Reaction Rates and Equilibrium

What is meant by the rate of a chemical reaction?

• Can also be explained as the speed of he reaction, it is the amount of time required for a chemical reaction to come to completion.

• Different reactions take different times– Burning– Aging– Ripening– Rusting

Collision Theory• Atoms, ions and molecules can react to form

products when they collide provided that the particles are orientated correctly and have enough kinetic energy.

• The relative orientations of the molecules during their collisions determine whether the atoms are suitably positioned to form new bonds.

• Particles lacking necessary kinetic energy to react bounce apart when they collide.

• Imagine clay balls…• The minimum amount of energy that particles must

have in order to react is called the activation energy

Activated Complex

Activation Energy

• Swedish Chemist Svante Arrhenius explored activation energy, Ea

• He found that most reaction rate data obeyed an equation based on three factors:– The fraction of molecules possessing the necessary

activation energy or greater– The number of collisions occurring per second – The fraction of collisions that have the appropriate

orientation

• Arrhenius equationk = Ae-Ea/RT

Factors that Affect Reaction Rates• Temperature (Alkaseltzer activity)

– Raise temperature, faster reaction rate– Lower temperature, slower reaction rate– Higher temperatures make molecules move faster

because they have more kinetic energy so reaction is more likely

• Particle Size/Surface Area (lycopodium demo)– The smaller the particle size the larger the surface area.– An increase in surface area, increases the amount of

reactant exposed, which increases the collision frequency.– Can be dangerous… coal powder or gas particles reacting

to our lungs.

• Catalyst (demo)– A substance that increases the rate of a reaction

without being used up itself during the reaction.– They help reactions to proceed at a lower energy

than is normally required.– Enzymes catalyze reactions in our body– An inhibitor interferes with the action of a catalyst

• Concentration – The number of reacting particles in a given

volume also affects the rate at which reactions occur.

– Cramming more particles into a fixed volume increases the concentration of reactants, the collision frequency and therefore, reaction rate.

Rate Law: Effect of Concentration on Rate

• One way of studying the effect of concentration on reaction rate is to determine the way in which the rate at the beginning of a reaction depends on the starting concentrations.

a A + b B c C + d D

Rate = k[A]m[B]n

• The exponents m and n are called reaction orders and depend on how the concentration of that reactant affects the rate of reaction.

Concentration and Rate

If we compare Experiments 1 and 2, we see that when [NH4

+] doubles, the initial rate doubles.

NH4+(aq) + NO2

−(aq) N2(g) + 2 H2O(l)

Concentration and Rate

Likewise, when we compare Experiments 5

and 6, we see that when [NO2−] doubles, the

initial rate doubles.

NH4+(aq) + NO2

−(aq) N2(g) + 2 H2O(l)

Concentration and Rate• This means

Rate [NH4+]

Rate [NO2−]

Rate [NH4+] [NO2

−]

which, when written as an equation, becomes

Rate = k [NH4+] [NO2

−]

• This equation is called the rate law, and k is the rate constant.

Therefore,

Rate Laws• A rate law shows the relationship between

the reaction rate and the concentrations of reactants.

• The exponents tell the order of the reaction with respect to each reactant.

• Since the rate law is

Rate = k [NH4+] [NO2

−]

the reaction is

First-order in [NH4+] and

First-order in [NO2−].

Rate Laws

Rate = k [NH4+] [NO2

−]

• The overall reaction order can be found by adding the exponents on the reactants in the rate law.

• This reaction is second-order overall.

Multistep Mechanisms

• In a multistep process, one of the steps will be slower than all others.

• The overall reaction cannot occur faster than this slowest, rate-determining step.

Slow Initial Step

• The rate law for this reaction is found experimentally to be

Rate = k [NO2]2

• CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration.

• This suggests the reaction occurs in two steps.

NO2 (g) + CO (g) NO (g) + CO2 (g)

Slow Initial Step• A proposed mechanism for this reaction is

Step 1: NO2 + NO2 NO3 + NO (slow)

Step 2: NO3 + CO NO2 + CO2 (fast)

• The NO3 intermediate is consumed in the second step.

• As CO is not involved in the slow, rate-determining

step, it does not appear in the rate law.

Fast Initial Step

• The rate law for this reaction is found to be

Rate = k [NO]2 [Br2]

• Because termolecular processes are rare, this rate law suggests a two-step mechanism.

2 NO (g) + Br2 (g) 2 NOBr (g)

Fast Initial Step

• A proposed mechanism is

Step 2: NOBr2 + NO 2 NOBr (slow)

Step 1 includes the forward and reverse reactions.

Step 1: NO + Br2 NOBr2 (fast)

Fast Initial Step

• The rate of the overall reaction depends upon the rate of the slow step.

• The rate law for that step would be

Rate = k2 [NOBr2] [NO]

• But how can we find [NOBr2]?

Fast Initial Step

• NOBr2 can react two ways:

– With NO to form NOBr

– By decomposition to reform NO and Br2

• The reactants and products of the first step are in equilibrium with each other.

• Therefore,

Ratef = Rater

Fast Initial Step

• Because Ratef = Rater ,

k1 [NO] [Br2] = k−1 [NOBr2]

• Solving for [NOBr2] gives us

k1

k−1

[NO] [Br2] = [NOBr2]

Fast Initial Step

Substituting this expression for [NOBr2] in the rate law for the rate-determining step gives

k2k1

k−1

Rate = [NO] [Br2] [NO]

= k [NO]2 [Br2]

Integrated Rate Laws

Using calculus to integrate the rate law for a first-order process gives us

ln[A]t

[A]0

= −kt

Where

[A]0 is the initial concentration of A, and

[A]t is the concentration of A at some time, t, during the course of the reaction.

First-Order Processes

Therefore, if a reaction is first-order, a plot of ln [A] vs. t will yield a straight line, and the slope of the line will be -k.

ln [A]t = -kt + ln [A]0

First-Order Processes

Consider the process in which methyl isonitrile is converted to acetonitrile.

CH3NC CH3CN

First-Order Processes

This data was collected for this reaction at 198.9 °C.

CH3NC CH3CN

First-Order Processes

• When ln P is plotted as a function of time, a straight line results.

• Therefore,– The process is first-order.– k is the negative of the slope: 5.1 10-5 s−1.

Second-Order Processes

Similarly, integrating the rate law for a process that is second-order in reactant A, we get

1[A]t

= kt +1

[A]0also in the form

y = mx + b

Second-Order Processes

So if a process is second-order in A, a plot of vs. t will yield a straight line, and the slope of that line is k.

1[A]t

= kt +1

[A]0

1[A]

Second-Order ProcessesThe decomposition of NO2 at 300°C is described by the equation

NO2 (g) NO (g) + O2 (g)

and yields data comparable to this:

Time (s) [NO2], M

0.0 0.01000

50.0 0.00787

100.0 0.00649

200.0 0.00481

300.0 0.00380

12

Second-Order Processes• Plotting ln [NO2] vs. t yields

the graph at the right.

Time (s) [NO2], M ln [NO2]

0.0 0.01000 −4.610

50.0 0.00787 −4.845

100.0 0.00649 −5.038

200.0 0.00481 −5.337

300.0 0.00380 −5.573

• The plot is not a straight line, so the process is not first-order in [A].

Second-Order Processes• Graphing ln

vs. t, however, gives this plot.

Time (s) [NO2], M 1/[NO2]

0.0 0.01000 100

50.0 0.00787 127

100.0 0.00649 154

200.0 0.00481 208

300.0 0.00380 263

• Because this is a straight line, the process is second-order in [A].

1[NO2]

Half-Life

• Half-life is defined as the time required for one-half of a reactant to react.

• Because [A] at t1/2 is one-half of the original [A],

[A]t = 0.5 [A]0.

Half-Life

For a first-order process, this becomes

0.5 [A]0

[A]0

ln = −kt1/2

ln 0.5 = −kt1/2

−0.693 = −kt1/2

= t1/2

0.693kNOTE: For a first-order

process, then, the half-life does not depend on [A]0.

Half-Life

For a second-order process, 1

0.5 [A]0

= kt1/2 + 1

[A]0

2[A]0

= kt1/2 + 1

[A]0

2 − 1[A]0

= kt1/2

1[A]0

=

= t1/2

1k[A]0

Temperature and Rate

• Generally, as temperature increases, so does the reaction rate.

• This is because k is temperature dependent.

Reversible Reactions• Until now, most of the reactions we have examined have gone

completely to products. Some reactions are reversible meaning the reaction from reactants to products also goes from products to reactants at the same time.

• Consider the following reaction:2SO2 + O2 2SO3

In this reaction, SO2 and O2 are placed in a container. Initially, the forward reaction proceeds and SO3 is produced. The rate of the forward reaction is much greater than the rate of the reverse reaction. As SO3 builds up, it starts to decompose into SO2 and O2. The rate of the forward reaction is decreasing and the rate of the reverse reaction is increasing. Eventually, SO3 decomposes to SO2 and O2 as fast as SO2 and O2 combine to form SO3 (see Fig. 19.10 from text). When this happens, the reaction has achieved chemical equilibrium.

• Chemical equilibrium is when the rate of the forward reaction = the rate of the reverse reaction. It says nothing about the amounts of reactants and products at equilibrium. Simulation

Equilibrium Constants • When a system reaches equilibrium there is a mathematical

relationship between the concentrations of the products and the concentrations of the reactants.

• Pure solids and pure liquids (including water) do not appear in the equilibrium expression

• When a reactant or product is preceded by a coefficient, its concentration is raised to the power of that coefficient in the equilibrium expression.

• aA + bB cC + dD

• Kc = [C]c * [D]d

[A]a * [B]b

Sample Exercise 15.1 Writing Equilibrium-Constant Expressions

Write the equilibrium expression for Kc for the following reactions:

Solution

Analyze: We are given three equations and are asked to write an equilibrium-constant expression for each.Plan: Using the law of mass action, we write each expression as a quotient having the product concentration terms in the numerator and the reactant concentration terms in the denominator. Each concentration term is raised to the power of its coefficient in the balanced chemical equation.

Practice Exercise

The Equilibrium Constant

Since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written

Kp =(PC

c) (PDd)

(PAa) (PB

b)

Relationship Between Kc and Kp

Plugging this into the expression for Kp for each substance, the relationship between Kc and Kp becomes

where

Kp = Kc (RT)n

n = (moles of gaseous product) - (moles of gaseous reactant)

Sample Exercise 15.2 Converting between Kc and Kp

In the synthesis of ammonia from nitrogen and hydrogen,

Kc = 9.60 at 300 °C. Calculate Kp for this reaction at this temperature.

Solution

Analyze: We are given Kc for a reaction and asked to calculate Kp.Plan: The relationship between Kc and Kp is given by Equation 15.14. To apply that equation, we must determine Δn by comparing the number of moles of product with the number of moles of reactants (Equation 15.15). Solve: There are two moles of gaseous products (2 NH3) and four moles of gaseous reactants (1 N2 + 3 H2). Therefore, Δn = 2 – 4 = –2. (Remember that Δ functions are always based on products minus reactants.) The temperature, T, is 273 + 300 = 573 K. The value for the ideal-gas constant, R, is 0.0821 L-atm/mol-K. Using Kc = 9.60, we therefore have

Practice Exercise

Equilibrium Can Be Reached from Either Direction

As you can see, the ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter what the initial concentrations of NO2 and N2O4 are.

Equilibrium Can Be Reached from Either Direction

This is the data from the last two trials from the table on the previous slide.

Equilibrium Can Be Reached from Either Direction

For the equation, N2 + H2 NH3

It doesn’t matter whether we start with N2 and H2 or whether we start with NH3: we will have the same proportions of all three substances at equilibrium.

What Does the Value of K Mean?

• If K>>1, the reaction is product-favored; since products are on top of the expression when they are bigger than the reactants the constant will be greater than 1.

What Does the Value of K Mean?

If K= 1, then the reactants and products are in equal amounts.

• If K<<1, the reaction is reactant-favored; since reactants are on the bottom of the expression when they are the constant will be less than 1. .

Manipulating Equilibrium Constants

The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.

Kc = = 0.212 at 100 C[NO2]2

[N2O4]N2O4 (g)

QuickTime™ and aPhoto - JPEG decompressor

are needed to see this picture. 2 NO2 (g)

Kc = = 4.72 at 100 C[N2O4][NO2]2

N2O4 (g)

QuickTime™ and aPhoto - JPEG decompressor

are needed to see this picture.2 NO2 (g)

Sample Exercise 15.4 Evaluating an Equilibrium Constant When an Equation is Reversed

The equilibrium constant for the reaction of N2 with O2 to form NO equals Kc = 1 × 10–30 at 25 °C:

Using this information, write the equilibrium constant expression and calculate the equilibrium constant for the following reaction:

Solution

Analyze: We are asked to write the equilibrium-constant expression for a reaction and to determine the value of Kc given the chemical equation and equilibrium constant for the reverse reaction.Plan: The equilibrium-constant expression is a quotient of products over reactants, each raised to a power equal to its coefficient in the balanced equation. The value of the equilibrium constant is the reciprocal of that for the reverse reaction.Solve:Writing products over reactants, we have

Both the equilibrium-constant expression and the numerical value of the equilibrium constant are the reciprocals of those for the formation of NO from N2 and O2:Comment: Regardless of the way we express the equilibrium among NO, N2, and O2, at 25 °C it lies on the side that favors N2 and O2. Thus, the equilibrium mixture will contain mostly N2 and O2, with very little NO present.

The Concentrations of Solids and Liquids Are Essentially ConstantThe concentrations of solids and liquids do not appear in the equilibrium expression. Therefore, when a solid is dissolved in water, the expression for the solubility product constant is as follows…

Ksp = [Pb2+] [Cl-]2

Low Ksp means low solubility, high Ksp means good solubility

PbCl2 (s) Pb2+ (aq) + 2 Cl-(aq)

As long as some CaCO3 or CaO remain in the system, the amount of CO2 above the solid will remain the same.

CaCO3 (s) CO2 (g) + CaO(s)

Sample Exercise 15.6 Writing Equilibrium-Constant Expressions for Heterogeneous Reactions

Write the equilibrium-constant expression for Kc for each of the following reactions:

Solution

Analyze: We are given two chemical equations, both for heterogeneous equilibria, and asked to write the corresponding equilibrium-constant expressions.Plan: We use the law of mass action, remembering to omit any pure solids, pure liquids, and solvents from the expressions.Solve:(a) The equilibrium-constant expression is

Because H2O appears in the reaction as a pure liquid, its concentration does not appear in the equilibrium-constant expression.

(b) The equilibrium-constant expression is

Because SnO2 and Sn are both pure solids, their concentrations do not appear in the equilibrium-constant expression.

Equilibrium Constants for Reactions with more than one step

• The equilibrium constant of a net reaction that has two or more steps is found by the product of the equilibrium constants for each of the steps…

K=K1+K2+K3…

Sample Exercise 15.5 Combining Equilibrium Expressions

Given the following information,

determine the value of Kc for the reaction

Solution

Analyze: We are given two equilibrium equations and the corresponding equilibrium constants and are asked to determine the equilibrium constant for a third equation, which is related to the first two.Plan: We cannot simply add the first two equations to get the third. Instead, we need to determine how to manipulate the equations to come up with the steps that will add to give us the desired equation.Solve: If we multiply the first equationby 2 and make the corresponding change to its equilibrium constant (raising to the power 2), we get

Sample Exercise 15.5 Combining Equilibrium Expressions

Solution (continued)

Reversing the second equation and again making the corresponding change to its equilibrium constant (taking the reciprocal) gives

Now we have two equations that sum to give the net equation, and we can multiply the individual Kc values to get the desired equilibrium constant.Practice Exercise

Sample Exercise 15.8 Calculating K When All Equilibrium Concentrations Are Known

A mixture of hydrogen and nitrogen in a reaction vessel is allowed to attain equilibrium at 472 °C. The equilibrium mixture of gases was analyzed and found to contain 7.38 atm H2, 2.46 atm N2, and 0.166 atm NH3. From these data, calculate the equilibrium constant Kp for the reaction

Solution

Analyze: We are given a balanced equation and equilibrium partial pressures and are asked to calculate the value of the equilibrium constant.Plan: Using the balanced equation, we write the equilibrium-constant expression. We then substitute the equilibrium partial pressures into the expression and solve for Kp.Solve:

An aqueous solution of acetic acid is found to have the following equilibrium concentrations at 25 °C: [HC2H3O2] = 1.65 × 10–2 M; [H+] = 5.44 × 10–4 M; and [C2H3O2

–] = 5.44 × 10–4 M. Calculate the equilibrium constant Kc for the ionization of acetic acid at 25 °C. The reaction is

Answer: 1.79 × 10–5

Practice Exercise

Sample Exercise 15.9 Calculating K from Initial and Equilibrium Concentrations

A closed system initially containing 1.000 × 10–3 M H2 and 2.000 × 10–3 M I2 at 448 °C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 × 10–3 M. Calculate Kc at 448 °C for the reaction taking place, which is

Solve: First, we tabulate the initial and equilibrium concentrations of as many species as we can. We also provide space in our table for listing the changes in concentrations. As shown, it is convenient to use the chemical equation as the heading for the table.

Second, we calculate the change in concentration of HI, which is the difference between the equilibrium values and the initial values:

Solution

Analyze: We are given the initial concentrations of H2 and l2 and the equilibrium concentration of HI. We are asked to calculate the equilibrium constant Kc forPlan: We construct a table to find equilibrium concentrations of all species and then use the equilibrium concentrations to calculate the equilibrium constant.

Sample Exercise 15.9 Calculating K from Initial and Equilibrium Concentrations

Solution (continued)

Third, we use the coefficients in the balanced equation to relate the change in [HI] to the changes in [H2] and [I2]:

Fourth, we calculate the equilibrium concentrations of H2 and I2, using the initial concentrations and the changes. The equilibrium concentration equals the initial concentration minus that consumed:

The completed table now looks like this (with equilibrium concentrations in blue for emphasis):Notice that the entries for the changes are negative when a reactant is consumed and positive when a product is formed.

Sample Exercise 15.9 Calculating K from Initial and Equilibrium Concentrations

Comment: The same method can be applied to gaseous equilibrium problems to calculate Kp, in which case partial pressures are used as table entries in place of molar concentrations.

Solution (continued)Finally, now that we know the equilibrium concentration of each reactant and product, we can use the equilibrium-constant expression to calculate the equilibrium constant.

Sulfur trioxide decomposes at high temperature in a sealed container:

Initially, the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm. At equilibrium the SO3 partial pressure is 0.200 atm. Calculate the value of Kp at 1000 K.Answer: 0.338

Practice Exercise

Factors Affecting Equilibrium: Le Chatelier’s Principle

• If a stress is applied to a system at equilibrium, the system changes to relieve the stress.– Concentration– Temperature– Pressure

Concentration• Increasing the concentration of a substance causes the reaction to

shift to remove some of the extra substance. • Decreasing the concentration of a substance causes the reaction to

shift to produce some of the substance that was removed. • For example: 2SO2 + O2 2SO3

• a) If [SO2] increases by adding more, what happens the concentrations of all of the substances?

[SO2] goes up; [O2] goes down; [SO3] goes up.• Adding SO2 causes that to go up. The system tries to get rid of that

extra SO2 so some of the additional will react with O2 causing the O2 to decrease. This speeding up of the forward reaction causes more SO3 to be produced.

• b) If [SO3] decreases by removing some, what happens the concentrations of all of the substances?

[SO2] goes down; [O2] goes down; [SO3] goes down.• Removing SO3 causes the reaction to shift to the right to try to

produce more.

Temperature

• Increasing the temp. causes the equilibrium position to shift in the direction that absorbs heat (favors the endothermic reaction). Decreasing the temp. causes the equilibrium position to shift in the direction that produces heat (favors the exothermic reaction).

• For example: 2SO2 + O2 2SO3 + heat• a) If temp increases it will cause the reaction to

shift left to try to remove the extra heat so: [SO2] goes up; [O2] goes up; [SO3] goes down.

Pressure• Affects systems containing gas particles. If

pressure is increased, the reaction will shift to the side that contains the fewest number of gas particles and vice-versa.

• For example: 2SO2(g) + O2 (g) 2SO3(g) + heat

• a) If pressure is increased, the reaction will shift to the side with the fewest gas particles so it will shift to the right and:

[SO2] goes down; [O2] goes down; [SO3] goes up.

Sample Exercise 15.13 Using Le Châtelier’s Principal to Predict shifts in Equilibrium

Solution

Analyze: We are given a series of changes to be made to a system at equilibrium and are asked to predict what effect each change will have on the position of the equilibrium.Plan: Le Châtelier’s principle can be used to determine the effects of each of these changes.Solve:(a) The system will adjust to decrease the concentration of the added N2O4, so the equilibrium shifts to the right, in the direction of products.(b) The system will adjust to the removal of NO2 by shifting to the side that produces more NO2; thus, the equilibrium shifts to the right.(c) Adding N2 will increase the total pressure of the system, but N2 is not involved in the reaction. The partial pressures of NO2 and N2O4 are therefore unchanged, and there is no shift in the position of the equilibrium.(d) If the volume is increased, the system will shift in the direction that occupies a larger volume (more gas molecules); thus, the equilibrium shifts to the right. (This is the opposite of the effect observed in Figure 15.13, where the volume was decreased.)(e) The reaction is endothermic, so we can imagine heat as a reagent on the reactant side of the equation. Decreasing the temperature will shift the equilibrium in the direction that produces heat, so the equilibrium shifts to the left, toward the formation of more N2O4. Note that only this last change also affects the value of the equilibrium constant, K.

Consider the equilibrium

In which direction will the equilibrium shift when (a) N2O4 is added, (b) NO2 is removed, (c) the total pressure is increased by addition of N2(g), (d) the volume is increased, (e) the temperature is decreased?

Sample Exercise 15.13 Using Le Châtelier’s Principal to Predict shifts in Equilibrium

For the reaction

in which direction will the equilibrium shift when (a) Cl2(g) is removed, (b) the temperature is decreased, (c) the volume of the reaction system is increased, (d) PCl3(g) is added?Answer: (a) right, (b) left, (c) right, (d) left

Practice Exercise

The Reaction Quotient (Q)

• Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.

• To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression.

If Q = K,

the system is at equilibrium.

If Q > K,there is too much product, and the

equilibrium shifts to the left.

If Q < K,there is too much reactant, and the

equilibrium shifts to the right.

Sample Exercise 15.10 Predicting the Direction of Approach to Equilibrium

At 448 °C the equilibrium constant Kc for the reaction

is 50.5. Predict in which direction the reaction will proceed to reach equilibrium at 448 °C if we start with 2.0 × 10–2 mol of HI, 1.0 × 10–2 mol of H2, and 3.0 × 10–2 mol of I2 in a 2.00-L container.Solution

Analyze: We are given a volume and initial molar amounts of the species in a reaction and asked to determine in which direction the reaction must proceed to achieve equilibrium.Plan: We can determine the starting concentration of each species in the reaction mixture. We can then substitute the starting concentrations into the equilibrium-constant expression to calculate the reaction quotient, Qc. Comparing the magnitudes of the equilibrium constant, which is given, and the reaction quotient will tell us in which direction the reaction will proceed.

Solve: The initial concentrations are

The reaction quotient is therefore

Because Qc < Kc, the concentration of HI must increase and the concentrations of H2 and I2 must decrease to reach equilibrium; the reaction will proceed from left to right as it moves toward equilibrium.

Sample Exercise 15.12 Calculating Equilibrium Concentrations from Initial Concentrations

A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448 °C. The value of the equilibrium constant Kc for the reaction

at 448 °C is 50.5. What are the equilibrium concentrations of H2, I2, and HI in moles per liter?Solution

Analyze: We are given the volume of a container, an equilibrium constant, and starting amounts of reactants in the container and are asked to calculate the equilibrium concentrations of all species.Plan: In this case we are not given any of the equilibrium concentrations. We must develop some relationships that relate the initial concentrations to those at equilibrium. The procedure is similar in many regards to that outlined in Sample Exercise 15.9, where we calculated an equilibrium constant using initial concentrations.Solve: First, we note the initial concentrations of H2 and I2 in the 1.000-L flask:

Second, we construct a table in whichwe tabulate the initial concentrations:

Sample Exercise 15.12 Calculating Equilibrium Concentrations from Initial Concentrations

Solution (continued)

Third, we use the stoichiometry of the reaction to determine the changes in concentration that occur as the reaction proceeds to equilibrium. The concentrations of H2 and I2 will decrease as equilibrium is established and that of HI will increase. Let’s represent the change in concentration of H2 by the variable x. The balanced chemical equation tells us the relationship between the changes in the concentrations of the three gases:

Fourth, we use the initial concentrationsand the changes in concentrations, as dictated by stoichiometry, to express the equilibrium concentrations. With all our entries, our table now looks like this:

Sample Exercise 15.12 Calculating Equilibrium Concentrations from Initial Concentrations

Fifth, we substitute the equilibriumconcentrations into the equilibrium-constant expression and solve for the unknown, x:

If you have an equation-solving calculator, you can solve this equation directly for x. If not, expand this expression to obtain a quadratic equation in x:

Solving the quadratic equation (Appendix A.3) leads to two solutions for x:

When we substitute x = 2.323 into the expressions for the equilibrium concentrations, we find negative concentrations of H2 and I2. Because a negative concentration is not chemically meaningful, we reject this solution. We then use x = 0.935 to find the equilibrium concentrations:

Solution (continued)

Sample Exercise 15.12 Calculating Equilibrium Concentrations from Initial Concentrations

Solution (continued)Check: We can check our solution by putting these numbers into the equilibrium-constant expression to assure that we correctly calculate the equilibrium constant:Comment: Whenever you use a quadratic equation to solve an equilibrium problem, one of the solutions will not be chemically meaningful and should be rejected.Practice Exercise

Free Energy and Equilibrium Constants

• The standard state free energy of a reaction can relates to the free energy of a reaction at any moment in time according to the following equations…

ΔG°= -RT ln K

ΔG = ΔG°+ RT ln Qwhere ΔG = free energy at any moment, ΔG° = standard-

state free energy, R=ideal gas constant (8.314 J/molK), T=temperature in Kelvin