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Chapter 5  Dynamic Equilibrium

Dynamic equilibrium is achieved in a reaction when the forward reaction rate equals the reverse reaction rate, r1 = r−1.

This results in a mixture of reactants and products whose concentrations are no longer changing even though the forward and reverse reactions are still occurring.

Because dynamic equilibrium is just r1 = r−1, all non-equilibrium mixtures of reactants and products are constantly moving toward the state of dynamic equilibrium to become an equilibrium mixture.

In the case of too many reactants in the mixture the forward reaction will be going fast due to the high concentration of reactants whereas the reverse reaction will be going slower due the low concentration of products.

As reactants are converted into products, the forward reaction will continually slow down, because the reactant concentrations are decreasing, while the reverse reaction will continually go faster, because the product concentrations are increasing, until the two reaction rates meet in the middle and become equal, thus establishing dynamic equilibrium.

This scenario is roughly depicted in the following graphic.

5.1  The Equilibrium Constant

Equilibrium mixtures are quantified by the ratio of the product concentrations to the reactant concentrations as given by the Law of Mass Action which is related to kinetics, the action of "mass".

This ratio of product and reactant concentrations is called the equilibrium constant, Kc.

For a generic reaction in dynamic equilibrium as indicated by the double reaction arrow,

a A + b B ⇋ d D + f F

The equilibrium constant is defined as:

Kc = 
[D]eqd [F]eqf
[A]eqa [B]eqb

The concentrations of all reactants and products must be equilibrium concentrations ([ ]eq) as found in the equilibrium mixture.

There is no accidental similarity in the use of stoichiometric coefficients as exponents as if they are behaving as individual reaction orders.

r1 = r−1 
k1 [A]a [B]b = k−1 [D]d [F]f 
[D]d [F]f
[A]a [B]b
Kc =?

All under the approximation that the forward and reverse reactions are elementary (1-step) so that the stoichiometric coefficients can equal the individual reaction orders.

In the context of the connection to kinetics is an appropriate time to note that the concentration of solid and solvent reactants and products do not enter into the equilibrium constant expression.

The concentration of solids (i.e. the density, g/mL, converted to mol/L) is not a factor in the rate of a reaction that can be accounted for in terms of the concentration raised to some exponent because the density, regardless of its units, is not a variable; as such any effect a solid’s density would have on a reaction rate would be absorbed into the rate constant, k.


What is the equilibrium constant for the following reaction?

Cr(OH)3 (s) ⇋ Cr3+ (aq) + 3 OH (aq)
Kc =  [Cr3+] [OH]3  


The effect of the concentration of the solvent (approx the solvent density in dilute solute solutions) is accounted for by the rate constant, k, instead of by the solvent concentration raised to some exponent because the reaction is never waiting for a reactant to collide with a solvent molecule because the solvent molecules are everywhere, making accounting for a solvent’s "concentration" a bit silly and relatively meaningless.

Just for giggles, the concentration of H2O in pure water is 55.5 M.

1.00 g H2O
mol H2O
18.0152 g H2O
1000 mL
 = 55.5 M H2O 

What is the equilibrium constant for the following reaction?

NH3 (aq) + H2O (ℓ) ⇋ NH4+ (aq) + OH (aq)
Kc = 
[NH4+] [OH]



What is Kc for H2 (g) + I2 (g) ⇋ 2 HI (g) if the equilibrium concentrations are 0.400 M for both H2 and I2 and 0.200 M for HI?

Kc = 
Kc = 
(0.200  M)2
(0.400  M)(0.400  M)
 = 0.25 

Note that equilibrium constants are always treated as unit-less, although in this reaction all units of molarity cancel across the fraction anyway.


Since Kc is a ratio of product to reactant concentrations, the magnitude of Kc indicates whether the equilibrium mixture is dominated by reactants (Kc << 1), products (Kc >> 1), or is an equal mixture of both (Kc ≈ 1).

A reaction is said to be reactant favored when Kc << 1 and product favored when Kc >> 1.

The evolution of reactant and product concentrations for both types of equilibrium mixtures is depicted in the following figures.

5.1.1  Kp

For gas phase reactions, instead of concentrations, equilibrium partial pressures (which must be in units of atm!) can be used to define the equilibrium constant.

a A (g) + b B (g) ⇋ d D (g) + f F (g)
Kp = 
(PD)d (PF)f
(PA)a (PB)b

Again, these partial pressures must be in units of atm as that is the common convention which is handy because the most common reference pressure is the pressure of the atmosphere which at sea level is defined to be 1 atm.

Any other pressure unit can give the wrong value, being off by some power of the conversion factor between that pressure unit and atm.

The only instance where the same value will be obtained for Kp regardless of the pressure unit used is if there are the same number of reactant gas molecules as product gas molecules in the chemical equation; this is because such a reaction would yield a truly unit-less Kp as the pressure unit used would cancel out of the numerator and denominator of the fraction entirely.

We can use the ideal gas law to derive a relationship between Kc and Kp for gas phase reactions.

PXV = nXRT → PX = 
RT → PX = [X]RT 

Adopting a simpler reaction for the sake of efficiency

a A (g) → d D (g)
Kp = 
 = Kc(RT)da 

Generally speaking, the conversion is written in terms of the change of moles of gas molecules going from reactants to products, Δ n = nproductnreactant.

  Kp = Kc(RT)Δ n     (1)

This equation is really easy to remember because it looks just like the ideal gas law.

P = MRT 

What is Kc at 723 K if Kp = 2.20 × 104 at 723 K for the following reaction?

2 NH3 (g) ⇋ N2 (g) + 3 H2 (g)
Δ n = (1 + 3) − (2) = 4 − 2 = 2 

Solving equation 5.1 for Kc gives

Kc = 
(RT)Δ n
2.20 × 104

0.08206 L · atm/mol · K × 723 K 
 = 6.25 

What’s interesting is you can see R in there converting Kp based in atm to Kc based in mol/L = M.


Note that Kp = Kc in the event that Δ n = 0.

5.2  Manipulating Reactions and K

A generic reaction, A ⇋ D, has Kc = [D]/[A].

  1. If you reverse a reaction, the new K is the reciprocal of the original K.

    D ⇋ A, has Kcnew = [A]/[D] = 1/Kc = (Kc)−1

  2. If you multiply a reaction by a factor, the new K is the original K raised to the power of that factor.

    2 A ⇋ 2 D, has Kcnew = [D]2/[A]2 = (Kc)2

  3. If you add (subtract) two reactions together, the K of the sum of the reactions is the product (quotient) of the K from each reaction.
     A ⇋ DKc = [D]/[A]
    +D ⇋ BKcother = [B]/[D]
     A + DD + BKcsum = Kc × Kcother = [D]/[A] × [B]/[D] = [B]/[A]

Calculate the equilibrium constant for the following reaction given the information in the table.

3 O2 (g) ⇋ 2 O3 (g)
O2 (g) ⇋ 2 O (g)K1 = 5.67 × 10−6
O3 (g) ⇋ O2 (g) + O (g)K2 = 2.35 × 10−3

Ultimately, we need to add the first reaction to the second, but we must manipulate the reactions in some way first before adding them together so that the O (g) atoms cancel each out.

The answer is to multiply the reverse of the second reaction by a factor of 2 before adding it to the first reaction which is left unchanged.

 O2 (g) ⇋ 2 O (g)K1 = 5.67 × 10−6
+2 O2 (g) + 2 O (g) ⇋ 2 O3 (g)K2new = ((K2)−1)2 = ((2.35 × 10−3)−1)2 = 1.81 × 105
 3 O2 (g) + 2 O (g)2 O (g) + 2 O3 (g)Ksum = K1 × K2new = 5.67 × 10−6 × 1.81 × 105 = 1.03


5.3  Calculating Kc from Initial Concentrations and 1 Equilibrium Concentration


Calculate Kc for the following reaction knowing that the initial concentrations you started with are 6 M H2, 4 M N2, and 0 M NH3 and once an equilibrium mixture was obtained you were able to measure the equilibrium concentration of H2 to be 0.024 M.

3 H2 (g) + N2 (g) ⇋ 2 NH3 (g)

We will solve this problem twice, once the "old" way and once using a new method that is faster and will be used throughout the rest of this book.

Solution #1

The equilibrium concentrations of N2 and NH3 can be deduced from the change in the concentration of H2 and the stoichiometry of the chemical equation.

 Δ [H2] = 0.024 M − 6 M = −5.976 M H2  

Stoichiometry is then used to convert Δ [H2] to Δ [N2] and Δ [NH3].

 Δ [N2] = Δ [H2] × 
mol N2
mol H2
 = −1.992 M N2 
 Δ [NH3] = Δ [H2] × 
mol NH3
mol H2
 = +3.984 M N2 

Note that Δ [NH3] must be positive because it is being produced as a product in the reaction, whereas Δ [N2] and Δ [H2] are both negative because they are being consumed (akin to their average rates of change from the kinetics chapter).

The equilibrium concentrations of N2 and NH3 are then calculated from their initial concentrations and their change in concentrations.

 [N2]eq = [N2]0 + Δ [N2] = 4 M + −1.992 M = 2.008 M N2  
 [NH2]eq = [NH3]0 + Δ [NH3] = 0 M + 3.984 M = 3.984 M NH3  

Now Kc can be calculated from the equilibrium concentrations.

Kc = 
[H2]eq3 [N2]eq
(3.984 M NH3)2
(0.024 M H2)3(2.008 M N2)
 = 5.72 × 105 
Solution #2

The RICE problem solving method (Reaction Initial Change Equilibrium) is based on a table that organizes all of the information.

The Initial row lists initial concentrations, the Change row is simply the stoichiometric coefficient multiplied by x further modified by sign that indicates whether the concentration will decrease (−) or increase (+) as the reaction proceeds towards equilibrium, and the Equilibrium row is just the sum of the Initial and Change rows.

Reaction3 H2 (g)+N2 (g)2 NH3 (g)
Initial6 M 4 M 0 M
Change−3x x +2x
Equilibrium6 − 3x 4 − x 2x

The variable x is a measure of how much change there will be in the reaction as it proceeds towards equilibrium.

Since there are 3 H2 in the 1 reaction, there will be 3x change in [H2] by the time the reaction gets to equilibrium.

The original problem gave us [H2]eq = 0.024 which we can set equal to the Equilibrium cell for H2 from the RICE table in order to solve for x.

6 − 3x = 0.024 
x = 1.992 

This solution for x is then used to calculate the other equilibrium concentrations from their entries in the RICE table.

 [N2]eq = 4 − x = 4 − 1.992 = 2.008 M N2  
 [NH3]eq = 2x = 2(1.992) = 3.984 M NH3  

Since these are the same equilibrium concentrations obtained in Solution #1, the value of Kc is the same, Kc = 5.72 × 105, as it should be.

All we have done is present a faster method to do the same math.


5.4  Reaction Quotient, Q

The reaction quotient, Q, is the exact same mathematical expression as the equilibrium constant, K, except that it accepts any reaction mixture’s concentrations be they equilibrium, initial, or anything else, as long as all the concentrations correspond to the same time.

The utility of Q is when it is compared to K.

If a reaction mixture’s current concentrations are used to solve for Q and it turns out that Q = K, then we know the reaction mixture is in dynamic equilibrium (r1 = r−1) and those concentrations are equilibrium concentrations.

If Q < K, the reaction mixture is not at equilibrium; specifically, the reactant concentrations must be too high relative to the product concentrations which is what makes Q smaller than K.

Q  or  K ≡ 

When Q < K, with an abundance of reactants, the forward reaction must be going faster than the reverse reaction, r1 > r−1; therefore it is stated that the reaction is going "forward" as it proceeds towards equilibrium.

In a RICE table, the condition of Q < K manifests as a subtraction of x from the reactants in the Change row and an addition of x to the products.

If Q > K, the reaction mixture is not at equilibrium; specifically, the product concentrations must be too high relative to the reactant concentrations which is what makes Q larger than K.

When Q > K, with an abundance of products, the reverse reaction must be going faster than the forward reaction, r1 < r−1; therefore it is stated that the reaction is going "backward" or "in reverse" as it proceeds towards equilibrium.

In a RICE table, the condition of Q > K manifests as an addition of x to the reactants in the Change row and a subtraction of x from the products.

Perceiving Q as a number line, with a special point K indicating equilibrium, can be a useful concept.

5.5  Solving for Equilibrium Concentrations using a Known Kc

In general, we won’t be calculating Kc from equilibrium concentrations, but the other way around.


Calculate the equilibrium concentrations of all species in the reaction at 600 K given that the initial concentrations are 0.300 M CO (g), 0.500 M Cl2, and 0.700 M COCl2 and Kc = 77.5 at 600 K.

CO (g) + Cl2 (g) ⇋ COCl2 (g)

Start with a calculation of Q to figure out which way the reaction needs to go to get to equilibrium.

Qc = 
[CO]0 [Cl2]0
0.700 M
(0.300 M)(0.500 M)
 = 4.67 

Qc = 4.67 < Kc = 77.5 indicates that the concentrations of reactants are too high relative to the amount of product so the reaction needs to go forward to reach equilibrium.

This means in the RICE table, we’ll be subtracting x from reactants and adding x to products.

RCO (g)+Cl2 (g)COCl2 (g)
I0.300 0.500 0.700
Cx x +x
E0.300 − x 0.500 − x 0.700 + x

Filling the equilibrium row values into the expression for Kc yields that following equation that needs to be solved.

Kc = 
[CO]eq [Cl2]eq
0.700 + x
(0.300 − x)(0.500 − x)
 = 77.5 

Time to use all the algebra you learned in high school.

(0.700 + x) = 77.5 (0.300 − x) (0.500 − x

First Outside Inside Last

(0.700 + x) = 77.5 (0.150 − 0.300x − 0.500x + x2

Collecting the two terms in x and distributing 77.5.

(0.700 + x ) = 11.625 − 62x + 77.5x2 

Solving for zero on the left hand side.

0 = 10.925 − 63x + 77.5x2 

This quadratic equation, ax2 + bx + c = 0, can be solved using the quadratic formula, x = −b ± √b2−4ac/2a.

x = 
63 ± 
(−63)2 − 4(77.5)(10.925)
x = 0.562  (the + answer) or  0.251  (the − answer) 

The quadratic formula returns two answers, only one of which is physically meaningful.

The correct value is x = 0.251 because it is the one that results in positive equilibrium concentration values.

 [CO]eq = 0.300 − x = 0.300 − 0.251 = 0.049 M CO  
 [Cl2]eq = 0.500 − x = 0.500 − 0.251 = 0.249 M Cl2  
 [COCl2]eq = 0.700 + x = 0.700 + 0.251 = 0.951 M COCl2  

An easy way to double check your answer is to plug them in and make sure you get roughly close to the given Kc = 77.5; because of significant figures and rounding you won’t get exactly the same value.

 = 77.9 



What are the equilibrium partial pressures of all species in the following reaction at 350 K if the initial partial pressures are 0.600 atm CH4, 0.600 atm CCl4, and 0.0 atm CH2Cl2 and Kp = 0.0952 at 350 K?

CH4 (g) + CCl4 (g) ⇋ 2 CH2Cl2 (g)

Q is zero because there are no products at initial time so the reaction must go forward to reach equilibrium.

RCH4 (g)+CCl4 (g)2 CH2Cl2 (g)
I0.600 0.600 0.0
Cx x +2x
E0.600 − x 0.600 − x 2x
Kp = 
(PCO)eq (PCl2)eq
(0.600 − x)(0.600 − x)
(0.600 − x)2
 = 0.0952 

While this is a quadratic equation, we do not need the quadratic formula to solve it because we are paying attention.

We just need to take the square root of both sides and turn this quadratic equation into an easy to solve linear equation.

(0.600 − x)2
(0.600 − x)
 = 0.309 
2x = 0.309(0.600 − x
2x = 0.185 − 0.309x 
2.309x = 0.185 
x = 0.0801 
(PCO)eq = 0.600 − x = 0.600 − 0.0801 = 0.520  atm CO 
(PCl2)eq = 0.600 − x = 0.600 − 0.0801 = 0.520  atm Cl2 
(PCH2Cl2)eq = 2x = 2(0.0801) = 0.160  atm CH2Cl2 

Check your answer!

 = 0.0947 


5.6  Relearning how to work limiting reactant problems with the RICE method

The RICE method can be used to work limiting reactant problems.

Limited only by a limiting reactant, limiting reactant problems assume all reactants will turn into products, which we’ve seen equilibrium does not allow.

However, given an equilibrium constant so huge, K ≈ ∞, some reactions are so product favored that there is no trace of reactant at equilibrium, making the limiting reactant algorithm a suitable way to solve the equilibrium problem.

Because it has a huge Kc = 5.72 × 105, we will use a reaction from earlier where we will solve the equilibrium problem using the limiting reactant algorithm using a RICE table.

Reaction3 H2 (g)+N2 (g)2 NH3 (g)
Initial6 M 4 M 0 M
Change−3x x +2x
Equilibrium6 − 3x 4 − x 2x
[H2]eq3 [N2]eq
 = 5.72 × 105 ≈ ∞ 

The first step to solve this problem using the limiting reactant algorithm is to just take the denominator and set it equal to zero, since dividing by zero is the fastest way to get to infinity.

(6−3x)3(4−x) = 0 

Any exponents don’t matter at all.

(6−3x)(4−x) = 0 

There are two solutions to this equation, having set each parenthetical equation equal to the 0 on the right hand side.

(6−3x) = 0  or  (4−x)=0 
x = 2  or  x = 4 

The solution to the limiting reactant problem is the smaller of the values obtained for x.

Recall that x is how far the reaction will go to get to equilibrium, hence why you must select the smaller value; the larger value will result in negative concentrations and moles.

So the limiting reactant solution to this problem indicates that H2 is the limiting reactant and N2 is the excess reactant.

 [H2]limiting reactant = 6 − 3x = 6 − 3(2) = 0 M H2  
 [N2]limiting reactant = 4 − x = 4 − 2 = 2 M N2  
 [NH3]limiting reactant = 2x = 2(2) = 4 M NH3  

You can compare these values to the true equilibrium concentrations where x = 1.992.

 [H2]eq = 6 − 3x = 6 − 3(1.992) = 0.024 M H2  
 [N2]eq = 4 − x = 4 − 1.992 = 2.008 M N2  
 [NH3]eq = 2x = 2(1.992) = 3.984 M NH3  

If we were to do this reaction in the lab, upon reaching equilibrium we would observe a 99.6% yield for the production of NH3.

percent yield = 
 × 100 = 
 × 100 = 99.6 

We can see that for beginning chemistry students, the limiting reactant approximation is a good approximation for reactions that have enormous (product favored) equilibrium constants.

Of course, if K is not a huge number, the solution to the equilibrium problem using the limiting reactant algorithm won’t be anywhere close to accurately representing the equilibrium reaction mixture because the limiting reactant algorithm always assumes that one of the reactant concentrations, the limiting reactant to be precise, will be 0 at equilibrium.

In this example I left the units in the RICE table in molarity, but it would be best to use moles for the limiting reactant algorithm that way reactants in solid phases can, and should, be included in the RICE table.

The assumption when working equilibrium problems is that there’s always enough solid around to satisfy the equilibrium needs of the reaction which is an assumption you don’t usually make when solving a limiting reactant problem as the solid reactant may be the limiting reactant.

5.7  Le Chatelier’s Principle

A chemical system in equilibrium, when disturbed, will respond in a way to reattain equilibrium.

Fancy way of saying the forward and reverse reactions never cease, regardless of their rates.

3 ways to disturb equilibrium

  1. change the amount of 1 chemical artificially
    DisturbanceH2 (g) + I2 (g) ⇋ 2 HI (g)
    + H2too much reactant: r1 > r−1; Q<K; rxn goes "forward"
    + HItoo much product: r1 < r−1; Q>K; rxn goes "backward"
    − I2too much product: r1 < r−1; Q>K; rxn goes "backward"
    + Arnothing changes: r1 = r−1; Q=K; rxn stays at equilibrium
  2. change the volume which changes all concentrations/partial pressures simultaneously

    Both concentration and partial pressure are inversely proportional to volume.

    [X] = 
                PX = 
    nX RT

    If the volume increases, concentration and partial pressure decrease.

    If the volume decreases, concentration and partial pressure increase.

    Depending on the reaction, these changes will affect the numerator or denominator of the equilibrium constant more than the other.

    2 N2 (g) + O2 (g) ⇋ 2 N2O (g)
    Kc = 
    [N2]2 [O2]
              Kp = 
    (PN2)2 (PO2)2

    Volume changes do not affect the equilibrium of reactions that have the same number of reactant as product gases.

    Both the forward and reverse reaction rates would increase (volume decrease) or decrease (volume increase) by the same amount leaving them still equal to each other.

    H2 (g) + I2 (g) ⇋ 2 HI (g)

    Both numerator and denominator are squared in [X] and PX so the volume change doesn’t affect the ratio between reactants and products.

    Volume changes are not limited to gas phase reactions; reactions in solution can have their volumes changed by changing the amount of solvent liquid.

    HA (aq) + H2O (l) ⇋ A (aq) + H3O+ (aq)

    for liquid phase reactions it is not the moles of gas but the moles of dissolved species that track the volume change.

    For the reaction of an acid (HA) with water, water is not included in the analysis because it is the solvent and not a dissolved species.

    Qc = 

    We can easily see that the numerator and denominator will be affected differently by changes in volume.

    An Increase in volume (by just pouring in more water) → Decrease in [X] → Numerator decreases more than denominator → Q<K → "Forward" reaction to make more products → r1 > r−1 which makes sense because the forward reaction rate only depends on how often an HA runs into a H2O molecule which is always happening because water is the solvent, whereas the reverse reaction requires the collision of A and H3O+ which depends on their two concentrations which just decreased with a volume increase and so the reverse reaction rate decreased!

    It now takes longer for an A to find an H3O+ to react with due to the increased volume.


    At equilibrium, PN2O4 = 0.280 atm and PNO2 = 1.100 atm at 350 K. If the volume of this gaseous reaction mixture is doubled at constant temperature, what is the new equilibrium partial pressures for each compound?

    The reaction relating these two species is straightforward and can be written either way.

    2 NO2 (g) ⇋ N2O4 (g)

    Written this way, the equilibrium constant can be calculated from the initial equilibrium partial pressures:

    Kp = 
     = 0.2314 

    Based on the doubling of the volume at constant temperature, these previously equilibrium pressures must fall as the container becomes larger.

    P1 V1
    n1 T1
    P2 V2
    n2 T2

    Since V2 = 2 V1, T1 = T2, n1 = n2, the pressure must fall according to:

    P2 = 
    P1 V1
    P1 V1
    PNO2 = 
     = 0.550  atm 
    PN2O4 = 
     = 0.140  atm 

    This gives rise to a new reaction quotient, Qp.

    Qp = 
     = 0.4628 

    We notice that the denominator decreased faster than the numerator giving a Q > K; the reaction needs to proceed backwards to reattain equilibrium.

    R2 NO2 (g)N2O4 (g)
    E1.100 0.280
    D double volume = half the pressures 
    I0.550 0.140
    C+2x x
    E0.550 + 2x 0.140 − x
    Kp = 
    0.140 − x
     = 0.2314 
    0.140 − x = 0.2314(0.550 + 2x)(0.550 + 2x
    0.140 − x = 0.2314(0.3025 + 2.2x + 4x2
    0.140 − x = 0.0699985 + 0.50908x + 0.9256x2 
    0 = −0.0700015 + 1.50908x + 0.9256x2 
    x+ = 
      or  x− = −1.68 
    PN2O4 = 0.140 − x = 0.140 − 0.0451 = 0.0949 = 0.095 
    PNO2 = 0.550 + 2x = 0.550 + 2(0.0451) = 0.6402 = 0.640 

    Check your answer!

    Q = 
     = 0.2319 ≈ K = 0.2314 

    Close enough! Are these the same equilibrium partial pressures we had before? NO.

    The resulting equilibrium partial pressures (and concentrations) depend on their initial values and the value of K.

    K does NOT depend on initial or equilibrium partial pressures or concentrations; K can be calculated from equilibrium values, but equilibrium does not establish the same equilibrium partial pressures or concentrations every time as they depend on the initial values.


  3. changing the temperature doesn’t change [X] (though it does change PX but we’re not going there), but it does change the value of the equilibrium constant itself (Kc and Kp change)!

5.8  Temperature Dependence of K

Kc =?

Rate constants are dependent on temperature → Reaction rates depend on rate constants → Establishing dynamic equilibrium depends on equal reaction rates → Equilibrium constants are dependent on temperature as the rates of reaction are dependent on temperature → The forward and reverse reaction activation energies are different so the forward and reverse reaction rates change by different amounts when the temperature is changed.

The plot below shows how the activation energies of the forward (Ea) and reverse reactions (Ea′) are different.

The difference between those activation energies is EaEa′ = Δ E ≈ Δ H; where E is the internal energy (E = PE + KE) and H is the enthalpy (H = E + PV).

In reality, EaEa′ = Δ PE = −Δ KE assuming an isolated system (which can’t exchange energy or matter with the surroundings) where Δ E = Δ PE + Δ KE = 0 by the First Law of Thermodynamics.

However, we are not assuming an isolated system but a closed system which can exchange energy (but not matter) through the exchange of heat, q (i.e. the exchange of kinetic energy), with the surroundings.

The surroundings are approximated as a cosmic heat bath with infinite heat capacity which will therefore have a constant temperature regardless of how much heat is exchanged in or out.

The exchange of heat (kinetic energy) will affect the temperature of the system with the goal of the system reaching the same temperature as the surroundings to reach a state of thermal equilibrium between them.

Remember that the temperature is a measure of the average kinetic energy of the system (NOT the total kinetic energy); temperature is an intensive property, not an extensive property.

Therefore, the change in PE, Δ PE, for a reaction corresponds to the change in KE, Δ KE, which is reflected by a change in temperature of the system, Δ T.

For endothermic reactions, the temperature of the system decreases as reactants are converted to products due to the net conversion of kinetic energy to potential energy.

Therefore, in order to maintain the previous temperature that would be in thermal equilibrium with the surroundings, the endothermic system must absorb heat from the surroundings to compensate for the kinetic energy lost to potential energy during the reaction.

The amount of energy that must be absorbed to keep the temperature constant, and therefore how much the internal energy of the system must change, is: q = Δ E = Δ PE = −Δ KE.

It is this last realization that allows us to understand why Δ PE in the plot is also the amount of energy that needs to be absorbed as heat, q = Δ E ≈ Δ H, to keep the temperature from changing.

For exothermic reactions, the temperature of the system increases as reactants are converted to products due to the net conversion of potential energy to kinetic energy.

Therefore, in order to maintain the previous temperature that would be in thermal equilibrium with the surroundings, the exothermic system must release heat to the surroundings to remove the excess kinetic energy gained from the conversion of potential energy to kinetic energy; the amount of energy released to maintain thermal equilibrium is again equal to the change in potential energy from reactants to products, q = Δ E = Δ PE = −Δ KE.

The substitution of EaEa′ = Δ H is made below when we express the equilibrium constant in terms of the rate constants’ exponential Arrhenius form.

Kc =?
A1 eEa/RT
A−1 eEa′/RT
 e−(Ea − Ea′)/RT = 
 e−Δ H/RT 

More clearly stated:

Kc = 
 e−Δ H/RT 

The rate constants have an exponential form involving Ea; it is no surprise that the equilibrium constants also have an exponential form, but involving Δ H.

The linear form of this relationship is:

ln(Kc) = −


+ ln


A plot of ln(Kc) versus 1/T gives a slope, m = −Δ H/R, just like a plot of ln(k) versus 1/T gave a slope, m = −Ea/R.

Measuring an equilibrium constant at two different temperatures can allow you to calculate Δ H for a reaction via the slope, m.

m = 
ln(K2) − ln(K1)
 = −

Rearranging this gives the 2 point form of the van’t Hoff equation for equilibrium constants (which looks a lot like the 2 point form the Arrhenius equation for rate constants).



= −



When Δ H > 0 (positive, like Ea), K increases when T increases (just like k) and vice versa.

This relationship can be seen in an Arrhenius type plot of ln(K) versus 1/T.

An example of an endothermic reaction is vaporization; energy is required to break the intermolecular interactions between molecules in the liquid phase.

H2O (l) ⇋ H2O (g) Δ Hvaporization = +44.0 kJ/mol > 0

This reaction has the simplest equilibrium constant ever being equal to the partial pressure of the water vapor, a quantity that already has a name in the context of dynamic equilibrium, the vapor pressure, P.

Kp = PH2O 

Substituting P for for K and using Δ Hvap (enthalpy/heat of vaporization) in the 2 point form of the van’t Hoff equation (equation 5.2) gives the Clausius-Clapeyron Equation which describes the variation of the vapor pressure with temperature.



= −
Δ Hvap



We all know vapor pressures, and therefore equilibrium constants of vaporization reactions, increase with increasing temperature.

If you remember, you have to get the temperature high enough to get to the boiling point which is just the temperature at which the vapor pressure equals the external/atmospheric pressure!

Increase T → Increase P = Kp → Increase K.


The vapor pressure of water is 23.76 torr at 25C with an enthalpy of vaporization, Δ Hvap = 44.0 kJ/mol. What is the normal boiling point of water?

The problem wants us to find the temperature at which Kp = P = Patmosphere = 1 atm = 760 torr since it’s asking for the normal (1 atm) boiling point.

With 25C = 298.15 K, substituting all this information into the Clausius-Clapeyron equation (equation 5.3) gives:


(760  torr)2
(23.76  torr)1

= −
44.0  kJ/mol
8.314 × 10−3  kJ/mol·K

298.15  K

Solving for T2 is kind of irritating but you should get T2 = 370.5 K = 97.4C; this higher temperature has the higher Kp = P = 760 torr.

We don’t get exactly 100C (the actual normal boiling point of water) because the enthalpy of vaporization is not a constant value (it is 40.1 kJ/mol at 100C), but it’s close enough to assume it’s constant for our purposes.


When Δ H < 0 (negative, NOT like Ea), K decreases when T increases and vice versa.

This relationship can be seen in an Arrhenius type plot of ln(K) versus 1/T.

An example of an exothermic reaction is dissolving a gas into a liquid.

Such a reaction is exothermic because in order for the gas molecules to form intermolecular interactions with the liquid molecules, they must release their kinetic energy so that they slow down enough to become a part of the liquid phase.

In short, if breaking bonds is endothermic, making bonds is exothermic; "bonds" here refers to both intramolecular bonding (covalent or ionic) and intermolecular bonding (hydrogen bonds, dipole-dipole, London dispersion forces, etc).

For example, dissolving carbon dioxide in water.

CO2 (g) ⇋ CO2 (aq) Δ Hdissolution = −20.3 kJ/mol < 0

This reaction has a really weird equilibrium constant because it has two phases, one best represented with partial pressure and the other with concentration, so we mix them.

K = 
 = kH 

kH is Henry’s Law constant; Henry’s Law states that the solubility of a gas in a liquid ([CO2] above) is directly proportional to the partial pressure of the gas (PCO2 above).

Rearranging the above equation we get the mathematical equivalent of Henry’s Law so that we can see the direct proportionality between the concentration of the gas CO2 in the liquid phase and the partial pressure of the gas that exists over the liquid phase; the higher the partial pressure the more of the gas will be dissolved in the liquid:

 Solubility of CO2 = [CO2]  = kH  PCO2  

Unlike other equilibrium constants, Henry’s Law constant, kH, has units of M/atm because it was originally just an experimental observation derived outside of the context of equilibrium constants.

At room temperature, 25C, kH = 3.4 × 10−2 M CO2/atm CO2 and kH = 58 M NH3/atm NH3, illustrating the importance of intermolecular forces in determining the solubility of gases; that is to say the solubility is not just determined by the partial pressure in the gas phase, but also on the identity of the gas.

Of course, there’s nothing about Henry’s Law and temperature in these equations, but kH is the equilibrium constant for an exothermic reaction so its value depends on the temperature.


In order to keep your 3.0 L soda from going flat after opening it and letting all of the CO2 gas they pumped into the void space out, should you keep it in the refrigerator or out on the shelf?

We want to make equilibrium be more product favored so that more CO2 stays in the water.

Since the dissolution of gas is exothermic, in order to obtain a more product favored equilibrium (higher K), we need to decrease the T to "trap" the reaction on the product side so that the CO2 can’t escape the liquid phase because it doesn’t have enough kinetic energy to escape at lower temperatures.

So, that’s why you store your soda in the refrigerator.

If you just wanted it to be cold when you drink it, you could just add ice.

But what we want, after opening it, is to keep as much CO2 in the soda as possible so it doesn’t go flat; we’re not refrigerating it to make it cold so it’s more refreshing, we’re refrigerating it to keep it cold so that the CO2 stays dissolved in solution.

We could say that the solubility of CO2 in water increases with decreasing temperature because kH increases with decreasing temperature.


The solubility of salts (ionic compounds) also has a temperature dependence which can be related to the enthalpy of their dissolution reactions.

Those that have positive enthalpies of dissolution (endothermic) see their solubility increase, because of increasing Kc, with increasing T.

Those with negative enthalpies of dissolution (exothermic) see their solubility decrease, because of decreasing Kc, with increasing T.

Of course, chemistry is the science of exceptions.

LiOH has an exothermic dissolution at Δ H = −23.6 kJ/mol but an increasing solubility with increasing temperature.

There’s obviously more to the story than assuming Δ H is constant, and that’s assuming Δ S is constant with changing temperature, which neither are.

You’ll have to skip ahead to Thermodynamics to get the rest of that story.

5.9  How to solve *some* problems quickly

If x appears once in the numerator, K is less than 1, and [reactants]0 > K, this might work:


Solve for the equilibrium concentrations of all species in the reaction if the initial concentration of H2S is 2.5 × 10−4 M with all others equal to zero.

2 H2S (g) ⇋ 2 H2 (g) + S2 (g) with Kc = 1.67 × 10−7 at 800C.
R2 H2S (g)2 H2 (g)+S2 (g)
I2.5 × 10−4 0 0
C−2x +2x +x
E2.4 × 10−4 − 2x 2x x
Kc = 
[H2]2 [S2]
1.67 × 10−7 = 
(2x)2 (x)
(2.5 × 10−4 − 2x)2
1.67 × 10−7 = 
(2.5 × 10−4 − 2x)2

All conditions are satisfied.

  1. x appears only once in the numerator.
  2. K = 1.67 × 10−7 < 1
  3. [H2S]0 = 2.5 × 10−4 > K = 1.67 × 10−7

The equation to solve the equilibrium problem is cubic which would require use of the cubic formula which is very cumbersome.

Instead, because K << 1, the distance to equilibrium, x, is also very small meaning the reaction mixture will be mostly made of reactants which we already have a bunch of since at initial time there are no products present.

Assuming x is negligible, x ≈ 0, allows us to make an approximation for the denominator.

(2.5 × 10−4 − 2x) ≈ (2.5 × 10−4

Substituting this approximation into the original equilibrium problem returns an equation with only 1 x term making it very easy to solve for x without having to use the cubic formula.

1.67 × 10−7 = 
(2.5 × 10−4)2
x = 
1.67 × 10−7(2.5 × 10−4)2
 = 1.377 × 10−5 

If we plug this value of x into the original equilibrium expression we get a rather large error in the known equilibrium constant.

Kc = 
4(1.377 × 10−5)3
(2.5 × 10−4 − 2(1.377 × 10−5))2
 = 2.11 × 10−7 

This is a 26.3% error from the known value so this x is not fantastic.

2.11 × 10−7 − 1.67 × 10−7
1.67 × 10−7
 = 0.263 

We can use the method of iteration to solve this problem.

We’ll take the solution obtained for x and substitute it for the x in the denominator and again solve for the x in the numerator.

1.67 × 10−7 = 
(2.5 × 10−4 − 2(1.377 × 10−5))2
x = 
1.67 × 10−7(2.5 × 10−4 − 2(1.377 × 10−5))2
 = 1.274 × 10−5 

Repeating the procedure again gives another iteration.

1.67 × 10−7 = 
(2.5 × 10−4 − 2(1.274 × 10−5))2
x = 
1.67 × 10−7(2.5 × 10−4 − 2(1.274 × 10−5))2
 = 1.282 × 10−5 

Repeating the pattern again, with the goal being that the x we put into the denominator is the same value for x that is returned by solving for the x in the numerator.

1.67 × 10−7 = 
(2.5 × 10−4 − 2(1.282 × 10−5))2
x = 
1.67 × 10−7(2.5 × 10−4 − 2(1.282 × 10−5))2
 = 1.281 × 10−5 

In this last iteration, we used x = 1.282 × 10−5 in the denominator and solving for the x in the numerator gave, x = 1.281 × 10−5.

To 3 significant figures, which is all we care about in this problem, the iterative method has converged giving x = 1.28 × 10−5.

We’ll use this method of iteration in the next chapter to solve quadratic equations because it’s simply faster than using the quadratic formula.
When this method doesn’t work:

  1. K ≈ [reactants]; instead of converging, the iterative solutions of x diverge farther and farther from each other; sometimes the averaged answers of diverging solutions gives a good solution but it is very slow
  2. K = ax3 + bx2 + cx + d/[reactants] − x; because there’s more than 1 x term in the numerator meaning you can’t just take a root to solve for x

Lastly, what if K >> 1?

Trying reversing the reaction, Kreverse = Kforward−1.

If Kforward > 1, then Kreverse < 1 and you’re set.


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