Chemical Equilibrium (I): Types of Equilibrium. Equilibrium Constant & Calculations on Equilibrium Constant
In this post, we shall be discussing the basic sub-topics on chemical equilibrium such as:
~Meaning of equilibrium
~Types of equilibrium
~Conditions for chemical equilibrium
~Equilibrium constant, K
~Calculations involving equilibrium constant
~Types of equilibrium
~Conditions for chemical equilibrium
~Equilibrium constant, K
~Calculations involving equilibrium constant
We will look at other areas such as Le Chatelier's Principle et al in our next/subsequent posts.
Meaning & Types of Equilibrium
Equilibrium is a state of rest of a body or system. It is the state when all the forces acting on a body counteract themselves, thereby making the net force on the body to be equal to zero. This is known as static equilibrium. Example of a body that exhibits static equilibrium is the seesaw, when two bodies of equal weights or masses sit on each of its arms.
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| Static Equilibrium |
In the saturated solution of sodium chloride, NaCl, the dissolved solutes are in equilibrium with undissolved solutes, as shown in the equation below:
NaCl(aq) <-----------------> NaCl(s)
dissolved solutes undissolved solutes
dissolved solutes undissolved solutes
Let us assume that the solubility of NaCl at room temperature is 36 gdm^-3, i.e, the maximum amount of the salt that can dissolve in 1 dm^3 of water at 25°C is 36 g. If you add 50 g of the salt to 1 dm^3 of water at 25°C, 14 g will remain undissolved because the solution is saturated. At that point, those 14 g keep going into and coming out of solution. The fact is, as the excess salt particles above its solubility limit undergo dissolution, the same amount is being precipitated out of solution. Thereby making it look as if there is no change.
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| Dynamic Equilibrium |
CaCO3(s) + heat <-------------> CaO(s) + CO2(g)
For that to be achieved, certain conditions which are necessary for chemical equilibrium to occur must be met.
Conditions Necessary for Chemical Equilibrium
1. The reaction must occur in a closed vessel or it must be a closed system. Here, none of the products (especially gases) will be able to escape.
2. The reaction must be reversible. A reversible reaction is one that proceeds in both directions under appropriate conditions. The inability of the carbon (IV) oxide to escape will force it to combine with the calcium oxide to form the reactant again, thereby, making the reaction reversible.
3. The rate of the forward reaction must equal the rate of the backward reaction. As the products continue to combine to form the reactants and vice versa, a point will be reached when the concentration of the reactants and products will be constant, though not necessarily the same. This is the point of equilibrium at which the rate of the formation of products from reactants equals the rate of the decomposition of products back to reactants.
Equilibrium Constant, K
According to the law of mass action, the rate of a chemical reaction is directly proportional to the active masses of the reactants. The active mass of a reactant is its concentration raised to its stoichiometric value in a given equation.
Consider the hypothetical reaction stated below:
aA + bB <----------------> cC + dD
If the rate of the forward reaction (aA + bB ----------> cC + dD) is r1 ,
and the rate of the backward reaction (cC + dD -----------> aA + bB) is r2;
then, according to the law of mass action:
r1 & [A]^a × [B]^b ..................(i)
r2 & [C]^c × [D]^d ..................(ii)
r2 & [C]^c × [D]^d ..................(ii)
where [A]^a, [B]^b, [C]^c and [D]^d are the active masses of A, B, C and D respectively
and, & = sign of proportionality.
and, & = sign of proportionality.
From equations (i) and (ii),
r1 = k1 × [A]^a × [B]^b .................(iii)
r2 = k2 × [C]^c × [D]^d .................(iv)
r2 = k2 × [C]^c × [D]^d .................(iv)
where k1 and k2 are the velocity constants for the forward and backward reactions respectively.
At equilibrium, r1 = r2
=> k1 × [A]^a × [B]^b = k2 × [C]^c × [D]^d ..........................(v)
Rearranging equation (v) gives
k1/k2 = ([C]^c × [D]^d)/([A]^a × [B]^b) .....................(vi)
The ratio k1/k2 is equal to K, the equilibrium constant of the reaction. Since k1 and k2 are affected by temperature, it also implies that K will be affected by temperature.
Therefore, equation (vi) can be rewritten as
Kc = ([C]^c × [D]^d)/([A]^a × [B]^b) .........................(vii)
Equation (vii) gives the expression for the equilibrium constant of concentration, Kc of the reaction. Generally, it gives the relationship between the concentrations of the reactants and that of the products in an equilibrium mixture.
In other words, equilibrium constant of concentration, Kc of the reversible reaction
mReactants <--------------> nProducts
is given as:
Kc = [Products] ^n/[Reactants] ^m
Equilibrium Constant of a Gaseous System, Kp
The equilibrium constant of pressure, Kp, applies to a gaseous system only (while Kc applies to both aqueous and gaseous systems). To determine Kp, the partial pressure of the gases in the system is always used.
Hence, for a given gaseous system
aA(g) + bB(g) <------------------> cC(g) + dD(g)
Kp = ((PC)^c × (PD)^d)/((PA)^a × (PB)^b) ....................(viii)
where PA, PB, PC and PD are the partial pressures of gases A, B, C and D respectively in the system, while a, b, c and d are the their stoichiometric values in the equation.
Please refer to our post on Dalton's Law of Partial Pressure for the calculation of partial pressure.
Relationship between Kp and Kc
Recall from the ideal gas equation that
PV = nRT ............................(ix)
where P, V, T and n represent pressure, volume, temperature, number of moles respectively for a given gas. R is the molar gas constant.
Also, the relationship between number of moles, concentration and volume of a substance is given as
n = CV ..................................(x)
If we divide both sides of equation (ix) by V, we will obtain:
P = nRT/V
which can be rewritten as
P = (n/V) RT ..........................(xi)
From equation (x), it can be seen that n/V = C. Hence, substituting this into equation (xi), we will have
P = C RT ...............................(xii)
Equation (xii) gives us the relationship between the pressure P, concentration C, and temperature T, of a given gas.
Now, let us substitute equation (xii) into equation (viii) as follows:
PA = [A].RT; PB = [B].RT; PC = [C].RT; PD = [D].RT
where [A], [B], [C] and [D] are the concentrations of gases A, B, C and D respectively.
Therefore, equation (viii) becomes
Kp = (([C].RT)^c × ([D].RT)^d)/(([A].RT)^a × ([B].RT)^b)
Evaluating the above equation gives:
Kp = ([C]^c × [D]^d × (RT)^c × (RT)^d)/([A]^a × [B]^b × (RT)^a × (RT)^b)
If we rearrange the above equation and apply the laws of indices, then
Kp = (([C]^c × [D]^d)/([A]^a × [B]^b)) × (RT)^(c+d))/(RT)^(a+b)
Kp = ([C]^c × [D]^d)/([A]^a × [B]^b) × (RT)^(c+d)-(a+b)
If we substitute equation (vii) into the above equation, we will obtain
Kp = Kc × (RT)^(c+d)-(a+b) ...................(xiii)
but
(c+d) - (a+b) = ∆n (change in number of moles)
Hence, equation (xiii) becomes
Kp = Kc × RT^∆n ...........................(xiv)
Equation (xiv) gives us the relationship between the equilibrium constant of pressure Kp, equilibrium constant of concentration Kc, temperature T, and number of moles n, of a gaseous system at equilibrium.
Calculations on Equilibrium Constants
Example 1
Write the expressions and units for the Kc and Kp for the following reactions:
Write the expressions and units for the Kc and Kp for the following reactions:
a) N2(g) + 3H2(g) <---------------> 2NH3(g)
b) 2SO2(g) + O2(g) <---------------> 2SO3(g)
c) CaCO3(s) <---------------> CaO(s) + CO2(g)
Answer
a) N2(g) + 3H2(g) <---------------> 2NH3(g)
i) Kc = [NH3]^2/([N2] × [H2]^3)
Since we are dealing with concentration, the unit will be moldm^-3. Therefore, the unit for Kc of the above reaction will be:
Kc = (moldm^-3)^2/((moldm^-3) × (moldm^-3)^3)
= (moldm^-3)^2/(moldm^-3)^4
= (moldm^-3)^-2
= dm^6 mol^-2
ii) Kp = (PNH3)^2/((PN2) × (PH2)^3)
The units of pressure include atm, Pa, Nm^-2, torr etc. For the sake of simplicity, we will use atm in this analysis. So, the unit of Kp for the above reaction is:
Kp = atm^2/(atm × atm^3)
= atm^2/atm^4
= 1/atm^2
= atm^-2
b) 2SO2(g) + O2(g) <---------------> 2SO3(g)
i) Kc = [SO3]^2/([SO2]^2 × [O2])
Unit
Kc = (moldm^-3)^2/((moldm^-3)^2 × (moldm^-3))
= (moldm^-3)^2/(moldm^-3)^3
= (moldm^-3)^-1
= dm^3 mol^-1
ii) Kp = (PSO3)^2/((PSO2)^2 × PO2)
Unit
Kp = atm^2/(atm^2 × atm)
= atm^2/atm^3
= 1/atm
= atm^-1
c) CaCO3(s) <---------------> CaO(s) + CO2(g)
i) Kc = [CaO] × [CO2]/[CaCO3]
The concentration of solids is always considered to be negligible (when compared to liquids or gases) and assigned a value of 1. Since CaCO3 and CaO are in the solid phase, their concentrations will reduce to 1.
Hence, the final expression for the Kc is:
Kc = [CO2]
Unit
Kc = moldm^-3
ii) Kp = PCO2
Since pressure has no effect on solid and liquid phases, CaCO3 and CaO exert no partial pressure on the system. This explains why CO2 alone appears in the Kp expression for the reaction.
Unit
Kp = atm
Example 2
Ethanol reacts with ethanoic acid to form ethyl ethanoate and water, according to the equation
CH3COOH(l) + C2H5OH(l) <-------------> CH3COOC2H5(l) + H2O(l)
250 cm^3 of the reaction mixture at equilibrium contained 0.135 mol of ethanoic acid and 0.0250 mol of ethanol together with 0.0820 mol of ethyl ethanoate and 0.0820 mol of water. Use this data to calculate a value of Kc for the reaction.
Answer
Here, the number of moles of the reactants and products and volume of mixture (or reaction vessel) are provided. From these, you can determine the equilibrium concentration of each species.
Here, the number of moles of the reactants and products and volume of mixture (or reaction vessel) are provided. From these, you can determine the equilibrium concentration of each species.
CH3COOH(l) + C2H5OH(l) <-------------> CH3COOC2H5(l) + H2O(l)
Volume of mixture, V = 250 cm^3 = 0.250 dm^3 (recall 1000 cm^3 = 1 dm^3)
Number of moles of CH3COOH, n = 0.135 mol
Concentration of CH3COOH at equilibrium, C = n/V
= 0.135/0.25
= 0.540 moldm^-3
Number of moles of C2H5OH, n = 0.0250 mol
Concentration of C2H5OH at equilibrium, C = n/V
= 0.0250/0.25
= 0.100 moldm^-3
Number of moles of CH3COOC2H5, n = 0.0820 mol
Concentration of CH3COOC2H5 at equilibrium, C = n/V
= 0.082/0.25
= 0.328 moldm^-3
Since the number of moles of ethyl ethanoate and water are equal, their equilibrium concentrations will also be equal.
Hence concentration of H2O at equilibrium = 0.328 moldm^-3
Therefore,
Kc = [CH3COOC2H5][H2O]/[CH3COOH][C2H5OH]
= (0.328 × 0.328)/(0.54 × 0.1)
= 0.107584/0.054
= 1.99
Since the stoichiometric values of the reactants and products are equal, the units will cancel out themselves. So, the value of the Kc for the reaction will have no unit.
Example 3
Propanone, CH3COCH3 reacts with hydrogen cyanide, HCN as follows:
CH3COCH3(l) + HCN(l) <----------------> CH3(OH)(CN)CH3(l)
A mixture of 0.0400 moldm^-3 propanone and 0.0400 moldm^-3 hydrogen cyanide is left to reach equilibrium at room temperature. At equilibrium, the concentration of the product is 0.0133 moldm^-3. Calculate Kc for this reaction.
Answer
Here, you are given the initial concentrations of the reactants and the equilibrium concentration of the product. From that, you are expected to calculate the equilibrium concentration of the reactants.
CH3COCH3 + HCN <----------------> CH3(OH)(CN)CH3
Initial concentrations of propanone and hydrogen cyanide = 0.0400 moldm^-3
If 0.0133 moldm^-3 of the product was formed at equilibrium, it means that same concentration of the reactants had decomposed.
Therefore, equilibrium concentration of propanone = 0.0400 - 0.0133 = 0.0267 moldm^-3
Similarly, the equilibrium concentration of hydrogen cyanide = 0.0267 moldm^-3
Hence,
Kc = [CH3(OH)(CN)CH3]/[CH3COCH3][HCN]
= 0.0133/(0.0267 × 0.0267)
= 18.66 (moldm^-3) ^-1
= 18.66 dm^3 mol^-1
Example 4
In the reaction
In the reaction
2SO2(g) + O2(g) <----------------> 2SO3(g)
The equilibrium partial pressures at constant temperature are 10 atm, 70 atm and 80 atm for SO2, O2 and SO3 respectively. What is the Kp for this reaction.
Answer
2SO2(g) + O2(g) <----------------> 2SO3(g)
Kp = (PSO3)^2/((PSO2)^2 × PO2)
where PSO3 = 80 atm, PSO2 = 10 atm and PO2 = 70 atm
Therefore, substituting the values into the above equation gives:
Kp = 80^2/(10^2 × 70)
= 6400/(100 × 70)
= 0.91 atm^-1 (check Example 1 to see how to determine the unit)
Example 5
Nitrogen reacts with hydrogen to form ammonia according to the equation:
N2(g) + 3H2(g) <---------------> 2NH3(g)
The pressure exerted by this mixture at constant temperature is 200 atm. Under these conditions, the partial pressure of nitrogen is 120 atm and that of hydrogen is 40 atm. Calculate the value of Kp and Kc for this reaction, assuming that it occurred at 25°C. (Take R = 0.082 atm dm^3 K^-1 mol ^-1)
Answer
Here, you are given the total pressure of the mixture of gases, and the partial pressure of the reactants. You are expected to determine the partial pressure of ammonia from the information provided.
N2(g) + 3H2(g) <---------------> 2NH3(g)
Total pressure of mixture = 200 atm
Partial pressure of nitrogen = 120 atm
Partial pressure of hydrogen = 40 atm
Partial pressure of ammonia = total pressure - (sum of partial pressures of nitrogen and hydrogen)
= 200 - (120 + 40)
= 200 - 160
= 40 atm
Kp = (PNH3)^2/((PN2) × (PH2)^3)
= 40^2/(120 × 40^3)
= 1/(120 × 40)
= 0.00021 atm^-2
but, Kp = Kc × (RT)^∆n
so, Kc = Kp/(RT)^∆n
where, ∆n = n(products) - n(reactants)
= 2 - (3 + 1)
= 2 - 4
= -2
T = absolute temperature = 25°C + 273 = 298 K
R = molar gas constant = 0.082 atm dm^3 K^-1 mol^-1
Substituting these values into the last equation gives Kc as:
Kc = 0.00021/(0.082 × 298)^-2
= 0.00021/(24.436)^-2
= 0.00021 × 597.118
= 0.125 dm^6 mol^-2 (check Example 1a for the unit)
Do These
Question 1
Write the equilibrium expressions for the Kc and Kp of the following reactions and state their units:
a) CO(g) + 2H2(g) <-------------> CH3OH(g)
b) H2(g) + I2(g) <---------------> 2HI(g)
c) N2(g) + O2(g) <----------------> 2NO(g)
d) 4HCl(g) + O2(g) <---------------> 2H2O(g) + 2Cl2(g)
Question 2
Calculate the value of Kc for the following reaction given the information below:
H2(g) + CO2(g) <---------------> H2O(g) + CO(g)
Initial concentration of H2(g) = 5.00 moldm^-3
Initial concentration of CO2(g) = 5.00 moldm^-3
Equilibrium concentration of CO(g) = 3.74 moldm^-3
Question 3
The reaction below was carried out at a pressure of 5.00 × 10^4 Pa and at constant temperature
N2(g) + O2(g) <----------------> 2NO(g)
The partial pressures of nitrogen and oxygen are both 2.42 × 10^4 Pa. Calculate the partial pressure of nitrogen (II) oxide, NO(g) at equilibrium.
Question 4
The information provided below gives the data for the reaction of hydrogen with iodine at 500°C.
H2(g) + I2(g) <---------------> 2HI(g)
The initial partial pressures and the partial pressures at equilibrium of hydrogen, iodine and hydrogen iodide are:
Partial Pressure (atm) H2 I2 HI
Initial 70 40 -
At equilibrium 34 - 77
Initial 70 40 -
At equilibrium 34 - 77
Assuming that the total pressure remained constant throughout the experiment,
a) Deduce the partial pressure of iodine at equilibrium
b) Calculate the values of Kp and Kc for this reaction, including the units. (R = 0.082 atm dm^3 K^-1 mol^-1)
b) Calculate the values of Kp and Kc for this reaction, including the units. (R = 0.082 atm dm^3 K^-1 mol^-1)
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