AP Chemistry 7.7-7.8: Equilibrium Calculations and Representations

ICE table

An ICE table organizes equilibrium calculations into Initial concentrations, Changes caused by the reaction, and Equilibrium concentrations. The equilibrium row is obtained by adding the initial and change rows.

How are concentration changes related to the coefficients in a balanced equilibrium equation?

The changes occur in the stoichiometric mole ratios of the balanced equation. For example, in $2\mathrm{NH_3}\rightleftharpoons\mathrm{N_2}+3\mathrm{H_2}$, if $[\mathrm{N_2}]$ increases by $x$, then $[\mathrm{H_2}]$ increases by $3x$ and $[\mathrm{NH_3}]$ decreases by $2x$.

What do positive and negative signs represent in the change row of an ICE table?

A positive change indicates that a species is formed, while a negative change indicates that it is consumed. The signs depend on the direction in which the reaction proceeds.

For $\mathrm{C_3H_8}+5\mathrm{O_2}\rightleftharpoons3\mathrm{CO_2}+4\mathrm{H_2O}$, what are the concentration changes if propane decreases by $x$?

$\Delta[\mathrm{C_3H_8}]=-x$, $\Delta[\mathrm{O_2}]=-5x$, $\Delta[\mathrm{CO_2}]=+3x$, and $\Delta[\mathrm{H_2O}]=+4x$.

For $2\mathrm{SO_2}+\mathrm{O_2}\rightleftharpoons2\mathrm{SO_3}$, what are the changes if $[\mathrm{O_2}]$ increases by $x$?

$\Delta[\mathrm{SO_2}]=-2x$, $\Delta[\mathrm{O_2}]=+x$, and $\Delta[\mathrm{SO_3}]=+2x$.

How is the equilibrium constant $K_c$ calculated from equilibrium concentrations?

Substitute equilibrium molar concentrations into the balanced expression: products are placed in the numerator and reactants in the denominator, with each concentration raised to its stoichiometric coefficient. Pure solids and pure liquids are omitted.

How is an equilibrium constant expression written for $\mathrm{I_2(aq)}+\mathrm{I^-(aq)}\rightleftharpoons\mathrm{I_3^-(aq)}$?

$K_c=\dfrac{[\mathrm{I_3^-}]}{[\mathrm{I_2}][\mathrm{I^-}]}$.

A reaction begins with $[\mathrm{I_2}]=[\mathrm{I^-}]=1.000\times10^{-3}\,\mathrm{M}$ and reaches $[\mathrm{I_2}]=6.61\times10^{-4}\,\mathrm{M}$. What are the equilibrium concentrations for $\mathrm{I_2}+\mathrm{I^-}\rightleftharpoons\mathrm{I_3^-}$?

The reaction change is $x=3.39\times10^{-4}\,\mathrm{M}$. Therefore, $[\mathrm{I_2}]_\mathrm{eq}=[\mathrm{I^-}]_\mathrm{eq}=6.61\times10^{-4}\,\mathrm{M}$ and $[\mathrm{I_3^-}]_\mathrm{eq}=3.39\times10^{-4}\,\mathrm{M}$.

Using $[\mathrm{I_2}]_\mathrm{eq}=[\mathrm{I^-}]_\mathrm{eq}=6.61\times10^{-4}\,\mathrm{M}$ and $[\mathrm{I_3^-}]_\mathrm{eq}=3.39\times10^{-4}\,\mathrm{M}$, calculate $K_c$.

$K_c=\dfrac{3.39\times10^{-4}}{(6.61\times10^{-4})(6.61\times10^{-4})}\approx7.76\times10^2$.

What is the reaction quotient $Q$ used for in an equilibrium calculation?

$Q$ has the same form as the equilibrium-constant expression but uses the current, not necessarily equilibrium, concentrations or pressures. Comparing $Q$ with $K$ predicts the direction of net reaction.

How does comparing $Q$ and $K$ determine the direction a reaction proceeds?

If $Q<K$, the reaction proceeds toward products; if $Q>K$, it proceeds toward reactants; if $Q=K$, the system is at equilibrium.

What does $Q=0$ imply when a system initially contains only reactants?

Because the product concentration is initially zero, $Q=0$. For a reaction with a positive $K$, the reaction initially proceeds forward to form products.

What four-step strategy can be used to calculate equilibrium concentrations from initial concentrations and $K$?

Determine the direction using $Q$ and $K$, construct an ICE table, solve for the concentration change and equilibrium concentrations, and substitute the results back into the equilibrium expression to verify them.

How are equilibrium concentrations represented in an ICE table for $\mathrm{PCl_5}\rightleftharpoons\mathrm{PCl_3}+\mathrm{Cl_2}$ when only $1.00\,\mathrm{M}$ of $\mathrm{PCl_5}$ is initially present?

If the reaction proceeds forward by $x$, the equilibrium concentrations are $[\mathrm{PCl_5}]=1.00-x$, $[\mathrm{PCl_3}]=x$, and $[\mathrm{Cl_2}]=x$.

For $\mathrm{PCl_5}\rightleftharpoons\mathrm{PCl_3}+\mathrm{Cl_2}$ with $K_c=0.0211$ and initial $[\mathrm{PCl_5}]=1.00\,\mathrm{M}$, what equation results from the ICE table?

$0.0211=\dfrac{x^2}{1.00-x}$, which rearranges to $x^2+0.0211x-0.0211=0$.

For the $\mathrm{PCl_5}$ equilibrium problem with $K_c=0.0211$ and initial $[\mathrm{PCl_5}]=1.00\,\mathrm{M}$, what are the equilibrium concentrations?

Solving the quadratic gives the physically meaningful root $x=0.135\,\mathrm{M}$. Thus, $[\mathrm{PCl_5}]=0.865\,\mathrm{M}$, $[\mathrm{PCl_3}]=0.135\,\mathrm{M}$, and $[\mathrm{Cl_2}]=0.135\,\mathrm{M}$.

Why must one reject a mathematically valid negative root when solving for an equilibrium concentration change?

A concentration change variable such as $x$ represents a physically occurring amount and must produce nonnegative concentrations. A root that makes a concentration negative or contradicts the chosen reaction direction is not physically meaningful.

Quadratic formula

For an equation $ax^2+bx+c=0$, the solutions are $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$. Equilibrium calculations may require this when the $x$ terms cannot be neglected.

When is the small-$x$ approximation appropriate in an equilibrium calculation?

If $x$ is much smaller than the initial concentration, the expression $C-x$ can be approximated as $C$. A common check is that $x$ is less than about 5% of $C$; the approximation must be verified after solving.

For $\mathrm{HCN(aq)}\rightleftharpoons\mathrm{H^+(aq)}+\mathrm{CN^-(aq)}$, how does the small-$x$ approximation simplify $K_a$ for an initial HCN concentration $C$?

The equilibrium concentrations are approximately $[\mathrm{H^+}]=[\mathrm{CN^-}]=x$ and $[\mathrm{HCN}]\approx C$, so $K_a\approx\dfrac{x^2}{C}$. Therefore, $x\approx\sqrt{K_aC}$.

A $0.15\,\mathrm{M}$ HCN solution has $K_a=4.9\times10^{-10}$. What are the approximate equilibrium concentrations of $\mathrm{H^+}$ and $\mathrm{CN^-}$?

$x\approx\sqrt{(4.9\times10^{-10})(0.15)}=8.6\times10^{-6}\,\mathrm{M}$. Thus, $[\mathrm{H^+}]=[\mathrm{CN^-}]\approx8.6\times10^{-6}\,\mathrm{M}$ and $[\mathrm{HCN}]\approx0.15\,\mathrm{M}$.

How can the validity of a small-$x$ approximation be tested?

Compare the calculated $x$ with the initial concentration. If $100x/C$ is approximately 5% or less, the approximation is generally acceptable; otherwise, solve the original equation without neglecting $x$.

How can a missing equilibrium concentration be calculated when $K_c$ and all other equilibrium concentrations are known?

Substitute the known concentrations into the equilibrium expression and algebraically isolate the unknown concentration. For a product with coefficient 2, the expression may require taking a square root.

For $\mathrm{N_2}+\mathrm{O_2}\rightleftharpoons2\mathrm{NO}$, $K_c=4.1\times10^{-4}$, $[\mathrm{N_2}]=0.036\,\mathrm{M}$, and $[\mathrm{O_2}]=0.0089\,\mathrm{M}$, what is $[\mathrm{NO}]$ at equilibrium?

Since $K_c=\dfrac{[\mathrm{NO}]^2}{[\mathrm{N_2}][\mathrm{O_2}]}$, $[\mathrm{NO}]=\sqrt{K_c[\mathrm{N_2}][\mathrm{O_2}]}=3.6\times10^{-4}\,\mathrm{M}$.

For $\mathrm{N_2}+3\mathrm{H_2}\rightleftharpoons2\mathrm{NH_3}$ with $K_c=6.00\times10^{-2}$, $[\mathrm{N_2}]=4.26\,\mathrm{M}$, and $[\mathrm{H_2}]=2.09\,\mathrm{M}$ at equilibrium, how is $[\mathrm{NH_3}]$ found?

Use $K_c=\dfrac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3}$, so $[\mathrm{NH_3}]=\sqrt{K_c[\mathrm{N_2}][\mathrm{H_2}]^3}=1.53\,\mathrm{M}$.

How are gas pressures used in equilibrium calculations instead of concentrations?

For gaseous equilibria, partial pressures can be substituted into an analogous expression to calculate $K_p$ or $Q_p$. Each partial pressure is raised to the stoichiometric coefficient, just as concentrations are in $K_c$.

How do nonzero initial product concentrations affect an ICE table?

They must be included in the initial row. The reaction direction is determined using $Q$; if the reaction proceeds in reverse, reactants increase and products decrease according to the stoichiometric coefficients.

How can calculated equilibrium concentrations be checked?

Substitute the calculated equilibrium values back into the appropriate $K$ expression. The result should agree with the given equilibrium constant within the expected rounding and significant-figure uncertainty.

For the reaction $\mathrm{CH_3CO_2H+C_2H_5OH\rightleftharpoons CH_3CO_2C_2H_5+H_2O}$, what does $K_c=4.0$ indicate qualitatively?

Products are favored at equilibrium relative to reactants, although a substantial amount of reactants remains. The numerical conclusion comes from comparing the initial reaction quotient with $K_c$ and solving the ICE table.

What does a particulate-level representation of a system at equilibrium show?

It shows reactant and product particles coexisting in the same system after equilibrium is reached. The particles continue reacting, but their macroscopic concentrations remain constant because the forward and reverse reaction rates are equal.

Does equilibrium require equal numbers or concentrations of reactant and product particles?

No. Equilibrium requires equal forward and reverse rates, not equal concentrations. The relative amounts of reactants and products depend on the value of $K$ and the balanced equation.

How should particle counts in an equilibrium diagram reflect a balanced equation?

The particles should be grouped in the molecular or ionic ratios specified by the balanced equation. For example, $2\mathrm{A}\rightleftharpoons\mathrm{B}$ requires two A particles to form one B particle.

How can a particulate diagram distinguish a system before equilibrium from one at equilibrium?

A nonequilibrium diagram represents a composition that is still changing toward equilibrium. An equilibrium diagram contains both sides of the reaction, with no net change in particle populations over time even though individual collisions and reactions continue.

What does the value of $K$ suggest about a particulate-level equilibrium representation?

A large $K$ corresponds to a representation dominated by products, while a small $K$ corresponds to one dominated by reactants. A value near 1 indicates appreciable amounts of both remain at equilibrium.