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AP Chemistry Notes

7.7.1 Predicting equilibrium amounts from initial conditions

AP Syllabus focus: ‘Given a balanced reaction, initial concentrations or partial pressures, and the appropriate equilibrium constant, the equilibrium composition of a system can be predicted.’

Predicting equilibrium composition connects measurable starting conditions to the final, constant mixture at equilibrium.

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This figure shows how reactant and product concentrations evolve with time and then level off once equilibrium is reached. The flat (constant) regions of the curves represent the equilibrium composition, even though forward and reverse reactions continue to occur at equal rates. It provides a visual bridge between the qualitative idea of “constant mixture” and the quantitative ICE-table method used to compute equilibrium amounts. Source

You will translate a balanced equation into an algebraic model, then use the equilibrium constant to determine equilibrium concentrations or partial pressures.

Core idea: model equilibrium with a single variable

For a reversible reaction, stoichiometry constrains how species change as the system relaxes to equilibrium. You represent the change using one unknown (often xx) and solve for the equilibrium amounts that satisfy the equilibrium constant.

ICE table: A bookkeeping setup that tracks Initial amounts, Change based on stoichiometry, and Equilibrium amounts in terms of a single variable.

An ICE table can be written using molar concentrations (for KcK_c) or partial pressures (for KpK_p), matching the form of the equilibrium constant provided.

Step-by-step procedure (no shortcuts)

1) Start from the balanced chemical equation

  • Confirm the equation is balanced and identify stoichiometric coefficients.

  • Decide whether the problem uses KcK_c (concentrations) or KpK_p (partial pressures); use the one given.

2) Write equilibrium amounts in terms of a reaction progress variable

Let a general reaction be: aA+bBcC+dDaA + bB \rightleftharpoons cC + dD.

  • Choose a variable (commonly xx) to represent the extent of the forward reaction.

  • Build change expressions using coefficients:

    • AA: decreases by axax

    • BB: decreases by bxbx

    • CC: increases by cxcx

    • DD: increases by dxdx

3) Convert given initial information into the right form

  • For KcK_c problems, express initial conditions as initial concentrations.

  • For KpK_p problems, express initial conditions as initial partial pressures.

  • If initial information is given as moles in a fixed volume, convert to concentration before building the ICE table.

  • If initial information is given as total pressure plus mole information, obtain partial pressures before building the ICE table.

4) Substitute equilibrium expressions into the equilibrium-constant expression

Kc=[C]<em>eq,c[D]</em>eq,d[A]<em>eq,a[B]</em>eq,bK_c=\dfrac{[C]<em>{eq}^{,c}[D]</em>{eq}^{,d}}{[A]<em>{eq}^{,a}[B]</em>{eq}^{,b}}

[,]eq[,]_{eq} = equilibrium molar concentration, mol,L1\mathrm{mol,L^{-1}}

[A]eq=[A]0ax[A]_{eq}=[A]_0-ax

[A]0[A]_0 = initial molar concentration of AA, mol,L1\mathrm{mol,L^{-1}}

[C]eq=[C]0+cx[C]_{eq}=[C]_0+cx

xx = change variable (extent in concentration units), mol,L1\mathrm{mol,L^{-1}}

After substitution, you will have one equation with one unknown (xx). The same substitution approach applies to KpK_p using partial pressures (with PeqP_{eq} terms instead of [,]eq[,]_{eq}).

Solving for equilibrium composition

Recognise the algebra you may need

  • If powers are 1 and the expression simplifies neatly, you may get a linear equation in xx.

  • Many common equilibria produce a quadratic (or higher-order) equation; solving may require the quadratic formula or numerical methods.

Apply chemical validity checks (required)

Once you obtain xx, compute each equilibrium amount and check:

  • Non-negativity: no equilibrium concentration/pressure can be negative.

  • Stoichiometric limits: xx cannot exceed what would consume a limiting reactant below zero (e.g., [A]0ax0[A]_0-ax\ge 0).

  • Reasonableness with powers: when coefficients create squared/cubed terms, verify the final equilibrium amounts satisfy the constant’s expression when substituted back in.

Optional approximation (only if justified)

If equilibrium shifts only slightly from initial conditions, you may test an approximation such as xx is small compared with an initial amount” to simplify algebra, but you must confirm the approximation is consistent with the final value of xx.

FAQ

Reject any root that makes an equilibrium concentration/pressure negative.

Also reject roots that violate a stoichiometric bound, e.g. $[A]_0-ax<0$.

It is typically justified only when $x$ is a small fraction of the relevant initial amount (often a few per cent).

After solving, check the ratio $\dfrac{x}{[,]_0}$ (or $\dfrac{x}{P_0}$) to confirm.

Nothing about the method changes: include the nonzero product amount in the ICE table.

This can reduce the magnitude of the net change needed to satisfy the equilibrium constant.

The multiplier comes directly from stoichiometric coefficients.

If the balanced equation forms $2$ moles (or pressure units) of a species per “reaction event,” its change term must be $+2x$.

Keep units consistent with the form used: $\mathrm{mol,L^{-1}}$ for concentrations and pressure units (e.g. atm) for $K_p$ work.

Carry extra digits during algebra, then round final equilibrium amounts appropriately (typically to the given data’s significant figures).

Practice Questions

Question 1 (2 marks) For A(g)B(g)A(g)\rightleftharpoons B(g), Kc=4.0K_c=4.0 at a fixed temperature. Initially, [A]<em>0=0.50,mol,L1[A]<em>0=0.50,\mathrm{mol,L^{-1}} and [B]0=0.10,mol,L1[B]0=0.10,\mathrm{mol,L^{-1}}. Write expressions for [A]eq[A]{eq} and [B]</em>eq[B]</em>{eq} in terms of xx, and write the equation you would solve for xx.

  • [A]<em>eq=0.50x[A]<em>{eq}=0.50-x and [B]</em>eq=0.10+x[B]</em>{eq}=0.10+x (1)

  • Kc=[B]<em>eq[A]</em>eq=0.10+x0.50x=4.0K_c=\dfrac{[B]<em>{eq}}{[A]</em>{eq}}=\dfrac{0.10+x}{0.50-x}=4.0 (1)

Question 2 (5 marks) For N2O4(g)2NO2(g)N_2O_4(g)\rightleftharpoons 2NO_2(g), Kp=0.15K_p=0.15 at a fixed temperature. Initially, PN2O4=1.00,atmP_{N_2O_4}=1.00,\mathrm{atm} and PNO2=0.00,atmP_{NO_2}=0.00,\mathrm{atm}. Calculate the equilibrium partial pressures.

  • Let PN2O4,eq=1.00xP_{N_2O_4,eq}=1.00-x and PNO2,eq=2xP_{NO_2,eq}=2x (1)

  • Correct KpK_p setup: 0.15=(2x)21.00x0.15=\dfrac{(2x)^2}{1.00-x} (1)

  • Rearrangement to solvable form, e.g. 4x2+0.15x0.15=04x^2+0.15x-0.15=0 (1)

  • Correct physically valid root: x0.174,atmx\approx 0.174,\mathrm{atm} (1)

  • PN2O4,eq0.826,atmP_{N_2O_4,eq}\approx 0.826,\mathrm{atm} and PNO2,eq0.348,atmP_{NO_2,eq}\approx 0.348,\mathrm{atm} (1)

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