Chapter 15:  Equilibrium

 

 

Section 15-1: Equilibrium Constant, Kc and Kp, Expressions

Section 15-2: Equilibrium Calculations and RICE Charts

Section 15-3: Reaction Quotients, Qc and Qp

Section 15-4: Heterogeneous Equilibria

Section 15-5: Le Chteliers Principle

Section 15-6: Converting Between Kc and Kp

Section 15-7: Relationship Between G and Equilibrium Constants

Section 15-8: The van't Hoff Equation - Effect of Temperature on Equilibrium Constants

Section 15-9: The Clausius-Clapeyron Equation - Effect of Temperature on Vapor Pressure

Chapter 15 Practice Exercises and Review Quizzes

 

 

 

Section 15-1:  Equilibrium Constant, Kc and Kp, Expressions

 

As soon as a chemical reaction is initiated, the amount of each reactant on the left side of the equation will begin to decrease, and the rate of the forward reaction (toward the right) will begin to decrease.  At the same time, the amount of each product on the right side of the equation will begin to increase, and the rate of the reverse reaction (toward the left) will begin to increase.  Eventually, the rates of the forward and reverse reactions will become equal and, therefore, the amounts of the reactants and products will remain constant.  A reaction that reaches this point is said to be at equilibrium, which can be indicated by replacing the forward arrow () in the chemical equation with a double equilibrium arrow ().

 

For the discussion that follows, the notation [X] in an equation indicates that the concentration of X in moles/liter or M should be inserted into the equation.  The notation PX in an equation indicates that the pressure of X in atm should be inserted into the equation.

 

For a reaction aA + bB cC + dD, where the lower case letters a through d represent coefficients, the amounts of the products relative to the reactants present at equilibrium will be indicated by the values Kc and Kp, known as equilibrium constants, according to the following expressions:

 

 

Note that the Kp expression will not apply if the reaction does not contain gases.

 

Sample Exercise 15A:

 

Write the Kc and Kp (only if applicable) expressions for each reaction:

 

(a) N2 (g) + 3 F2 (g) 2 NF3 (g)

(b) Sn2+ (aq) + 2 Fe3+ (aq) Sn4+ (aq) + 2 Fe2+ (aq)

 

Solution:

 

 

Once we have written a Kc expression, we can substitute the equilibrium molarities for each reactant and product to calculate the numerical value of Kc.  Similarly, once we have written a Kp expression, we can substitute the equilibrium pressures in atm for each reactant and product to calculate the numerical value of Kp:

 

Sample Exercise 15B:

 

For reaction (a) above, calculate the value of Kp given the following equilibrium pressures:

 

nitrogen gas = 0.125 atm

fluorine gas = 0.85 atm

nitrogen trifluoride gas = 0.40 atm

 

Solution:

 

Equilibrium constant values are typically expressed without units, so we substitute the pressure in atm of each reactant and product, but without including units, into the Kp expression:

 

 

If we know the value of Kp and all but one pressure in the expression, we can calculate the unknown pressure.  Similarly, if we know the value of Kc  and all but one molarity in the expression, we can calculate the unknown molarity:

 

Sample Exercise 15C:

 

For reaction (b) above, Kc = 37.  Calculate the equilibrium molarity of iron(III) ions given the following equilibrium molarities:

 

tin(II) = 0.12 M

tin(IV) = 0.28 M

iron(II) = 0.36 M

 

Solution:

 

 

Note that we must always include a unit when solving for an equilibrium molarity or pressure.

 

 

If a reaction is written in the reverse direction, the equilibrium constant for the new reaction will be the reciprocal of the original equilibrium constant:

 

A B     equilibrium constant = K

B A     equilibrium constant = 1/K

 

If the coefficients of a reaction are multiplied by an integer or fraction x, the equilibrium constant for the new reaction will have a value equal to the original equilibrium constant raised to the x power:

 

A B     equilibrium constant = K

xA xB     equilibrium constant = (K)x

 

Sample Exercise 15D:

 

For the reaction N2 (g) + 3 F2 (g) 2 NF3 (g), the value of Kp = 2.1.  Calculate the value of Kp for the following reaction:

 

NF3 (g) 1/2 N2 (g) + 3/2 F2 (g)

 

Solution:

 

To obtain the new reaction, the original reaction was reversed and also the coefficients were multiplied by 1/2.  Therefore, we can either take the reciprocal of the original Kp value and then raise the result to the 1/2 power, or we can raise the original Kp value to the 1/2 power and then take the reciprocal of the result.  The same answer will be achieved using either method:

 

 

 

If two or more reactions are added, the equilibrium constant for the new reaction will the be product of the equilibrium constants for each reaction that was added:

 

A B     equilibrium constant = K1

C D     equilibrium constant = K2

A + C B + D     equilibrium constant = (K1)(K2)

 

 

 

Section 15-2:  Equilibrium Calculations and RICE Charts

 

In the numerical examples above, the given molarities could be substituted into the Kc expression and the given pressures could be substituted into the Kp expression because they represented the amounts of reactants and products at equilibrium.  If instead we are given initial molarities or pressures before the reaction has begun, we must use stoichiometry to determine how the molarity or pressure of each reactant and product changes as the reaction proceeds until equilibrium is reached.  Only after we have determined the final molarity or pressure of each reactant and product at equilibrium can we substitute those values into the Kc or Kp expression.

 

A Reaction-Initial-Change-Equilibrium or RICE chart is a convenient way to organize equilibrium stoichiometry calculations.  The four lines of a RICE chart are as follows:

 

R = Write the reaction.

 

I = Write the initial molarities (for calculations involving Kc) or pressures in atm (for calculations involving Kp) for each reactant and product, but units may be omitted on the RICE chart.

 

C = Write the change in the molarity or pressure of each reactant and product as the reaction proceeds.  We will begin completing this line by representing the change in the molarity or pressure of one reactant or product as +x if the change is known to be an increase or –x if the change is known to be decrease.  We will then use the coefficients of the reaction and stoichiometric ratios to deduce the change in all other reactants and products in terms of x.

 

E = Write the molarity or pressure of each reactant and product at equilibrium in terms of x.  Each value on this line will simply be the sum of the values on the Initial and Change lines (I + C = E).

 

After completing a RICE chart, if the molarity or pressure of one reactant or product at equilibrium has been given, we can determine the numerical value of x.  Once x is known, we can deduce the molarities or pressures of all other reactants and products at equilibrium, after which we can substitute these values into the Kc or Kp expression to find the numerical values of Kc or Kp:                 

 

Sample Exercise 15E:

 

(a) Write the Kc expression for the reaction 2 H2S (g) + CH4 (g) CS2 (g) + 4 H2 (g).

(b) H2S gas with an initial concentration of 0.37 M and methane gas with an initial concentration of 0.42 M are placed in an empty container and a reaction is initiated.  When equilibrium is reached, the concentration of hydrogen gas is found to be 0.52 M.  Calculate the concentrations of each gas at equilibrium and also the value of Kc for the reaction.

 

Solution:

 

(a)

 

(b) We begin by creating a RICE chart showing the initial molarities of each reactant and product.  Note that the initial molarities of only H2S and CH4 were specified, so we can assume that CS2 and H2 each have an initial molarity equal to zero.  Since the initial molarities of CS2 and H2 are zero and, therefore, must increase as the reaction proceeds, we make the signs of the changes positive for CS2 and H2.  If the signs of the changes on the right side are positive, then the signs of the changes on the left side must be negative to represent a decrease in the molarities of H2S and CH4 as the reaction proceeds:

 

R

      2 H2S (g)     +          CH4 (g)               CS2 (g)      +        4 H2 (g)

I

0.37

0.42

0

0

C

-

-

+

+

E

 

 

 

 

 

If we define the change in molarity of CS2 as +x, then according to the 4:1 stoichiometric ratio from the coefficients of the reaction, the change in molarity of H2 will be +4x.  Using the same method, the change in molarity of H2S will be -2x and the change in molarity of CH4 will be –x.  After completing the C line, we add the I and C lines to obtain the values in the E line in terms of x:

 

R

      2 H2S (g)     +          CH4 (g)               CS2 (g)      +        4 H2 (g)

I

0.37

0.42

0

0

C

-2x

-x

+x

+4x

E

0.37-2x

0.42-x

x

4x

 

From the information given in the problem, we know that the equilibrium molarity of H2 is 0.52 M, which allows us to determine the numerical value of x and equilibrium molarities of the other gases as follows:

 

[H2] = 0.52 M = 4x

x = 0.13 M = [CS2]

[H2S] = 0.37-2x  =  0.37-2(0.13) = 0.11 M

[CH4] = 0.42-x = 0.42-0.13 = 0.29 M

 

Finally, we substitute the equilibrium molarities of each gas into the Kc expression to obtain the value of Kc:

 

 

 

If we are given the value of Kc or Kp in a RICE chart problem rather than any of the equilibrium molarities or pressures, we can solve the Kc or Kp expression for x:

 

Sample Exercise 15F:

 

(a) Write the Kp expression for the reaction CO (g) + 2 H2 (g) CH3OH (g).

(b) Carbon monoxide gas with an initial pressure of 38 atm and hydrogen gas with an initial pressure of 56 atm are placed in an empty container and a reaction is initiated.  If Kp = 9.5 x 10-3 for this reaction, calculate the equilibrium pressure of each gas.

(c) Calculate the total pressure at equilibrium.

 

Solution:

 

(a)

 

(b) First, we complete the RICE chart in terms of x:

 

R

        CO (g)        +        2 H2 (g)           CH3OH (g)     

I

38

56

0

C

-x

-2x

+x

E

38-x

56-2x

x

 

We then substitute the equilibrium pressures in terms of x into the Kp expression and solve for x, after which we can determine the equilibrium pressures of each gas:

 

 

The equation above can be solved for x using a graphing calculator, keeping in mind that the value of x must be always be a real number greater than zero and also that the value of x cannot be so large as to result in a negative equilibrium molarity or pressure for any reactant or product.  For the equation being solved above, since x > 0 and 56-2x > 0, we know that the correct root must have 0 < x < 28.

 

(c) To find the total pressure at equilibrium, we simply add together the equilibrium pressures of all the gases:

 

Ptotal = 16 atm + 12 atm + 22 atm = 50. atm

 

 

If one or more of the reactants on the left side of an equation have an initial molarity or pressure equal to zero, then the reaction will proceed in the reverse direction and, thus, the RICE chart changes will be positive for each of the reactants on the left while being negative for each of the products on the right:

 

Sample Exercise 15G:

 

(a) Write the Kc expression for the reaction Sn4+ (aq) + 2 Cr2+ (aq) Sn2+ (aq) + 2 Cr3+ (aq).

(b) A reaction mixture is prepared with the following initial concentrations:

 

chromium(II) = 0.46 M

tin(II) = 0.65 M

chromium(III) = 0.72 M

 

If Kc = 0.40 for this reaction, calculate the equilibrium molarity of each ion.

 

Solution:

 

(a)

 

(b)

R

       Sn4+ (aq)    +    2 Cr2+ (aq)          Sn2+ (aq)       +     2 Cr3+ (aq)

I

0

0.46

0.65

0.72

C

+x

+2x

-x

-2x

E

x

0.46+2x

0.65-x

0.72-2x

 

 

 

(x > 0 and 0.72-2x > 0, so 0 < x < 0.36)

 

 

Section 15-3:  Reaction Quotients, Qc and Qp

 

For a reaction aA + bB cC + dD, if none of the reactants or products have an initial molarity or pressure equal to zero and if we are not given the molarity or pressure of any reactant or product at equilibrium, we cannot determine by simple inspection whether the reaction proceeds in the forward or reverse direction to reach equilibrium.  In these cases, we must first calculate a reaction quotient Qc or Qp by entering the initial (i) molarities or pressures of each reactant and product into one of the following expressions:

 

 

After Qc or Qp has been calculated, we can determine the direction that the reaction must proceed to reach equilibrium as follows:

 

If Qc < Kc or Qp < Kp, the reaction must proceed in the forward direction to reach equilibrium.  Therefore, the RICE chart changes for the products on the right will be positive while the changes for the reactants on the left will be negative.

 

If Qc > Kc or Qp > Kp, the reaction must proceed in the reverse direction to reach equilibrium.  Therefore, the RICE chart changes for the reactants on the left side will be positive while the changes for the products on the right side will be negative.

 

Note that if Qc = Kc or Qp = Kp, then the reaction is already at equilibrium and, therefore, the initial molarities or pressures given are equal to the equilibrium molarities or pressures.

 

Sample Exercise 15H:

 

(a) For the reaction 2 SO2 (g) + O2 (g) 2 SO3 (g), Kp = 0.85.  In which direction must this reaction proceed to reach equilibrium starting with the following initial pressures?:

 

sulfur dioxide gas = 0.66 atm

oxygen gas = 0.51 atm

sulfur trioxide gas = 0.98 atm

 

(b) Calculate the equilibrium pressure of each gas.

(c) Calculate the total pressure at equilibrium.

 

Solution:

 

(a) We first calculate Qp and then compare to Kp:

 

 

Since Qp > Kp, the reaction will proceed in the reverse direction to reach equilibrium.

 

(b)

R

      2SO2 (g)      +         O2 (g)                2 SO3 (g)     

I

0.66

0.51

0.98

C

+2x

+x

-2x

E

0.66+2x

0.51+x

0.98-2x

 

 

 

(x > 0 and 0.98-2x > 0, so 0 < x < 0.49)

 

(c) Ptotal = 0.94 atm + 0.65 atm + 0.70 atm = 2.29 atm

 

 

Section 15-4:  Heterogeneous Equilibria

 

Reactions that involve only a single state of matter are said to be homogeneous.  All of the reactions above that involve only gaseous or only aqueous reactants and products are homogeneous.  Reactions that involve more than one state of matter are said to be heterogeneous.  When Kc, Kp, Qc, and Qp expressions for heterogeneous reactions are written, solids and liquids are omitted from the expressions:

 

Sample Exercise 15I:

 

(a) Write the Kc and Kp expressions for the reaction 4 KO2 (s) + 2 CO2 (g) 2 K2CO3 (s) + 3 O2 (g).

(b) Calculate Kp for this reaction if the equilibrium pressure of carbon dioxide gas is 13 atm and the equilibrium pressure of oxygen gas is 17 atm.

 

Solution:

 

(a)

(b)

 

 

 

On a RICE chart for a heterogeneous reaction, the columns under solids and liquids will be left blank.  However, we can still calculate changes in the amounts of solids and liquids as the reaction proceeds using standard stoichiometry with mole ratios:

 

Sample Exercise 15J:

 

(a) Write the Kp expression for the decomposition of solid ammonium carbonate to form ammonia gas, carbon dioxide gas, and liquid water: 

(NH4)2CO3 (s) 2 NH3 (g) + CO2 (g) + H2O (l)

(b) After a 9.85 gram sample of solid ammonium carbonate was placed in a 75.0 mL container and heated to 335C, the total pressure at equilibrium was found to be 49.8 atm.  Calculate the equilibrium pressure of each gas and Kp for this reaction.

(c) Calculate the number of water molecules and the mass of solid ammonium carbonate present at equilibrium. 

 

Solution:

 

(a)

 

(b)

R

 (NH4)2CO3 (s)        2 NH3 (g)    +         CO2 (g)      +          H2O (l)

I

 

0

0

 

C

+2x

+x

E

2x

x

 

 

 

(c) We first use the Ideal Gas Law to convert the equilibrium pressure of ammonia to find the moles of ammonia formed in the reaction.  Then we can use a mole ratio from the balanced equation to find the moles of water formed in the reaction, after which we use Avogadros number to convert to water molecules present at equilibrium.  Next, we again use a mole ratio to find the moles of ammonium carbonate that reacted to make the products, after which we use the molar mass to convert to grams.  Finally, we subtract the mass of ammonium carbonate reacted from the initial mass to find the final mass of ammonium carbonate present at equilibrium:

 

 

Note that we could have arrived at the same answers by first finding the moles of carbon dioxide gas formed in the reaction rather than first finding the moles of ammonia gas formed in the reaction.

   

 

 

 

Section 15-5:  Le Châteliers Principle

 

A variety of factors may cause a reaction at equilibrium to temporarily fall out of equilibrium.  Le Châteliers Principle essentially states that a reaction which has fallen out of equilibrium will respond by proceeding at unequal forward and reverse rates until equilibrium has been reestablished.  A reaction that proceeds faster in the forward direction is said to shift to the right.  When a reaction shifts to the right, the amount of products on the right side of the equation will increase and the amount of reactants on the left side of the equation will decrease.  A reaction that proceeds faster in the reverse direction is said to shift to the left.  When a reaction shifts to the left, the amounts of reactants on the left side of the equation will increase and the amount of products on the right side of the equation will decrease.  According to Le Châteliers Principle, the following guidelines predict what will occur when a variety of changes are made to a reaction at equilibrium:

 

1a. (g) or (aq) reactant is added = shifts right

1b. (g) or (aq) product is added = shifts left

 

2a. (g) or (aq) reactant is removed = shifts left

2b. (g) or (aq) product is removed = shifts right

 

3a. (s) or (l) reactant or product is added = no shift

3b. (s) or (l) reactant or product is removed = no shift

 

4a. Volume of reaction container is increased = shifts in direction having more moles of (g) [sum of coefficients for (g) only is greater]

4b. Volume of reaction container is decreased = shifts in direction having fewer moles of (g) [sum of coefficients for (g) only is less]

 

5a. Water is added to increase volume of solution = shifts in direction having more moles of (aq) [sum of coefficients for (aq) only is greater]

5b. Water is evaporated to decrease volume of solution = shifts in direction having fewer moles of (aq) [sum of coefficients for (aq) only is less]

 

6. Catalyst is added = both forward and reverse rates increase but remain equal = no shift

 

7a. Inert (unreactive) gas (for example:  noble gases helium, neon, argon) added with no change in volume of reaction container = no shift

7b. Inert gas added with no change in pressure = volume must increase = shifts in direction having more moles of (g)

 

8a. Temperature is increased = shifts in endothermic direction (H > 0)

8b. Temperature is decreased = shifts in exothermic direction (H < 0)

 

Note that the H value listed after a reaction refers to the forward direction.  Therefore, for the reaction A B   H < 0, the reaction is exothermic in the forward direction and endothermic in the reverse direction. 

 

Of the changes listed above, only a change in temperature will change the value of Kc and Kp.  If the reaction shifts right due to a temperature change, then Kc and Kp will increase.  If the reaction shifts left due to a temperature change, Kc and Kp will decrease.

 

Sample Exercise 15K:

 

Consider the reaction:

2 H2O (g) + 3 S (s) 2 H2S (g) + SO2 (g)   H > 0

 

(a) State whether the amount of H2S gas present at equilibrium will increase, decrease, or remain unchanged when each of the following occurs:

 

i. Sulfur dioxide gas is added.

ii. Water vapor is removed.

iii. Solid sulfur is removed.

iv. The container volume is increased.

v. A catalyst is added.

vi. Argon gas is added at constant volume.

vii. Helium gas is added at constant pressure.

viii. The temperature is decreased.

 

(b) Of the changes above, which will change the value of Kc and Kp, and will Kc and Kp increase or decrease?

 


Solution:

 

(a)

i. Add (g) product  = shifts left = amount of H2S decreases (guideline 1b).

ii. Remove (g) reactant = shifts left = amount of H2S decreases (guideline 2a).

iii. Remove (s) reactant = no shift = amount of H2S unchanged (guideline 3b).

iv. Volume of container increases = shifts to side with more (g) moles.  Although total moles is larger on left (5 3), there are more (g) moles on the right (2 3), so shifts right = amount of H2S increases (guideline 4a). 

v. Catalyst is added = no shift = amount of H2S unchanged (guideline 6).

vi. Inert gas added with no volume change = no shift = amount of H2S unchanged (guideline 7a).

vii. Inert gas added with no pressure change = volume increases = shifts to side with more (g) moles = shifts right = amount of H2S increases (guideline 7b).

viii. Temperature decreases = shifts in exothermic direction = shifts left = amount of H2S decreases (guideline 8b).

 

(b) Only a temperature change will change the value of Kc and Kp.  Reaction shifts left when temperature is decreased, so Kc and Kp decrease.

 

 

 

Section 15-6:  Converting Between Kc and Kp


We can use the following equation to convert between Kc and Kp:

 

Kp = Kc(RT)n of gas

 

When using the equation above, we will not show units in the calculation, but we must use R = 0.0821 and the temperature T in K.  The exponent “n of gas” is found by subtracting the sum of only (g) coefficients for reactants on the left side of the equilibrium reaction from the sum of only (g) coefficients for products on the right side of the equilibrium reaction:

 

Sample Exercise 15L:


If Kp = 7.5 x 104 at 65°C for the reaction S (s) + 2 HI (g)  I2 (s) + H2S (g), calculate Kc for the reaction at 65°C.  

 

Solution:

 

Do not include solids: 

 

7.5 x 104 = Kc(0.0821 x 338)1 - 2 = -1

Kc = 2.1 x 106

 

We will demonstrate converting from Kc to Kp in Sample Exercise 15M.

 

 

 

Section 15-7:  Relationship Between G° and Equilibrium Constants


A chemical reaction with a more negative G° will have larger values of Kc and Kp and, therefore, larger amount of products relative to reactants at equilibrium.  A chemical reaction with a more positive G° will have smaller values of Kc and Kp and, therefore, smaller amount of products relative to reactants at equilibrium.  Although Kc and Kp for a given reaction are known as equilibrium constants, note that their values will change when the temperature is changed.

 

We can use the following equation to convert between G° and an equilibrium constant:

 

G° = -RTlnKeq

 

When using the equation above, we must use R = 8.31 J/mol•K and the temperature T in K.  For reactions with at least one (aq) reactant or product but no (g), Kc should be substituted for Keq in the equation.  For reactions with at least one (g) reactant or product but no (aq), Kp should be substituted for Keq in the equation.  We will not focus on reactions that involve both (aq) and (g).

 

Sample Exercise 15M:


If Kc = 0.00155 at 175°C for the reaction NO2 (g) NO (g) + 1/2 O2 (g), calculate G° for the reaction at 175°C.  

 

Solution:

 

Since the reaction involves (g) but no (aq), first convert Kc to Kp and then substitute Kp for Keq to calculate G°:

 

Kp = 0.00155(0.0821 x 448)(1 + 1/2) - 1 = 1/2 = 0.00940

G° = -8.31 J/mol•K(448 K)ln(0.00940) = 1.74 x 104 J/mol = 17.4 kJ/mol 

 

Sample Exercise 15N:


If G° = 15.6 kJ/mol at 25°C for the reaction LiF (s) Li+ (aq) + F- (aq), calculate Kc for the reaction at 25°C.  

 

Solution:

 

Since the reaction involves (aq) but no (g), substitute Kc for Keq.  Also, use G° in J/mol:

 

1.56 x 104 J/mol = -8.31 J/mol•K(298 K)lnKc

Kc = 0.00184 

 

 

Section 15-8:  The van’t Hoff Equation – Effect of Temperature on Equilibrium Constants


Given that G° = -RTlnKeq and G° = H° - TS°, then -RTlnKeq = H° - TS°.  Dividing both sides of this equation by -RT yields the following equation relating the equilibrium constant to the temperature in K (with R = 8.31 J/mol•K and assuming that both H° and S° remain constant as the temperature changes):

 

lnKeq = -H°/R (1/T) + S°/R

 

At a given temperature T1, ln(Keq)1 = -H°/R (1/T1) + S°/R.  At a different temperature T2, ln(Keq)2 = -H°/R (1/T2) + S°/R.  By subtracting the two equations and rearranging, we obtain the van’t Hoff equation:

 

ln(Keq)2 - ln(Keq)1 = -H°/R (1/T2) + S°/R - [-H°/R (1/T1) + S°/R]

 

 

If we know the equilibrium constant for a reaction at one temperature as well as H°, we can use the van’t Hoff equation to estimate the equilibrium constant at a second temperature as follows:

 

Sample Exercise 15O:


Given that Kp = 1.1 x 10-6 at 675°C for the reaction below, estimate Kp for the reaction at 575°C:

 

3 H2 (g) + CO (g) H2O (g) + CH4 (g)     H° = -206 kJ/mol

 

Solution:

 

Keq = Kp

 

 

Section 15-9:  The Clausius-Clapeyron Equation – Effect of Temperature on Vapor Pressure


Vapor pressure is the pressure of a gas in equilibrium with the liquid state of the same substance:

 

X (l) X (g)     Keq = Kp = Px = vapor P and H° = H°vaporization

 

Substituting vapor pressure for Keq yields a special case of the van’t Hoff equation known as the Clausius-Clapeyron equation:

 

 

If we know the vapor pressure of a substance at one temperature as well as H°vaporization, we can use the Clausius-Clapeyron equation to estimate the vapor pressure at a second temperature.

 

A substance will boil when its vapor pressure equals the atmospheric pressure.  The normal boiling point of a substance is the temperature at which the vapor pressure equals exactly 1 atm or 760 mmHg.  If we know the normal boiling point and H°vaporization for a substance, we can use the Clausius-Clapeyron equation to estimate the vapor pressure at a second temperature as follows:  

 

Sample Exercise 15P:


The normal boiling point of aniline is 184°C.  If H°vaporization = 55.0 kJ/mol for aniline, estimate the vapor pressure of aniline in mmHg at 165°C.

 

Solution:

 

At normal boiling point 184°C, vapor pressure of aniline = 760 mmHg:

 

 

 

 

Chapter 15 Practice Exercises and Review Quizzes:

 

15-1) (a) Write the Kc and Kp expressions for the reaction Hg (l) + 1/2 O2 (g) HgO (s).

           (b) If Kp = 9.5 for this reaction, calculate the equilibrium pressure of oxygen gas.

 

Click for Solution

 

15-1) (a) Heterogeneous reaction = omit solids and liquids from Kc and Kp expressions:

 

 

          (b)

 

 

 

15-2) For the reaction Hg (l) + 1/2 O2 (g) HgO (s), the value of Kc = 47.  Calculate the value of Kc for the following reaction:

 

2 HgO (s) 2 Hg (l) + O2 (g)

 

Click for Solution

 

15-2) Reverse original reaction and multiply coefficients by 2 = either take reciprocal of original Kc and then raise result to the 2 power or raise original Kc to the 2 power and then take reciprocal of result:

 

 

 

 

15-3) (a) Write the Kp expression for the reaction 4 HCl (g) + O2 (g) 2 Cl2 (g) + 2 H2O (g).

           (b) HCl gas with an initial pressure of 0.17 atm and oxygen gas with an initial pressure of 0.58 atm are placed in a container along with chlorine gas with an initial pressure of 0.92 atm and water vapor with an initial pressure of 0.83 atm.  When equilibrium is reached, the pressure of HCl gas is found to be 0.81 atm.  Calculate the equilibrium pressures of each gas, the total pressure at equilibrium, and Kp for the reaction.

 

Click for Solution

 

15-3) (a)

 

           (b) Since the pressure of HCl has increased from 0.17 atm to 0.81 atm, then the reaction must have proceeded in the reverse direction to reach equilibrium.  Therefore, the changes for the reactants on the left side of the RICE chart will be positive while the changes for the products on the right side will be negative:

 

R

       4 HCl (g)    +         O2 (g)               2 Cl2 (g)      +       2 H2O (g)

I

0.17

0.58

0.92

0.83

C

+4x

+x

-2x

-2x

E

0.17+4x

0.58+x

0.92-2x

0.83-2x

 

 

 

 

15-4) For the reaction 2 Al (s) + 3 Sn4+ (aq) 2 Al3+ (aq) + 3 Sn2+ (aq), Kc = 0.0011.  Solid aluminum metal is added to a solution containing the initial concentrations 0.18 M Sn4+, 0.59 M Al3+, and 0.97 M Sn2+.  Calculate the molarity of each ion at equilibrium.

 

Click for Solution

 

15-4) Heterogeneous reaction = omit solid from Qc expression, RICE chart, and Kc expression:

 

 

Since Qc > Kc, the reaction will proceed in the reverse direction to reach equilibrium:

 

R

        2 Al (s)      +     3 Sn4+ (aq)        2 Al3+ (aq)     +     3 Sn2+ (aq)

I

 

0.18

0.59

0.97

C

+3x

-2x

-3x

E

0.18+3x

0.59-2x

0.97-3x

 

 

 

 

 

15-5) (a) Write the Kc expression for the decomposition of solid sodium bicarbonate to form solid sodium oxide, carbon dioxide gas, and water vapor: 

2 NaHCO3 (s) Na2O (s) + 2 CO2 (g) + H2O (g)

           (b) An 8.2 gram sample of solid sodium bicarbonate was partially decomposed in a 225 mL container.  If Kc = 0.0040 for the decomposition reaction, calculate the equilibrium molarity of each gas and the mass of each solid present at equilibrium. 

 

Click for Solution

 

15-5) (a) Kc = [CO2]2[H2O]

           (b)

R

 2 NaHCO3 (s)         Na2O (s)     +        2CO2 (g)     +          H2O (g)

I

 

 

0

0

C

+2x

+x

E

2x

x

 

0.0040 = (2x)2(x)

x = 0.10 M = [H2O]

[CO2] = 2(0.10 M) = 0.20 M

 

 

 

15-6) Consider the reaction:

4 H2 (g) + Fe3O4 (s) 4 H2O (g) + 3 Fe (s)   H > 0

 

(a) State whether the amount of water vapor present at equilibrium will increase, decrease, or remain unchanged when each of the following occurs:

 

i. Solid iron metal is added.

ii. A catalyst is added.

iii. The volume of the container is decreased.

iv. Neon gas is added at constant volume.

v. The temperature is increased.

vi. Hydrogen gas is removed.

vii. Argon gas is added at constant pressure

 

(b) Of the changes above, which will change the value of Kc and Kp, and will Kc and Kp increase or decrease?

 

Click for Solution

 

15-6)

(a)

i. Add (s) product = no shift = amount of water vapor unchanged (guideline 3a).

ii. Add catalyst = no shift = amount of water vapor unchanged (guideline 6).

iii. Volume of container decreases = shifts to side with fewer (g) moles.  Although total moles is less on left (5 7), there are equal (g) moles on both sides (4 4), so no shift = amount of water vapor unchanged (guideline 4b).

iv. Inert gas added with no volume change = no shift = amount of water vapor unchanged (guideline 7a).  

v. Temperature increases = shifts in endothermic direction = shifts right = amount of water vapor increases (guideline 8a).

vi. Remove (g) reactant = shifts left = amount of water vapor decreases (guideline 2a).

vii. Inert gas added with no pressure change = volume increases = shifts to side with more (g) moles = no shift because equal (g) moles on both sides = amount of water vapor unchanged (guideline 7b).

 

(b) Only a temperature change will change the value of Kc and Kp.  Reaction shifts right when temperature increases, so Kc and Kp increase.

 

 

 

15-7) If Kp = 0.18 at 975°C for the reaction CO (g)  1/2 CO2 (g) + 1/2 C (s), calculate Kc for the reaction at 975°C.

Click for Solution

 

15-7) Do not include solid:

 

0.18 = Kc(0.0821 x 1248)1/2 - 1 = -1/2

Kc = 1.8  

 

 

 

15-8) If Kc = 0.00645 at 675°C for the reaction 3H2 (g) + CO (g) H2O (g) + CH4 (g), calculate G° for the reaction at 675°C.

Click for Solution

15-8) Since the reaction involves (g) but no (aq), first convert Kc to Kp and then substitute Kp for Keq to calculate G°:

 

Kp = 0.00645(0.0821 x 948)(1+1) - (3+1) = -2 = 1.06 x 10-6

G° = -8.31 J/mol•K(948 K)ln(1.06 x 10-6) = 1.08 x 105 J/mol = 108 kJ/mol 

 

 

 

15-9) If G° = 10.6 kJ/mol at 25°C for the reaction HClO2 (aq) H+ (aq) + ClO2- (aq), calculate Kc for the reaction at 25°C. 

Click for Solution

15-9) Since the reaction involves (aq) but no (g), substitute Kc for Keq.  Also, use G° in J/mol:

 

1.06 x 104 J/mol = -8.31 J/mol•K(298 K)lnKc

Kc = 0.0138

 

 

 

15-10) Given that Kp = 6.4 x 104 at 425°C for the reaction below, estimate the temperature in °C at which Kp = 1.3 x 104:

 

O2 (g) + 2 SO2 (g) 2 SO3 (g)     H° = -198 kJ/mol

Click for Solution

15-10)

Keq = Kp

 

 

15-11) The vapor pressure of cyclohexane at 66°C is 463 mmHg.  If H°vaporization = 33.0 kJ/mol for cyclohexane, determine the normal boiling point of cyclohexane in °C.  

Click for Solution

15-11)

At normal boiling point, vapor pressure of cyclohexane = 760 mmHg:

 

 

 

Click for Review Quiz 1

Click for Review Quiz 1 Answers