Quick questions on Static vs dynamic equilibrium explained: HSC Chemistry Module 5

Static equilibrium is a state of balance in which no net change is occurring and no process is active at the molecular level. A pencil balanced on a table is in static equilibrium. Two unreactive gases mixed in a sealed flask sit in static equilibrium because nothing is reacting.

<!-- Diagram: dynamic equilibrium rate vs time | reviewed 2026-05-21 --> <svg class="fig" viewBox="0 0 480 300" role="img" aria-labelledby="dyn-t dyn-d"> <title id="dyn-t">Forward and reverse reaction rates approaching dynamic equilibrium</title> <desc id="dyn-d">A plot of reaction rate against time. The forward rate starts high and falls. The reverse rate starts at zero and rises. The two curves meet at a constant value and remain equal beyond that point, which is dynamic equilibrium.</desc> <g> <line x1="60" y1="260" x2="440" y2="260" stroke="var(--ink)" stroke-width="0.9"/> <line x1="60" y1="40" x2="60" y2="260" stroke="var(--ink)" stroke-width="0.9"/> <text x="50" y="44" text-anchor="end" font-size="12" class="var">rate</text> <text x="440" y="278" text-anchor="end" font-size="12" class="var">t</text> <path d="M 70 60 C 110 100 150 140 200 160 L 410 160" fill="none" stroke="var(--ink)" stroke-width="1.6"/> <text x="80" y="80" font-size="11" font-weight="700">forward</text> <path d="M 70 260 C 110 220 150 180 200 160 L 410 160" fill="none" stroke="var(--accent)" stroke-width="1.6"/> <text x="80" y="244" font-size="11" font-weight="700" class="accent">reverse</text> <line x1="200" y1="40" x2="200" y2="260" stroke="var(--muted)" stroke-width="1" stroke-dasharray="5 3"/> <text x="202" y="52" font-size="11" class="muted">equilibrium reached</text> <text x="240" y="288" text-anchor="middle" font-size="11" class="muted">At equilibrium, forward and reverse rates are equal but non-zero.</text> </g> </svg>

Distinguish between static equilibrium and dynamic equilibrium, giving one example of each. [3 marks]

A sealed flask contains $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ with $[N_2O_4] = 0.20$ mol L$^{-1}$ and $[NO_2] = 0.10$ mol L$^{-1}$. The forward rate equals $1.0 \times 10^{-3}$ mol L$^{-1}$ s$^{-1}$. State the reverse rate and justify.

A student watches a saturated $NaCl$ solution sitting over excess solid for 24 hours. (a) Identify the equilibrium present. (b) State the macroscopic and molecular observations.