13.4 The Equilibrium Constant

The chemical equilibrium of ammonia production from hydrogen and nitrogen gases can evaluated and assessed by introduction the equilibrium constants called reaction quotient Q.

N₂(g)    +  3 H₂(g)          2 NH₃(g)

Reaction quotient Qc    = [ NH (g) ] ² / [ N2(g)] x [H2(g) ³ ]

The subscript c is denoting the molar concentrations and designated by the large brackets [ …… ].

Also, the reaction quotient can be expressed in Qp when the reactants and the products are gaseous and their partial pressures P are used:

reaction quotient Qp = ( P NH₃(g) ) ² / ( P N₂(g) ) x ( P H₂(g) ) ³

The videos below illustrate the concept of the reaction quotient Q p and Q c:

Some extra examples:

Find the Qc for the following chemical equilibria:

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

• HCl(g)    +    H2O(l)       H3O ⁺ (aq)  + Cl ^– (aq)

• NH4OH (s) +    H2O(l)     NH4 + (aq)      +      OH ^– (aq)

• CO2 (g)     +    H2O(l)         H2CO3 (aq)

Solutions:

1. Q c =  [ CO2 ]

Note that all solids (s) and liquids (l) phases are discarded from the reaction quotient Qc

expression.

Only aqueous (aq) and gases (g) phases are considered.

This chemical equilibrium is considered as heterogenous equilibrium.

• Q C = { [ H₃O ⁺ ] x [ Cl ^– ] } / [ HCl ]

This chemical equilibrium is considered as homogeneous equilibrium. The solvent is water

• Q C = [ NH4 + ] x [ OH ^– ]

This chemical equilibrium is considered as homogeneous equilibrium. The solvent is water

• Qc = [ H₂CO₃ ] / [ CO₂ ]

In general Qc has a large value when the concentrations of the products are high and the reactants concentrations are low. The reaction is said to favor the products

Qc has a small value when the concentrations of the products are low and the reactants concentrations are large. The reaction is said to favor the reactants.

This chemical equilibrium is considered as homogeneous equilibrium. The solvent is water.

When the chemical equilibrium is at equilibrium, then the equilibrium constant K will equal the reaction quotient Q.

Examples:

1. Consider the following chemical equilibrium: CO₂ (g)    +    H₂O(l)        H2CO3 (aq)

If the 0.550 moles of CO2 is added in 1.50 L flask at room temperature. At the equilibrium, it was found that CO2 concentration to be 0.250 M and the H2CO3 concentration at equilibrium was found to be 0.125 M.

Calculate:

1. What is the value of the reaction quotient before any reaction occurs?

• (b) What is the value of the equilibrium constant for the reaction?

Solution:

1. The reaction quotient before any reaction occurs:

Before the reaction occurs [ H2CO3 ] = 0 [No product is produced]

Concentration of CO2 before the reaction occurs =  [ 0.550 mol / 1.50 L ] = 0.367 mol / L

Qc = [ H2CO3 ] / [ CO2 ] = [ 0 ] / [ 0.367 mol / L ] = 0

• At equilibrium K C = Q C  = [ H2CO3 ] / [ CO2 ] = [ 0.125 M ] / [ 0.250 M ] = = 0.500 = 5.00 x 10 ^{– 1}

• Consider the following reaction:

HCl(g)    +    H2O(l)        H3O ⁺ (aq)  + Cl ^– (aq)

If the 0.150 moles of HCl is added in 1.00 L flask at room temperature. At the equilibrium, it was found that HCl concentration to be 0.0.075 M and the H3O ⁺ and Cl ^– concentrations at equilibrium were found to be 0.065 M for both products

Calculate:

1. What is the value of the reaction quotient before any reaction occurs?

• What is the value of the equilibrium constant for the reaction?

Solution:

1. What is the value of the reaction quotient before any reaction occurs?

Before the reaction occurs, then the concentrations of the products H3O ⁺ and Cl ^– are zero. HCl concentration = [ 0.150 mol / 1.00 L ] = 0.150 mol / L = 0.150 M

Q C =  { [ H3O ⁺ ] x [ Cl ^– ] } / [ HCl ]  = { [ 0 ] x [ 0 ] } / 0.150 M = 0

• At equilibrium K C = Q C  = { [ H3O ⁺ ] x [ Cl ^– ] } / [ HCl ]  = { [ 0.065] x [ 0.065] } / [ 0.075 ]  = 0.0563

Prediction of the direction of the chemical equilibrium:

The general Rule is given below:

K C > Q C The system has gone beyond the equilibrium. The ratio of concentrations is high. Products must be converted back into reactants to establish the equilibrium.

KC  < Q C  The system is below the equilibrium. The ratio of concentrations is small.

KC  = Q C  The system is in equilibrium. It means that the initial concentrations are equilibrium concentrations

Examples:

1. The reaction:     A(aq) + 2 B(aq)⇌ 2C(aq) + D(aq) has K C = 14.

At a particular moment in time, [A] = 0.40 M, [B] = 0.50 M, [C] = 1.1 M, and [D] = 1.4 M. Predict the relationship between K C and Q C and the direction of the chemical equilibrium.

Q C. = { [ C ]² x [ D ] } / { [ A ] x [ B ]² } = { [ 1.1 ]² x [ 1.4 ] } / { [ 0.40 ] x [ 0.50 ]² ] = 16.94 = 17

K C is less than Q C                                              K C < Q C

It means that the system is below the equilibrium. The ratio of concentrations is small and Q C > K C and the reaction is proceeding to the left (to the reactants’ side).

• The reaction below:

P4S6 (g) + 2 S2 (g) ⇌ P4S10 (g)

is carried at 600 K, which has K C = 6.2 x 10³ and ∆H^o = – 254 kJ, 1.0 mol of each substance of the chemical equilibrium is introduced into a 1.0 L vessel in the presence of a catalyst and allowed to reach equilibrium at 600 K. Predict the relationship between K C and Q C and the direction of the chemical equilibrium.

Q C = [ P4S10 ] / { [ P4S6 ] x [ S2 ] ² } = [ 1.0 mol / 1.0 L] / { [ 1.0 mol / 1.0 L ] x [ 1.0 mol / 1.0 L ] ² } = 1.0

KC is bigger than Qc                                             K C > QC

The system has gone beyond the equilibrium. The ratio of concentrations is high. Products must be converted back into reactants to establish the equilibrium.

Relationship of Kp and Kc

In the gas phase solutions, the chemical equilibrium constant can be expressed in the partial pressures of the reactants and it is designated by the symbol K P. Similarly, in the aqueous solutions, the equilibrium constant can be expressed in the molar concentrations and it is designated by K C.

Relationship of K p and K C can be utilized using the ideal gas equations and the molarity definition. P V = n R T

Divide both side by V:

P = (n / V) R T

Molarity M is defined as number of moles divided by the volume of the solution ( n / V )

P = M R T or P = X R T and hence [X] = P / RT

The relationship between K C and K P is given below:

Reference: https://slideplayer.com/slide/3268383/ Page 742

Examples:

1. When the following reactions come to equilibrium, does the equilibrium mixture contain mostly reactants or mostly products?

N₂ (g) +O₂ (g)      2 NO (g) KC = 1.5 x 10 ^{– 10}

KC is <<<< 1, mostly reactants. Reaction favors reactants.

• When the following reactions come to equilibrium, does the equilibrium mixture contain mostly reactants or mostly products?

1. 2 SO2 (g)   +  O2 (g)            2 SO2 ( g)     K P  = 2.5 x 10 ⁹

KP is >>>> 1, mostly products. Reaction favors products

• Which of the following reactions lies to the right, favoring the formation of products, and which lies to the left, favoring formation of reactants? Explain your choice.

2 NO (g)  + O₂ (g)       2 NO₂ (g)                          KP = 5.0 x 10 ¹²

KP is >>>> 1, mostly products. Reaction favors products

• 2 HBr(g)         H2 (g)  + Br2  (g)                     KC = 5.8 x 10 ^{– 18}

KC is <<<< 1, mostly products. Reaction favors reactants.

• Calculate KC at 303 K for SO2 (g)  + Cl2 (g)     ßà    SO2Cl2 (g) if KP = 34.5 at this temperature.

KP = KC [ R x T ] ∆n    = 34.5 = KC [ 0.08206 x 303] ∆n = 1-1-1 = -1

34.5 = KC [ 0.08206 x 303] – 1

Cross / Multiply

KC = [ 34.5 / ( [ 0.08206 x 303) ]  = 857.81 = 858

• Consider the following equilibrium:

1. 2 H2 ( g) + S2 (g)     2 H2S (g)                             KC = 1.08 x 10 7 at 700 °C

Calculate KP

KP = KC [ R x T ] ∆n    =  [ 1.08 x 10 7 ] x [0.08206 x 973] ∆n = 2-2-1 = -1

Note that the Temperature is converted from ^oC into K (Kelvin)

KP = [ 1.08 x 10 ⁷ ] x [0.08206 x 973] -1 = 1.35 x 10 5

• Does the chemical equilibrium mixture contain mostly H2 or S2 or H2S. Explain

KC and KP are >>>> 1. Therefore, the chemical equilibrium mixture contains mostly H2S and the chemical equilibrium is said to favor the products.

• If the chemical equilibrium is re-written as follow: H2 ( g) + ½ S2 (g)           ßà    H2S (g)

What is the value of KP at 700 °C?

Kp ^’ = [ 1.35 x 10 5  ] ∆n = 1-1-1/2 = -1/2  = 2.72 x 10 – 3

• Consider the following equilibrium, for which Kp = 0.0752 at 480 °C: 2 Cl2 (g)             + 2 H2O (g)             4 HCI (g)     +    O2 (g)
  • What is the value of Kp for the reaction:

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

Kp  =  1 / KP’ = 1 / 0.0752 = 13.2978 = 13.3

• What is the value of Kp for the following reaction:

Cl2 (g)    +    H2O (g)                  2 HCI (g)    +    ½ O2 (g)

Kp^’    = [ KP ] ∆n = 2+ ½ -1 – 1= ½ = [ 0.0752 ] ½ = 0.274

• Calculate KC in part b.

KP = KC [ R x T ] ∆n    = 0.274

0.2742 = KC [0.08026 x 753] ∆n = 2 + ½ -1 -1 = + ½

Note that the 480 °C is converted into Kelvin. Cross x Multiply

KC = 02742 / [[0.08026 x 753] + ½ = 0.0349