2-3\mathrm{x}$ .

$K_c=\frac{(2x{)}^2}{(1-x)(2-3x{)}^3}=0.5$

Approximate: $x$  is small, $1-x\approx \mathrm{1,2}-3x\approx 2$ . Then:

$\frac{4x^2}{8}\approx 0.5,\mathrm{}{x}^2\approx 1,\mathrm{}x\approx 1,\mathrm{}\left[{\mathrm{N}\mathrm{H}}_3\right]\approx 2\mathrm{M}$

Problem 3: A saturated solution of AgCl has 0.00135 g in 1 L at 298 K . Calculate the solubility product ( $K_{sp}$ ). 

Solution: Molar mass of $\mathrm{A}\mathrm{g}\mathrm{C}\mathrm{l}=143.5\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}$ . Solubility:

$\begin{array}{c}s=\frac{0.00135}{143.5}\approx 9.41\times {10}^{-6}\mathrm{m}\mathrm{o}\mathrm{l}/\mathrm{L}\\ K_{sp}=\left[{\mathrm{A}\mathrm{g}}^{+}\right]\left[{\mathrm{C}\mathrm{l}}^{-}\right]=s^2\approx {\left(9.41\times {10}^{-6}\right)}^2\approx 8.85\times {10}^{-11}\end{array}$

Problem 4: For $2{\mathrm{S}\mathrm{O}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{S}\mathrm{O}}_3\left(\mathrm{g}\right)$ , initial: $0.5\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{S}\mathrm{O}}_2,0.2\mathrm{m}\mathrm{o}\mathrm{l}$ ${\mathrm{O}}_2$  in 1 L . At equilibrium, $0.3\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{S}\mathrm{O}}_3$   Find $K_c$ . 

Solution: Initial: $\left[{\mathrm{S}\mathrm{O}}_2\right]=0.5\mathrm{M},\left[{\mathrm{O}}_2\right]=0.2\mathrm{M}$ . At equilibrium: $\left[{\mathrm{S}\mathrm{O}}_3\right]=0.3\mathrm{M},\left[{\mathrm{S}\mathrm{O}}_2\right]=0.5$ .

$-0.15=0.35\mathrm{M},\left[{\mathrm{O}}_2\right]=0.2-0.075=0.125\mathrm{M}$

$K_c=\frac{{\left[{\mathrm{S}\mathrm{O}}_3\right]}^2}{{\left[{\mathrm{S}\mathrm{O}}_2\right]}^2\left[{\mathrm{O}}_2\right]}=\frac{(0.3{)}^2}{(0.35{)}^2\cdot 0.125}\approx 5.88$

Equilibrium processes are pivotal for JEE Main, bridging physical and chemical systems. NCERT's Section 6.1 (p. 169-171) lays the groundwork, enriched by JEEstyle problems. Refer to NCERT's Fig. 6.1 (p. 170) and Fig. 6.2 (p. 172) for clarity. Consistent practice and error correction ensure exam success.