Think of a sealed bottle of fizzy water: while it stays closed, the pressure never changes and nothing seems to happen. Yet at molecular level, CO₂ molecules are constantly leaving the liquid while others dissolve back in from the gas. What stays constant is not the amount of change but the balance between two opposite processes that never stop.
Dynamic equilibrium
Heat hydrogen and iodine vapour in a sealed flask at 450 °C. The purple colour fades for a while, then stops changing, yet it never disappears because some iodine always remains in the flask. Why does the change stop before the reactants are used up?
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Solid calcium carbonate is placed in a closed reactor, and the following heterogeneous equilibrium is established at $1073\text{ K}$:…
Solid calcium carbonate is placed in a closed reactor, and the following heterogeneous equilibrium is established at $1073\text{ K}$: $\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}$.
a) Write the expressions for the equilibrium constants $K_c$ and $K_p$ for this reaction, explaining with reasoning why pure solid species are not included.
b) Explain the physical meaning of "dynamic equilibrium" as opposed to a static equilibrium, stating whether the decomposition of $\mathrm{CaCO_3(s)}$ stops once this state is reached. Assume that the two pure solid phases remain present.
For the synthesis of ammonia according to the equation $\mathrm{N_2(g) + 3\,H_2(g) \rightleftharpoons 2\,NH_3(g)}$, the value of the equilibrium constant $K_{c,1}$ at…
For the synthesis of ammonia according to the equation $\mathrm{N_2(g) + 3\,H_2(g) \rightleftharpoons 2\,NH_3(g)}$, the value of the equilibrium constant $K_{c,1}$ at $673\text{ K}$ is equal to $0.50$.
a) Determine the value of the equilibrium constant $K_{c,2}$ at the same temperature for the reverse reaction: $2\,\mathrm{NH_3(g) \rightleftharpoons N_2(g) + 3\,H_2(g)}$.
b) Calculate the value of the constant $K_{c,3}$ at $673\text{ K}$ corresponding to the decomposition of one mole of ammonia: $\mathrm{NH_3(g) \rightleftharpoons \frac{1}{2}\,N_2(g) + \frac{3}{2}\,H_2(g)}$. For the calculations, use the ideal model and the teaching values provided.
At a temperature of $500\text{ K}$, the equilibrium constant for the synthesis of sulfur trioxide according to the equation…
At a temperature of $500\text{ K}$, the equilibrium constant for the synthesis of sulfur trioxide according to the equation $\mathrm{2\,SO_2(g) + O_2(g) \rightleftharpoons 2\,SO_3(g)}$ is $K_c = 400$.
Data: gas constant $R = 0.082\text{ atm}\times\text{L}\times\text{mol}^{-1}\times\text{K}^{-1}$.
a) Deduce the general mathematical relationship between $K_p$ and $K_c$ for a mixture of ideal gases and calculate the numerical value of $K_p$ for this reaction at $500\text{ K}$.
b) If, in a closed flask at that temperature, the partial pressures at equilibrium are $p_{\mathrm{SO_2}} = 0.10\text{ atm}$ and $p_{\mathrm{O_2}} = 0.20\text{ atm}$, calculate the partial pressure of $\mathrm{SO_3}$ at equilibrium and the total pressure of the mixture. For the calculations, use the ideal model and the teaching values provided.
Put arbitrary amounts of reactants and products into a closed flask at constant temperature. The mixture will change, but how can we tell beforehand whether reactants will be used up or products will break down?
Reading a graph of rates or concentrations
A graph of rate against time (or concentration against time) shows equilibrium as the point where the curves stop changing. For rates, the forward and reverse curves meet and stay equal ($v_f = v_r$). For concentrations, the curves of reactants and products become flat. To find $K_c$, read the concentrations on the flat part of the graph and substitute them into the expression. If the graph shows concentration against time, the rate of a reaction at any moment is the slope of the tangent to the curve. Steep slopes mean fast change; at equilibrium the slopes are zero, because the net rate is zero ($v_f = v_r$). Two things to remember: the rate of the forward reaction is highest at the start, when only reactants are present, and the rate of the reverse reaction starts at zero and rises as products build up.
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The reversible elementary reaction $\mathrm{A(g) \rightleftharpoons B(g)}$ is studied kinetically in the gas phase in a closed isothermal vessel of constant volume at…
The reversible elementary reaction $\mathrm{A(g) \rightleftharpoons B(g)}$ is studied kinetically in the gas phase in a closed isothermal vessel of constant volume at $300\text{ K}$, starting only from the reactant $\mathrm{A}$ at an initial concentration of $1.00\text{ mol}\times\text{L}^{-1}$. The attached graph monitors the variation with time of the rate of the forward reaction ($v_d$) and of the reverse reaction ($v_i$).
Look at the following representative figure:
a) Explain, using the graph, why the final state reached after a sufficiently long time, within experimental precision, corresponds to a dynamic equilibrium and not to a static cessation of reactivity.
b) If the forward rate constant is $k_d = 0.030\text{ s}^{-1}$ and at equilibrium the common rate shown is $v_{\mathrm{eq}} = 0.012\text{ mol}\times\text{L}^{-1}\times\text{s}^{-1}$, calculate the equilibrium concentration of $\mathrm{A}$, the equilibrium concentration of $\mathrm{B}$ and the value of the equilibrium constant $K_c$. For the calculations, use the ideal model and the teaching values provided.
Mix 1 mol of each reactant in a closed 1 L vessel. Because the reaction is reversible, not everything is used up, so limiting-reagent stoichiometry alone cannot give the final amounts. How can we organise the data to find the equilibrium composition?
Quadratic equations
Often the equation for $x$ is not a perfect square. Expand, write it as $ax^2 + bx + c = 0$ and use the quadratic formula. Two roots appear, and only one can be physically sensible. Core examples are limited to simple quadratics like the one below; harder ones are extension.
Extension: degree of dissociation
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Dinitrogen tetroxide dissociates thermally into nitrogen dioxide according to the equilibrium $\mathrm{N_2O_4(g) \rightleftharpoons 2\,NO_2(g)}$. In a closed vessel of…
Dinitrogen tetroxide dissociates thermally into nitrogen dioxide according to the equilibrium $\mathrm{N_2O_4(g) \rightleftharpoons 2\,NO_2(g)}$. In a closed vessel of fixed volume $V = 10.0\text{ L}$ at $320\text{ K}$, a mass of $18.4\text{ g}$ of pure $\mathrm{N_2O_4}$ is introduced. Once chemical equilibrium has been established at that temperature, the total pressure recorded on the manometer is $0.650\text{ atm}$.
a) Calculate the degree of dissociation $\alpha$ of $\mathrm{N_2O_4(g)}$ under these conditions.
b) Deduce the expression for $K_p$ in terms of $\alpha$ and the total pressure $P_t$, and determine its numerical value at $320\text{ K}$. For the calculations, use the ideal model and the teaching values provided.
Back to the start: the fizzy water stays unchanged because CO₂ leaving the liquid and CO₂ dissolving back happen at the same rate. That is dynamic equilibrium.
Summary
Let's practice! 2 medium exercises in this section