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Home
JoVE Core
Chemistry
Calculating Equilibrium Concentrations
Video Quiz
Calculating Equilibrium Concentrations
JoVE Core
Chemistry
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JoVE Core Chemistry
Calculating Equilibrium Concentrations

14.6: Calculating Equilibrium Concentrations

55,235 Views
02:05 min
September 24, 2020

Overview

Being able to calculate equilibrium concentrations is essential to many areas of science and technology—for example, in the formulation and dosing of pharmaceutical products. After a drug is ingested or injected, it is typically involved in several chemical equilibria that affect its ultimate concentration in the body system of interest. Knowledge of the quantitative aspects of these equilibria is required to compute a dosage amount that will solicit the desired therapeutic effect.

A more challenging type of equilibrium calculation can be one in which equilibrium concentrations are derived from initial concentrations and an equilibrium constant. For these calculations, a four-step approach is typically useful:

  1. Identify the direction in which the reaction will proceed to reach equilibrium.
  2. Develop an ICE table.
  3. Calculate the concentration changes and, subsequently, the equilibrium concentrations.
  4. Confirm the calculated equilibrium concentrations.

Calculation of Equilibrium Concentrations

Under certain conditions, the equilibrium constant Kc for the decomposition of PCl5(g) into PCl3(g) and Cl2(g) is 0.0211. The above procedure can be used to determine the equilibrium concentrations of PCl5, PCl3, and Cl2 in a mixture that initially contained only PCl5 at a concentration of 1.00 M.

Step 1. Determine the direction the reaction proceeds.

The balanced equation for the decomposition of PCl5 is

Thermodynamics, entropy change formula ΔS=qrev/T, diagram explaining energy transfer concepts.

Because only the reactant is present initially, Qc = 0, and the reaction will proceed to the right.

Step 2. Develop an ICE table.

PCl5 (g) PCl3 (g) Cl2 (g)
Initial Concentration (M) 1.00 0 0
Change (M) −x +x +x
Equilibrium Concentration (M) 1.00 − x x x

Step 3. Solve for the change and the equilibrium concentrations.

Substituting the equilibrium concentrations into the equilibrium constant equation gives

Chemical structure diagram, schematic of a molecular compound with multi-ring configuration.

An equation of the form ax2 + bx + c = 0 can be rearranged to solve for x:

Equilibrium of forces diagram, ΣF=0; illustrates static balance with vectors and forces labeled.

In this case, a = 1, b = 0.0211, and c = −0.0211. Substituting the appropriate values for a, b, and c yields:

Static equilibrium equations, ΣFx=0, ΣFy=0, diagram with force vectors for educational use.

The two roots of the quadratic are, therefore,

DNA transcription process, sequence of bases diagram, genetic information flow, biological mechanisms.

For this scenario, only the positive root is physically meaningful (concentrations are either zero or positive), and so x = 0.135 M. The equilibrium concentrations are

Balanced chemical equation, CaO + H₂O → Ca(OH)₂, exothermic reaction, chemical formula.

Step 4. Confirm the calculated equilibrium concentrations.

Substitution into the expression for Kc (to check the calculation) gives

Lewis structure diagram, carbonatoborate anion, CO3–Bo3, bond arrangement visualization, chemistry study.

The equilibrium constant calculated from the equilibrium concentrations is equal to the value of Kc given in the problem (when rounded to the proper number of significant figures).

This text has been adapted from Openstax, Chemistry 2e, Section 13.4 Equilibrium Calculations.

Transcript

Unknown equilibrium concentrations can be determined from the initial concentration of the reactants and the equilibrium constant, K, for the reaction.

Consider the reaction where 0.30 molar nitrogen gas and 0.40 molar oxygen gas react to produce nitric oxide gas and where K is 0.10. To calculate the equilibrium concentrations, the known values are tabulated in an ICE table.

For the change in concentration, the increase or decrease for each product or reactant, respectively, is denoted by x times its stoichiometric coefficient. The sum or difference is used to find the equilibrium concentrations, which are then substituted into the equilibrium expression.

To solve for x, the expression is expanded and all the terms are put on one side to convert it into the form: ax2 + bx + c. This equation can be solved using the quadratic formula.

Solving results in two values for x: 0.047 and −0.065. As the negative concentration of a substance is impossible, the value can be rejected.

Using 0.047 for x, the equilibrium concentrations of nitrogen, oxygen, and nitric oxide equal 0.25, 0.35, and 0.094 molar, respectively.

The perfect square condition is one situation where a shortcut can be used to avoid the quadratic formula.

For example, if the initial concentrations of nitrogen and oxygen in the above reaction were 0.30 molar each, the equation becomes a perfect square. In such cases, the equation can be simplified by taking the square root of both sides to solve for x.

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Equilibrium ConcentrationsInitial ConcentrationReactantsEquilibrium ConstantKReactionNitrogen GasOxygen GasNitric Oxide GasICE TableChange In ConcentrationStoichiometric CoefficientEquilibrium ExpressionXQuadratic FormulaSolving For XValues Of XEquilibrium Concentrations Of NitrogenOxygenNitric OxidePerfect Square Condition

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