Learning Objectives
By the end of this module, you will be able to:
- Write equations representing changes in concentration and pressure for chemical species in equilibrium systems
- Use algebra to perform various types of equilibrium calculations
We know that at equilibrium, the value of the reaction quotient of any reaction is equal to its equilibrium constant. Thus, we can use the mathematical expression for Q to determine a number of quantities associated with a reaction at equilibrium or approaching equilibrium. While we have learned to identify in which direction a reaction will shift to reach equilibrium, we want to extend that understanding to quantitative calculations. We do so by evaluating the ways that the concentrations of products and reactants change as a reaction approaches equilibrium, keeping in mind the stoichiometric ratios of the reaction. This algebraic approach to equilibrium calculations will be explored in this section.
Changes in concentrations or pressures of reactants and products occur as a reaction system approaches equilibrium. In this section we will see that we can relate these changes to each other using the coefficients in the balanced chemical equation describing the system. We use the decomposition of ammonia as an example.
On heating, ammonia reversibly decomposes into nitrogen and hydrogen according to this equation:
2NH3(g)⇌N2(g)+3H2(g)
If a sample of ammonia decomposes in a closed system and the concentration of N2 increases by 0.11 M, the change in the N2 concentration, Δ[N2], the final concentration minus the initial concentration, is 0.11 M. The change is positive because the concentration of N2 increases.
The change in the H2 concentration, Δ[H2], is also positive—the concentration of H2 increases as ammonia decomposes. The chemical equation tells us that the change in the concentration of H2 is three times the change in the concentration of N2, because for each mole of N2 produced, 3 moles of H2 are produced.
Δ[H2]=3×Δ[N2]
=3×(0.11M)=0.33M
The change in concentration of NH3, Δ[NH3], is twice that of Δ[N2]; the equation indicates that 2 moles of NH3 must decompose for each mole of N2 formed. However, the change in the NH3 concentration is negative because the concentration of ammonia decreases as it decomposes.
Δ[NH3]=−2×Δ[N2]=−2×(0.11M)=−0.22M
We can relate these relationships directly to the coefficients in the equation
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Note that all the changes on one side of the arrows are of the same sign and that all the changes on the other side of the arrows are of the opposite sign.
If we did not know the magnitude of the change in the concentration of N2, we could represent it by the symbol x.
Δ[N2]=x
The changes in the other concentrations would then be represented as:
Δ[H2]=3×Δ[N2]=3x
Δ[NH3]=−2×Δ[N2]=−2x
The coefficients in the Δ terms are identical to those in the balanced equation for the reaction.
The simplest way for us to find the coefficients for the concentration changes in any reaction is to use the coefficients in the balanced chemical equation. The sign of the coefficient is positive when the concentration increases; it is negative when the concentration decreases.
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Determining Relative Changes in ConcentrationComplete the changes in concentrations for each of the following reactions. Answer Check Your LearningComplete the changes in concentrations for each of the following reactions: Answer
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Calculations Involving Equilibrium Concentrations
When to Simplify and how to know: One thing I should have clarified on this video, both examples yield results that differ in the hundredth's place. The first example with a starting value of 1.00M, a difference in the hundredth's place is 1% of the starting value. That is a significant enough difference to not simplify. In the second example with a starting concentration of 100.M, a difference in the hundredth's place is 0.01% of the starting value. That is not a significant enough difference to avoid simplifying. When in doubt, use the 5% rule. x/initial concentration * 100 = a percent value. If that percent value is 5 or less, you are in the clear to use the simplification.
Because the value of the reaction quotient of any reaction at equilibrium is equal to its equilibrium constant, we can use the mathematical expression for Qc (i.e., the law of mass action) to determine a number of quantities associated with a reaction at equilibrium. It may help if we keep in mind that Qc = Kc (at equilibrium) in all of these situations and that there are only three basic types of calculations:
- Calculation of an equilibrium constant. If concentrations of reactants and products at equilibrium are known, the value of the equilibrium constant for the reaction can be calculated.
- Calculation of missing equilibrium concentrations. If the value of the equilibrium constant and all of the equilibrium concentrations, except one, are known, the remaining concentration can be calculated.
- Calculation of equilibrium concentrations from initial concentrations. If the value of the equilibrium constant and a set of concentrations of reactants and products that are not at equilibrium are known, the concentrations at equilibrium can be calculated.
A similar list could be generated using QP, KP, and partial pressure. We will look at solving each of these cases in sequence.
Calculation of an Equilibrium Constant
Since the law of mass action is the only equation we have to describe the relationship between Kc and the concentrations of reactants and products, any problem that requires us to solve for Kc must provide enough information to determine the reactant and product concentrations at equilibrium. Armed with the concentrations, we can solve the equation for Kc, as it will be the only unknown.
Example 2 showed us how to determine the equilibrium constant of a reaction if we know the concentrations of reactants and products at equilibrium. The following example shows how to use the stoichiometry of the reaction and a combination of initial concentrations and equilibrium concentrations to determine an equilibrium constant. This technique, commonly called an ICE chart—for Initial, Change, and Equilibrium—will be helpful in solving many equilibrium problems. A chart is generated beginning with the equilibrium reaction in question. Underneath the reaction the initial concentrations of the reactants and products are listed—these conditions are usually provided in the problem and we consider no shift toward equilibrium to have happened. The next row of data is the change that occurs as the system shifts toward equilibrium—do not forget to consider the reaction stoichiometry as described in a previous section of this chapter. The last row contains the concentrations once equilibrium has been reached.
Example |
Calculation of an Equilibrium ConstantIodine molecules react reversibly with iodide ions to produce triiodide ions. I2(aq)+I−(aq)⇌I3−(aq) If a solution with the concentrations of I2 and I− both equal to 1.000 × 10−3M before reaction gives an equilibrium concentration of I2 of 6.61 × 10−4M, what is the equilibrium constant for the reaction? Answer We will begin this problem by calculating the changes in concentration as the system goes to equilibrium. Then we determine the equilibrium concentrations and, finally, the equilibrium constant. First, we set up a table with the initial concentrations, the changes in concentrations, and the equilibrium concentrations using −x as the change in concentration of I2. Since the equilibrium concentration of I2 is given, we can solve for x. At equilibrium the concentration of I2 is 6.61 × 10−4M so that 1.000×10−3−x=6.61×10−4 x=1.000×10−3−6.61×10−4 =3.39×10−4M Now we can fill in the table with the concentrations at equilibrium. We now calculate the value of the equilibrium constant. Check Your LearningEthanol and acetic acid react and form water and ethyl acetate, the solvent responsible for the odor of some nail polish removers. C2H5OH+CH3CO2H⇌CH3CO2C2H5+H2O When 1 mol each of C2H5OH and CH3CO2H are allowed to react in 1 L of the solvent dioxane, equilibrium is established when ⅓ mol of each of the reactants remains. Calculate the equilibrium constant for the reaction. (Note: Water is not a solvent in this reaction.) Answer Kc = 4
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1. The reaction may be written as N2(g)+3H2(g)⇌2NH3(g) The equilibrium constant for the reaction is 3. The decomposition of PCl5 to PCl3 and Cl2 is given as
5. The equilibrium equation is 7. Because the decomposition must generate the same pressure of HCl as NH3, 1.75 atm of HCl must be present. |
Calculation of a Missing Equilibrium Concentration
If we know the equilibrium constant for a reaction and know the concentrations at equilibrium of all reactants and products except one, we can calculate the missing concentration.
Example |
Calculation of a Missing Equilibrium ConcentrationNitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperatures. At 2000 °C, the value of the equilibrium constant for the reaction, N2(g)+O2(g)⇌2NO(g), is 4.1 × 10−4. Find the concentration of NO(g) in an equilibrium mixture with air at 1 atm pressure at this temperature. In air, [N2] = 0.036 mol/L and [O2] 0.0089 mol/L. Answer We are given all of the equilibrium concentrations except that of NO. Thus, we can solve for the missing equilibrium concentration by rearranging the equation for the equilibrium constant. Thus [NO] is 3.6 × 10−4 mol/L at equilibrium under these conditions. We can check our answer by substituting all equilibrium concentrations into the expression for the reaction quotient to see whether it is equal to the equilibrium constant. The answer checks; our calculated value gives the equilibrium constant within the error associated with the significant figures in the problem. Check Your LearningThe equilibrium constant for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is 6.00 × 10−2. Calculate the equilibrium concentration of ammonia if the equilibrium concentrations of nitrogen and hydrogen are 4.26 M and 2.09 M, respectively. Answer 1.53 mol/L |
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1. Write the equilibrium constant expression and solve for [NH3]. [NH3]2 = 1.2 × (0.24)3 × 0.50 = 0.0083 [NH3] = 9.1 × 10-2M 3. Write the equilibrium constant expression and solve for [CO]. 5. Calculate Q based on the calculated concentrations and see if it is equal to Kc. Because Q does equal 4.32, the system must be at equilibrium. |
Calculation of Changes in Concentration
If we know the equilibrium constant for a reaction and a set of concentrations of reactants and products that are not at equilibrium, we can calculate the changes in concentrations as the system comes to equilibrium, as well as the new concentrations at equilibrium. The typical procedure can be summarized in four steps.
- Determine the direction the reaction proceeds to come to equilibrium.
- Write a balanced chemical equation for the reaction.
- If the direction in which the reaction must proceed to reach equilibrium is not obvious, calculate Qc from the initial concentrations and compare to Kc to determine the direction of change.
- Determine the relative changes needed to reach equilibrium, then write the equilibrium concentrations in terms of these changes.
- Define the changes in the initial concentrations that are needed for the reaction to reach equilibrium. Generally, we represent the smallest change with the symbol x and express the other changes in terms of the smallest change.
- Define missing equilibrium concentrations in terms of the initial concentrations and the changes in concentration determined in (a).
- Solve for the change and the equilibrium concentrations.
- Substitute the equilibrium concentrations into the expression for the equilibrium constant, solve for x, and check any assumptions used to find x.
- Calculate the equilibrium concentrations.
- Check the arithmetic.
- Check the calculated equilibrium concentrations by substituting them into the equilibrium expression and determining whether they give the equilibrium constant.
- Sometimes a particular step may differ from problem to problem—it may be more complex in some problems and less complex in others. However, every calculation of equilibrium concentrations from a set of initial concentrations will involve these steps.
- In solving equilibrium problems that involve changes in concentration, sometimes it is convenient to set up an ICE table, as described in the previous section.
Example |
Calculation of Concentration Changes as a Reaction Goes to EquilibriumUnder certain conditions, the equilibrium constant for the decomposition of PCl5(g) into PCl3(g) and Cl2(g) is 0.0211. What are the equilibrium concentrations of PCl5, PCl3, and Cl2 if the initial concentration of PCl5 was 1.00 M? Answer This equation contains only one variable, x, the change in concentration. We can write the equation as a quadratic equation and solve for x using the quadratic formula. Essential Mathematics shows us an equation of the form ax2 + bx + c = 0 can be rearranged to solve for x: In this case, a = 1, b = 0.0211, and c = −0.0211. Substituting the appropriate values for a, b, and c yields: Hence or Quadratic equations often have two different solutions, one that is physically possible and one that is physically impossible (an extraneous root). In this case, the second solution (−0.156) is physically impossible because we know the change must be a positive number (otherwise we would end up with negative values for concentrations of the products). Thus, x = 0.135 M. The equilibrium concentrations are Step 4: Check the arithmetic.Substitution into the expression for Kc (to check the calculation) gives 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). Thus, the calculated equilibrium concentrations check. Check Your LearningAcetic acid, CH3CO2H, reacts with ethanol, C2H5OH, to form water and ethyl acetate, CH3CO2C2H5. CH3CO2H+C2H5OH⇌CH3CO2C2H5+H2O The equilibrium constant for this reaction with dioxane as a solvent is 4.0. What are the equilibrium concentrations when a mixture that is 0.15 M in CH3CO2H, 0.15 M in C2H5OH, 0.40 M in CH3CO2C2H5, and 0.40 M in H2O are mixed in enough dioxane to make 1.0 L of solution? Answer [CH3CO2H] = 0.36 M, [C2H5OH] = 0.36 M, [CH3CO2C2H5] = 0.17 M, [H2O] = 0.17 M Check Your LearningA 1.00-L flask is filled with 1.00 moles of H2 and 2.00 moles of I2. The value of the equilibrium constant for the reaction of hydrogen and iodine reacting to form hydrogen iodide is 50.5 under the given conditions. What are the equilibrium concentrations of H2, I2, and HI in moles/L? H2(g)+I2(g)⇌2HI(g) Answer [H2] = 0.06 M, [I2] = 1.06 M, [HI] = 1.88 M |
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There is an error in the above video: x=0.020M is given
1.The changes in concentrations (or pressure, if requested) are as follows:
3. Activities of pure crystalline solids equal 1 and are constant; however, the mass of Ni does change. |
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1. (a) Write the starting conditions, change, and equilibrium constant in tabular form.
Since K is very small, ignore x in comparison with 0.129 M. The equilibrium expression is The concentrations are:
(b) Percent error = . The change in concentration of N2O4 is far less than the 5% maximum allowed. 3. (a) Write the balanced equation, and then set up a table with initial pressures and the changed pressures using x as the change in pressure. The simplest way to find the coefficients for the x values is to use the coefficient in the balanced equation.
Since the value of Kp is much smaller than 0.824, the 2x term in 0.824 − 2x is deemed negligible and, therefore, can be dropped.
Final equilibrium pressures:
(b) The 2x is dropped from the equilibrium calculation because 0.014 is negligible when subtracted from 0.824. The percent error associated with ignoring 2x is , which is less than allowed by the “5% test.” The error is, indeed, negligible. 5. As all species are gases and are in M concentration units, a simple Kc equilibrium can be solved using the balanced equation: As the value of Kc is substantial when compared with 2.00 M, the x terms in 2.00 − x cannot be disregarded. Thus, x must be solved by using the quadratic formula. Begin by arranging the terms in the form of the quadratic equation:
Next, solve for x using the quadratic formula. = 0.195 M or −0.216 M The process of dissociation renders only positive quantities. Thus, x must be a positive value when factored into the solution so as to guarantee a realistic result. The final equilibrium concentrations are: [PCl3] = 2.00 − x = 2.00 − 0.195 = 1.80 M; [PC3] = [Cl2] = x = 0.195 M; [PCl3] = [Cl2] = x = 0.195 M. |
