Chapter 1: Chemical Equilibrium Part 10 SABIS Grade 11 (Level M) Chemistry

1.7 Quantitative Aspects of Equilibrium

Le Chatelier’s Principle permits the chemist to make qualitative predictions about the equilibrium state. Despite the usefulness of such predictions, they represent far less than we wish to know. It is a help to know that raising the pressure will favor production of NH3. But how much will the pressure change favor NH3 production? To control a reaction, we need quantitative information about equilibrium. Experiments show that quantitative predictions are possible and they can be explained in terms of our view of equilibrium on the molecular level.

1.7.1 The Equilibrium Constant

By means of colorimetric determination in the laboratory you measured the concentration of FeSCN2+, which we shall designate [FeSCN2+], in solutions containing iron (III) and thiocyanate ions, Fe3+ and SCN–. The reaction is:

Fe3+(aq) + SCN–(aq) ⇌ FeSCN2+(aq)

From [FeSCN2+] and the initial values of [Fe3+] and [SCN–] you calculated the values of [Fe3+] and [SCN–] at equilibrium. You then made calculations for various combinations of these values. Many experiments just like these show that the ratio

[FeSCN2+][Fe3+][SCN−]${\displaystyle \frac{[{\mathrm{F}\mathrm{e}\mathrm{S}\mathrm{C}\mathrm{N}}^{2+}]}{[{\mathrm{F}\mathrm{e}}^{3+}][{\mathrm{S}\mathrm{C}\mathrm{N}}^{-}]}}$

comes closest to being a fixed value. Note that this ratio is the quotient of the equilibrium concentration of the single substance produced in the reaction divided by the product of the equilibrium concentrations of the reactants.

Colorimetric analysis based on visual estimation is not very exact. Some more accurate data on the H2, I2, HI system at equilibrium are shown in Table 1.1 The reaction is:

2HI(g) ⇌ H2(g) + I2(g)

The data have been expressed in concentrations, although pressure units are more usual for a reaction involving gases.

Let’s work with this data. Using results from your own laboratory data, let us compute the value of the ratio:

[H2][I2][HI]2${\displaystyle \frac{[{\mathrm{H}}_2][{\mathrm{I}}_2]}{[\mathrm{H}\mathrm{I}{]}^2}}$

Table 1.1 Equilibrium concentration at 698.6K of hydrogen, iodine, and hydrogen iodide.

*Values above the line were obtained by heating hydrogen and iodine together: values below the line, by heating pure hydrogen iodide.

The ratio is the product of the equilibrium concentrations of the substances produced in the reaction [H2] × [I2], divided by the square of the concentration of the reacting substance, [HI]2. In this ratio, the power to which we raise the concentration of each substance is equal to its coefficient in the reaction.

The results imply that with a fair degree of accuracy we can write:

[H2][I2][HI]2${\displaystyle \frac{[{\mathrm{H}}_2][{\mathrm{I}}_2]}{[\mathrm{H}\mathrm{I}{]}^2}}$ = a constant = 1.835 × 10–2 at 698.6 K

Look at this ratio in terms of the reaction:

2HI(g) ⇌ H2(g) + I2(g)

For the experiments 4 and 5 in Table, [H2] = [I2] since H2 and I2

•  have the same number of atoms = 2.
•  have the same coefficient.
•  are both elements.

answer :have the same coefficient.

1.7.2 The Law Of Chemical Equilibrium

Let us summarize what we have learned. For the reaction:

Fe3+(aq) + SCN–(aq) ⇌ FeSCN2+(aq)        (1)

We found that at equilibrium, the concentrations of the molecules involved have a simple relationship:

[FeSCN2+][Fe3+][SCN−]=a constant${\displaystyle \frac{[{\mathrm{F}\mathrm{e}\mathrm{S}\mathrm{C}\mathrm{N}}^{2+}]}{[{\mathrm{F}\mathrm{e}}^{3+}][{\mathrm{S}\mathrm{C}\mathrm{N}}^{-}]}=\mathrm{a}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}}$ (2)

Then we considered precise equilibrium data for the reaction

2HI(g)               ⇌                  H2(g)               +           I2(g)           (3)

The equilibrium concentrations of the molecules appearing in the reaction were found to have a simple relationship:

[H2][I2][HI]2=a constant${\displaystyle \frac{[{\mathrm{H}}_2][{\mathrm{I}}_2]}{[\mathrm{H}\mathrm{I}{]}^2}=\mathrm{a}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}}$       (4)

In each of our simple relationships 2 and 4 the concentrations of the products appear in the numerator, the concentrations of reactants appear in the denominator. In reaction (3), two molecules of hydrogen iodide react. This influences expression (4) because it is necessary to square the concentration of hydrogen iodide, [HI], in order to obtain a constant ratio.

These observations and many others like them lead to the generalization known as the Law of Chemical Equilibrium. For the reaction:

aA + bB ⇌ eE + fF

when equilibrium exists, there will be a simple relation between the concentrations of products, [E] and [F], and the concentrations of reactants, [A] and [B]:

[E]e[F]f[A]a[B]b=K=a constant at constant temperature${\displaystyle \frac{[\mathrm{E}{]}^{\mathrm{e}}[\mathrm{F}{]}^{\mathrm{f}}}{[\mathrm{A}{]}^{\mathrm{a}}[\mathrm{B}{]}^{\mathrm{b}}}=\mathrm{K}=\mathrm{a}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{a}\mathrm{t}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{t}\mathrm{e}\mathrm{m}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{e}}$

The expression  Q=[E]e[F]f[A]a[B]b${\displaystyle \mathrm{Q}=\frac{[\mathrm{E}{]}^{\mathrm{e}}[\mathrm{F}{]}^{\mathrm{f}}}{[\mathrm{A}{]}^{\mathrm{a}}[\mathrm{B}{]}^{\mathrm{b}}}}$ at that instant, called the mass action expression, or the reaction quotient, has a value equal to the equilibrium constant K at equilibrium. If its value is not equal to the constant (at constant temperature) then equilibrium does not exist.

The law of equilibrium can be stated as such: At equilibrium, the reaction quotient (Q) is equal to the equilibrium constant (K).

In this generalized equation, we see that again the numerator is the product of the equilibrium concentrations of the substances formed, each raised to the power equal to the number of moles of that substance in the chemical equation. The denominator is again the product of the equilibrium concentrations of the reacting substances, each raised to a power equal to the number of moles of the substance in the balanced chemical equation. The quotient of these two remains constant at equilibrium. The constant K is called the equilibrium constant. This generalization is one of the most useful in all of chemistry. From the equation for any chemical reaction one can immediately write an expression for Q, in terms of the concentrations of reactants and products, that will be constant at any given temperature at equilibrium. If this constant is measured (by measuring all of the concentrations in a particular equilibrium solution), then it can be used in calculations for any other equilibrium solution at that same temperature.

Table 1.2 lists some reactions along with the equilibrium law relation of concentrations and the numerical values of the equilibrium constants. First, let’s verify the forms of the equilibrium law relation among the concentrations.

Table 1.2 Some equilibrium constants.

The very first has an unexpected form. For the reaction

Cu(s) + 2Ag+(aq) = Cu2+(aq) + 2Ag(s)

you do not find:

[Cu2+][Ag]2[Ag+]2[Cu]=K${\displaystyle \frac{[{\mathrm{C}\mathrm{u}}^{2+}][\mathrm{A}\mathrm{g}{]}^2}{[{\mathrm{A}\mathrm{g}}^{+}{]}^2[\mathrm{C}\mathrm{u}]}=\mathrm{K}}$      (5)

but rather, you find:

[Cu2+][Ag+]2=K${\displaystyle \frac{[{\mathrm{C}\mathrm{u}}^{2+}]}{[{\mathrm{A}\mathrm{g}}^{+}{]}^2}=\mathrm{K}}$ (6)

This is because the concentrations of solid copper and solid silver are incorporated into the equilibrium constant. The concentration of solid copper is fixed by the density of the metal—it cannot be altered either by the chemist or by the progress of the reaction. The same is true of the concentration of solid silver. Since neither of these concentrations varies, no matter how much solid is added, there is no need to write them each time an equilibrium calculation is made.

* Another K of unexpected form applies to the reaction

H2O(l) ⇌ H+(aq) + OH–(aq)

For this reaction we might have written

[H+][OH−][H2O]=K${\displaystyle \frac{[{\mathrm{H}}^{+}][{\mathrm{O}\mathrm{H}}^{-}]}{[{\mathrm{H}}_2\mathrm{O}]}=\mathrm{K}}$

Instead, Table 1.2 lists the above expression as

[H+][OH–] = K

The concentration of water, [H2O], does not appear in the denominator of the above expression. This is usually done in treating aqueous reactions that consume or produce water. It is justified because the variation in the concentration of water during reaction is so slight in dilute aqueous solutions. We can treat [H2O] as a concentration that does not vary. Hence, [H2O] can be incorporated in the equilibrium constant.

In summary, the concentrations of solids and the concentrations of solvent (usually water) can be and usually are incorporated in the equilibrium constant, so they do not appear in the equilibrium law relation.

1.7.3 Large And Small Equilibrium Constants

Look at the numerical values of the equilibrium constants. The constants (K) listed in Table 1.2 range from 10+15 to 10–16, so we see there is a wide variation. We want to acquire a sense of the relation between the size of the equilibrium constant and the state of equilibrium.

A large value of Keq

Determine the number of moles produced given the number of moles of reactant.

Let us look at the reaction between copper metal and silver ions at 25°C:

Cu(s) + 2Ag+(aq) ⇌ Cu2+(aq) + 2Ag(s)        Keq = 2 × 1015

Suppose we place excess copper in a liter of solution containing 1 mole of silver ions. Initially the concentration of copper ions is zero. Let us say that by the time equilibrium has been reached, x moles (per liter) of Ag+(aq) would have reacted. According to the balanced equation, x/2 moles of Cu2+(aq) would have been produced:

Cu(s)   +   2Ag+(aq)     ⇌     Cu2+(aq)   +   2Ag(s)

Initial concentration                                             1.0 M                   ——
Part that reacts                                                      x                        x/2
Equilibrium concentration                                    (1 – x)                  x/2

To find the final concentrations if Keq is large, given initial concentrations.

The law of equilibrium for this equation is:

K=2×1015=[Cu2+][Ag+]2${\displaystyle \mathrm{K}=2\times {10}^{15}=\frac{[{\mathrm{C}\mathrm{u}}^{2+}]}{[\mathrm{A}{\mathrm{g}}^{+}{]}^2}}$

Substituting gives: 2×1015=[Cu2+][Ag+]2=x/2(1−x)2${\displaystyle 2\times {10}^{15}=\frac{[{\mathrm{C}\mathrm{u}}^{2+}]}{[\mathrm{A}{\mathrm{g}}^{+}{]}^2}=\frac{\mathrm{x}/2}{(1-\mathrm{x}{)}^2}}$     

2×1015=x/2(1−x)2${\displaystyle 2\times {10}^{15}=\frac{\mathrm{x}/2}{(1-\mathrm{x}{)}^2}}$       (1)

Since x is less than 1, the numerator, x/2 is less than 0.5. The only way for the above fraction to be so large is for (1– x) to be very small. So x must be almost equal to 1, and x/2 must equal to 0.5 (to one significant figure). Expression 1 above can now be rewritten as:

 2×1015=0.5(1−x)2${\displaystyle 2\times {10}^{15}=\frac{0.5}{(1-\mathrm{x}{)}^2}}$       (1)

Rearranging:   (1 − x)2 = 2.5 × 10−16

                        (1 − x) = 2.5 × 10⁻⁸

                        x = 1 − 1.6 × 108 = 1 (to one significant figure)

This confirms the fact that we can assume that x has the value of 1.0, to two significant figures.

A large K means that the reaction proceeds to the right practically completely.

We conclude that a large value of K must mean that at equilibrium there are much larger concentrations of products present than of reactants.

In this case, when equilibrium is finally reached, the concentration of Cu2+ ion, [Cu2+], is very much greater than the concentration of [Ag+].

Look at the following equation and choose the correct answer:

Ag+(aq)  +  2 NH3(aq)   ⇌   Ag(NH3)2+(aq)      K = 1.7 × 107 at 25°C

Excess ammonia (NH3) is added to 0.1 M solution of Ag+(aq). At equilibrium, the [Ag(NH3)2+(aq)] is expected to be:

0.1 M

0.2 M

unknown, information given is not enough to approximate it.

very small

answer : 0.2 M

A small value of Keq

Case 1:

Let us look at the reverse of the reaction considered above.

Cu2+(aq) + 2Ag(s) ⇌ Cu(s) + 2Ag+(aq),      K’eq at 25°C = 1/(2.0 × 1015)

Its constant is the reciprocal of the constant of the original equation, so it is very small. We have seen that at equilibrium, the concentration of Cu2+(aq) is large and the concentration of Ag+(aq) is small, so when the equilibrium constant is small a very small amount of products will be found at equilibrium.

Case 2:

To calculate [H+] in pure water, given Kw

Let us look at the dissociation of water at 25°C:

H2O(l) ⇌ H+(aq) + OH–(aq)       Keq = 1 × 10–14

Let us say that x moles of water dissociate in every liter (x mol/liter)

H2O(l)      ⇌      H+(aq)    +    OH−(aq)

Initial concentrations                                                             0                   0

Part that reacts                                             x

Equilibrium concentration                                                      x                   x

The law of equilibrium for this equation is:

[H+][OH–] = 1 × 10–14

x. x = 1 × 10–14

x = 1 × 10–7 M

This means that in one liter (dm3) of water, only 10–7 moles of water dissociate into ions. Since one liter of water contains 55.5 moles of water (1000g/18g/mol), the percentage decomposition is:

(1 × 10–7 / 55.5) × 100 = 2 × 10–7 %. One of 55,000,000 molecules dissociates!

We generalize that a small value of K for a given reaction implies that very little of the products have to be formed from the reactants before equilibrium is attained.

Summary

If Keq ≤ 1, then position of equilibrium is to the left.

If Keq ≥ 1, then position of equilibrium is to the right.

If Keq ≈ 1, there will be appreciable concentration of both products and reactants present at equilibrium.

Equilibrium constants are given for several systems below. In which case does the reaction as written occur to the greatest extent?

CdS(s)  ⇌  Cd2+(aq)  +  S2−(aq)                  K = 7.1 × 10−28

H+(aq)  +  HS−(aq)   ⇌   H2S(aq)                 K =  1.0 × 107

CH3COOH(aq)  ⇌  H+(aq)  +  CH3COO−(aq)                 K = 1.8 × 10−5

answer :

H+(aq)  +  HS−(aq)   ⇌   H2S(aq)                 K =  1.0 × 107 1.7.4 Equilibrium Concentrations Is a system at equilibrium? Example 1 Consider the following reaction, at 490ºC, (Keq = 45.9)                                         H2(g)      +      I2(g)        ⇌       2HI(g) Initial concentrations     0.005M          0.004M                0.01M  At a certain instant in a reaction container maintained at 490ºC, the following concentrations were determined: [HI] = 0.01 mole/dm3, [H2] = 0.005 mole/dm3, [I2] = 0.004 mole/dm3. (a) Is the system at equilibrium? (b) If not, which concentrations are increasing and which are decreasing? Challenge Problem (c) Find the final, equilibrium concentrations of all species. Solution (a) Is the system at equilibrium? If the mass action expression Q is equal to Keq then the system is at equilibrium Q=[HI]2[H2][I2]=[0.01]2[0.005][0.004]=5${\displaystyle \mathrm{Q}=\frac{[\mathrm{H}\mathrm{I}{]}^2}{[{\mathrm{H}}_2][{\mathrm{I}}_2]}=\frac{[0.01{]}^2}{[0.005][0.004]}=5}$ Since Q < Keq, the system is not at equilibrium  (b) If not, which concentrations are increasing and which are decreasing? As the system proceeds towards equilibrium, Q should approach Keq and eventually become equal to it. For this to happen, the fraction [HI]2[H2][I2]${\displaystyle \frac{[\mathrm{H}\mathrm{I}{]}^2}{[{\mathrm{H}}_2][{\mathrm{I}}_2]}}$ should increase in value until it becomes 45.9. For this to happen the numerator must increase in value and the denominator decrease in value, which means that the concentration of HI should increase and the concentrations of each of H2 and I2 should decrease. Therefore the reaction proceeds to the right. (c) Find the final, equilibrium concentrations of all species.                                              H2(g)      +      I2(g)      ⇌      2HI(g) Initial concentrations           0.005M         0.004M             0.01M Part that reacts                        x                    x                    2x Equilibrium concentration   0.005−x        0.004−x            0.01+2x Substituting in the law of equilibrium: Keq=[HI]2[H2][I2]=45.9${\displaystyle {\mathrm{K}}_{\mathrm{e}\mathrm{q}}=\frac{[\mathrm{H}\mathrm{I}{]}^2}{[{\mathrm{H}}_2][{\mathrm{I}}_2]}=45.9}$, gives: [0.01+2x]2[0.005−x][0.004−x]=45.9${\displaystyle \frac{[0.01+2\mathrm{x}{]}^2}{[0.005-\mathrm{x}][0.004-\mathrm{x}]}=45.9}$ Solving for x gives two solutions, one of them should be ruled out. A negative value for x means that products form reactants. x1 = 8.52 × 10–3 rejected, x2 = 2.29 x 10–3 accepted, [H2] = 0.005– x = 2.71 × 10–3 M, [I2] = 0.004– x = 1.71 × 10–3 M, [HI] = 0.01 + 2x = 0.0146 M Consider the following reaction, at 490°C, (Keq = 45.9):                                                         H2(g)    +    I2(g)   ⇌   2HI(g) Initial concentrations                     0.005M      0.004M       0.01M At a certain instant in a reaction, the container is maintained at 490°C, the following concentrations were determined: [HI] = 0.01 mol/dm3 [H2] = 0.005 mol/dm3 [I2] = 0.004 mol/dm3 Is the system at equilibrium?  Yes No answer : No

Consider the following reaction, at 490°C, (Keq = 45.9):

                                                H2(g)    +    I2(g)   ⇌    2HI(g)

Initial concentrations             0.005M      0.004M        0.01M

At a certain instant in a reaction, the container is maintained at 490°C, the following concentrations were determined:

[HI] = 0.01 mol/dm3

[H2] = 0.005 mol/dm3

[I2] = 0.004 mol/dm3

Which concentrations are increasing and which are decreasing?

[H2]

             Choose...             is decreasing.             is increasing.         

[I2]

             Choose...             is decreasing.             is increasing.         

[HI]

             Choose...             is decreasing.             is increasing.         

 answer :

[H2] = is decreasing. [I2] = is decreasing. [HI] = is increasing. Example 2 1.0 mole of each of H2(g) and I2(g) are injected into a 2.0 dm3 container at 490ºC. Find the equilibrium concentration of each species. Solution                                              H2(g)      +      I2(g)      ⇌      2HI(g) Initial concentrations           0.50M            0.50M               0.0M Part that reacts                       − x                − x                  + 2x Equilibrium concentration   0.50−x           0.50−x                2x The law of equilibrium can be written as shown below. Substitute Keq=[HI]2[H2][I2]=45.9${\displaystyle {\mathrm{K}}_{\mathrm{e}\mathrm{q}}=\frac{[\mathrm{H}\mathrm{I}{]}^2}{[{\mathrm{H}}_2][{\mathrm{I}}_2]}=45.9}$ Gives: [2x]2[0.5−x][0.5−x]=45.9${\displaystyle \frac{[2\mathrm{x}{]}^2}{[0.5-\mathrm{x}][0.5-\mathrm{x}]}=45.9}$ Solving for x gives x = 0.39 or 0.71. The second solution is imposible. [H2] = 0.5 − x = 0.11 M; [I2] = 0.5 − x = 0.11 M; [HI] = 2x = 0.78 M Summary If Q < Keq, a net forward reaction (to the right) can occur. If Q > Keq, a net backward reaction (tj the left) can occur. If Keq ≈ Q, the system is at equilibrium and do no net reaction occur. 2 moles each of H2 and I2 and 20 moles of HI were injected into a 2 dm3 cylinder at 490°C. To find the equilibrium concentration of HI in the cylinder, answer the following questions. H2 + I2  ⇌  2HI                  Keq = 45.9 [HI] =              Choose...             100             10 M             9.26 M             1.0 M          [I2] =              Choose...             100             10 M             9.26 M             1.0 M          [H2] =              Choose...             100             10 M             9.26 M             1.0 M          The value of Q =              Choose...             100             10 M             9.26 M             1.0 M          At equilibrium, [HI] =              Choose...             100             10 M             9.26 M             1.0 M          Keq for the reaction SO2(g) + NO2(g)  ⇌  SO3(g) + NO(g) at 500°C is 3.0. Into a 10 dm3 flask at 500°C 2 moles each of SO3(g) and NO(g) were introduced. Calculate the concentration of all 4 gases when equilibrium is reached. [SO2] =              Choose...             0.073 M             0.13 M          [SO3] =              Choose...             0.073 M             0.13 M          [NO2] =              Choose...             0.073 M             0.13 M          [NO] =              Choose...             0.073 M             0.13 M          answer : [SO2] = = 0.073 M [SO3] = = 0.13 M [NO2] = = 0.073 M [NO] = = 0.13 M 2.0 moles HI are injected into a 2.0 dm3 container at 490°C. Determine the equilibrium concentration of each species. H2(g)     +   I2(g)  ⇌   2HI(g)          Keq = 45.9 [H2] =              Choose...             0.11 M             0.78 M          [I2] =              Choose...             0.11 M             0.78 M          [HI] =              Choose...             0.11 M             0.78 M           answer : [H2] = = 0.11 M [I2] = = 0.11 M [HI] = = 0.78 M Next part