13Chemical equilibrium

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RATE OF REACTION & ACTIVE MASS
▢ Rate of Reaction: Change in concentration of reactant per unit time = amount of substance reacting per unit time.
▢ Concentration Change:
  • •
    Reaction progresses → reactant concentration decreases
  • •
    Reaction progresses → product concentration increases
▢ Active Mass: Reactant concentration expressed in moles per litre in chemical equilibrium.
▢ Active Mass Formula:
  • •
    \(a = fc\)
  • •
    \(f =\) activity coefficient
  • •
    \(c =\) molar concentration
  • •
    Dilute solution / ideal gas: \(f = 1\), therefore \(a = c\)
  • •
    \(\text{Active mass} = \dfrac{\text{number of moles}}{\text{volume in litres}} = \text{mol L}^{-1}\)
Active mass of phases
  • •
    Gas / liquid active mass = molar concentration
  • •
    Solid active mass = always unity, irrespective of quantity
  • •
    Pure liquid active mass = unity
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REVERSIBLE AND IRREVERSIBLE REACTIONS
▢ Reversible Reaction:
❖ Definition: Products reconverted into reactants; reaction proceeds in both directions: forward + backward; carried out in closed container.
❖ Characters:
  • •
    Can start from either side
  • •
    Never complete
  • •
    Tendency to attain equilibrium
  • •
    Slow process
  • •
    Maximum work done: \(W*{reversible} > W*{irreversible}\)
  • •
    Driving force and opposing force differ by infinitesimally small amount
❖ Examples:
  • •
    Weak acid/base neutralisation: \(CH_3COOH + NaOH \rightleftharpoons CH_3COONa + H_2O\)
  • •
    Salt hydrolysis: \(FeCl_3 + 3H_2O \rightleftharpoons Fe(OH)_3 + 3HCl\)
  • •
    Thermal decomposition: \(PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)\)
  • •
    \(CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)\)
  • •
    \(2HI(g) \rightleftharpoons H_2(g) + I_2(g)\)
  • •
    Esterification: \(CH_3COOH + C_2H_5OH \rightleftharpoons CH_3COOC_2H_5 + H_2O\)
  • •
    Evaporation in closed vessel: \(H_2O(l) \rightleftharpoons H_2O(g)\)
▢ Irreversible Reaction:
❖ Definition: Products cannot be converted back into reactants; reaction proceeds only in one direction: forward.
❖ Characters:
  • •
    Completes
  • •
    Driving force and opposing force differ by large amount
  • •
    Fast process
  • •
    Equilibrium state not achieved
❖ Examples:
  • •
    Strong acid + strong base: \(NaOH + HCl \rightarrow NaCl + H_2O\)
  • •
    Double decomposition: \(BaCl_2(aq) + H_2SO_4(aq) \rightarrow BaSO_4(s)\downarrow + 2HCl(aq)\)
  • •
    \(AgNO_3 + NaCl \rightarrow AgCl\downarrow + NaNO_3\)
  • •
    Thermal decomposition: \(2KClO_3 \xrightarrow{MnO_2,\Delta} 2KCl(s) + 3O_2\uparrow\)
  • •
    \(2Pb(NO_3)_2 \xrightarrow{heat} 2PbO + 4NO_2 + O_2\uparrow\)
  • •
    \(NH_4NO_2 \xrightarrow{heat} N_2\uparrow + 2H_2O\uparrow\)
  • •
    Redox: \(SnCl_2(aq) + 2FeCl_3(aq) \rightarrow SnCl_4(aq) + 2FeCl_2(aq)\)
❖ Cause of Irreversibility: One or more products separate as insoluble or volatile species, e.g. \(AgCl\), \(HCl\).
Free energy change
  • •
    Reversible reaction: \(\Delta G = 0\)
  • •
    Irreversible reaction: \(\Delta G < 0\)
Neutralisation point
Weak acid/base neutralisation = reversible; strong acid + strong base neutralisation = irreversible.
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PHYSICAL EQUILIBRIUM
▢ Definition: Different physical states of a substance in equilibrium.
▢ Examples:
  • •
    Solid ⇌ liquid: \(H_2O(ice) \rightleftharpoons H_2O(liquid)\)
  • •
    Liquid ⇌ gas: \(H_2O(liq) \rightleftharpoons H_2O(gas)\)
  • •
    Solid ⇌ gas: \(I_2(s) \rightleftharpoons I_2(g)\)
▢ Important Point: Water and its vapour at equilibrium have same temperature and same kinetic energy.
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CHEMICAL EQUILIBRIUM / DYNAMIC EQUILIBRIUM
▢ Definition: In closed vessel, reversible reaction reaches a stage where forward and backward reactions proceed with same rate.
▢ Alternative Definition: State of reversible reaction where concentrations of reactants and products do not change with time.
▢ Example: \(N_2 + 3H_2 \rightleftharpoons 2NH_3\)
▢ Characteristics:
  • •
    Reached only in closed vessels
  • •
    Dynamic equilibrium: reaction appears stopped but occurs in both directions at same speed
  • •
    Approachable from both sides
  • •
    At equilibrium, both reactants and products present
  • •
    Concentrations do not change with time
  • •
    Catalyst does not affect equilibrium state; only hastens approach
  • •
    Change in pressure / temperature / concentration shifts equilibrium position
  • •
    At equilibrium: \(\Delta G = 0\)
  • •
    Measurable properties constant at equilibrium
  • •
    Reactant and product concentrations need not be equal
📚
FACTORS AFFECTING EQUILIBRIUM STATE

Table 1: Factors and effects

Factor
Change
Effect
Concentration
Reactant concentration increased
Forward reaction favoured
Concentration
Product added
Backward reaction favoured
Concentration
Product removed
Forward reaction favoured
Temperature
Temperature increased
Endothermic reaction favoured
Temperature
Temperature decreased
Exothermic reaction favoured
Pressure
\(\Delta n = 0\)
No effect
Pressure
\(\Delta n > 0\)
Higher pressure favours backward reaction
Pressure
\(\Delta n < 0\)
Higher pressure favours forward reaction
Catalyst
Catalyst added
No change in equilibrium position; equilibrium attained quickly
Inert gas
Constant pressure / volume with \(\Delta n = 0\)
No effect
Inert gas
Constant volume
No effect whatever \(\Delta n\)
Inert gas
Constant pressure + \(\Delta n > 0\)
Forward reaction favoured
Inert gas
Constant pressure + \(\Delta n < 0\)
Backward reaction favoured
Inert gas examples
  • •
    Constant pressure addition decreases formation of \(NH_3\), \(SO_3\), \(PCl_5\)
  • •
    Constant pressure addition increases dissociation of \(PCl_5\)
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LAW OF CHEMICAL EQUILIBRIUM / LAW OF MASS ACTION
▢ Statement: At given conditions, rate of chemical reaction is directly proportional to product of active masses of reacting substances.
▢ Given By: Guldberg and Waage
▢ For General Reaction: \(aA + bB \rightleftharpoons cC + dD\)
▢ Forward Rate:
  • •
    \(r_f \propto [A]^a[B]^b\)
  • •
    \(r_f = K_f[A]^a[B]^b\)
  • •
    \(K_f =\) forward rate constant / forward velocity constant
▢ Backward Rate:
  • •
    \(r_b \propto [C]^c[D]^d\)
  • •
    \(r_b = K_b[C]^c[D]^d\)
  • •
    \(K_b =\) backward rate constant / backward velocity constant
▢ At Equilibrium:
  • •
    \(r_f = r_b\)
  • •
    \(K_f[A]^a[B]^b = K_b[C]^c[D]^d\)
  • •
    \(\dfrac{K_f}{K_b} = \dfrac{[C]^c[D]^d}{[A]^a[B]^b} = K_c\)
  • •
    \(K_c = \dfrac{[Product]}{[Reactant]}\)
Applicability
Law of mass action is applicable only to reversible reactions at constant temperature.
📚
EQUILIBRIUM CONSTANT
▢ Concentration Form: \(K_c = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}\)
▢ Pressure Form: \(K_p = \dfrac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}\)
▢ Use:
  • •
    \(K_c\) = equilibrium constant in terms of concentration
  • •
    \(K_p\) = equilibrium constant in terms of partial pressure
  • •
    For gaseous reactions, active mass may be replaced by partial pressure
▢ Relation between \(K_p\) and \(K_c\):
  • •
    \(PV = nRT\)
  • •
    \(P = \dfrac{n}{V}RT = CRT\)
  • •
    \(K_p = K_c(RT)^{\Delta n}\)
  • •
    \(\Delta n =\) moles of gaseous products − moles of gaseous reactants
  • •
    \(R =\) gas constant
  • •
    \(T =\) absolute temperature

Table 1: Relation between \(K_p\) and \(K_c\)

\(\Delta n\)
Condition
Relation
\(0\)
Moles of gaseous products = moles of gaseous reactants
\(K_p = K_c\) [IOM 2004]
\(>0\)
Products have more gaseous moles / reaction with increase in moles
\(K_p > K_c\)
\(<0\)
Reactants have more gaseous moles / reaction with decrease in moles
\(K_p < K_c\)

Table 2: Examples

Reaction
\(\Delta n\)
Relation
\(N_2(g) + O_2(g) \rightleftharpoons 2NO(g)\)
0
\(K_p = K_c\) [IOM 2004]
\(H_2(g) + I_2(g) \rightleftharpoons 2HI(g)\)
0
\(K_p = K_c\) [IOM 2008]
\(PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)\)
+1
\(K_p > K_c\)
\(2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)\)
-1
\(K_p < K_c\)
\(N_2 + 3H_2 \rightleftharpoons 2NH_3 + Heat\)
-2
\(K_p = K_c(RT)^{-2}\) [IOM 2006]
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UNITS OF \(K_P\) AND \(K_C\)
▢ General:
  • •
    Equilibrium constant has no fixed units
  • •
    Units vary from reaction to reaction
  • •
    Partial pressure in atm → unit of \(K_p = (atm)^{\Delta n}\)
  • •
    Concentration in mole/litre → unit of \(K_c = (mol\,L^{-1})^{\Delta n}\)

Table 1: Units according to \(\Delta n\)

\(\Delta n\)
Relation
Unit of \(K_p\)
Unit of \(K_c\)
0
\(K_p = K_c\)
No unit
No unit
\(>0\)
\(K_p > K_c\)
\((atm)^{\Delta n}\)
\((mol\,L^{-1})^{\Delta n}\)
\(<0\)
\(K_p < K_c\)
\((atm)^{-\Delta n}\)
\((L\,mol^{-1})^{\Delta n}\)
📚
CHARACTERISTICS OF EQUILIBRIUM CONSTANT \((K)\)
▢ Independent Of:
  • •
    Original concentration of reactants
  • •
    Volume
  • •
    Doubling/tripling concentration of reactants
  • •
    Presence of catalyst
  • •
    Change of pressure
  • •
    Presence of inert materials
  • •
    Direction from which equilibrium is attained
  • •
    Nature and number of reaction steps if stoichiometry unchanged
▢ Temperature Dependence:
  • •
    Definite value at given temperature
  • •
    Changes with temperature
  • •
    Endothermic reversible reaction: \(K\) increases with rise in temperature
  • •
    Endothermic reversible reaction: \(K\) decreases with fall in temperature
  • •
    Exothermic reversible reaction: \(K\) decreases with rise in temperature
  • •
    Exothermic reversible reaction: \(K\) increases with fall in temperature
▢ Van't Hoff Equation:
  • •
    \(\log\dfrac{K_2}{K_1} = \dfrac{\Delta H}{2.303R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)\)
  • •
    \(\log\dfrac{K_2}{K_1} = \dfrac{\Delta H}{2.303R}\left(\dfrac{T_2 - T_1}{T_1T_2}\right)\)
  • •
    \(K_1 =\) equilibrium constant at \(T_1\)
  • •
    \(K_2 =\) equilibrium constant at \(T_2\)
  • •
    \(\Delta H =\) enthalpy change

Table 1: Temperature and \(K\)

Condition
Result
\(\Delta H = 0\)
\(K_2 = K_1\); equilibrium constant not affected by temperature
\(\Delta H = +ve\)
\(K_2 > K_1\); \(K\) increases with temperature
\(\Delta H = -ve\)
\(K_2 < K_1\); \(K\) decreases with temperature
▢ Magnitude of K:
  • •
    Greater \(K\) → reaction more toward product side
  • •
    \(K > 1\) → forward direction favoured
  • •
    \(K < 1\) → forward direction less favoured
  • •
    Greater \(K\) → greater thermodynamic stability of products
  • •
    Greater \(K\) → greater instability of reactants
▢ Forward and Backward Constants:
  • •
    For \(A + B \rightleftharpoons C + D\)
  • •
    \(K_f = \dfrac{1}{K_b}\)
  • •
    \(K_b = \dfrac{1}{K_f}\)
  • •
    \(K_fK_b = 1\)
  • •
    At equilibrium: \(r_f = r_b\), but \(K_f > K_b\), \(K_f = K_b\), or \(K_f < K_b\) all possible
📚
EQUATION MANIPULATION AND EQUILIBRIUM CONSTANT

Table 1: Effect of changing equation on \(K\)

Equation operation
New equilibrium constant
Reverse reaction
\(K*{new} = \dfrac{1}{K}\)
Equation multiplied by \(m\)
\(K*{new} = K^m\)
Equation divided by \(n\)
\(K*{new} = K^{1/n}\)
Equation written in two steps
\(K = K_1 \times K_2\)
Step reversed / divided
\(K = \dfrac{K_1}{K_2}\)
▢ Examples:
  • •
    \(A_2 + B_2 \rightleftharpoons 2AB\) has \(K_1\)
  • •
    \(\dfrac{1}{2}A_2 + \dfrac{1}{2}B_2 \rightleftharpoons AB\) has \(K_2 = \sqrt{K_1}\)
  • •
    For \(N_2 + 2O_2 \rightleftharpoons 2NO_2\): \(K = \dfrac{[NO_2]^2}{[N_2][O_2]^2}\)
  • •
    If step 1: \(N_2 + O_2 \rightleftharpoons 2NO\), \(K_1\)
  • •
    If step 2: \(2NO + O_2 \rightleftharpoons 2NO_2\), \(K_2\)
  • •
    Overall: \(K = K_1K_2\)
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EQUILIBRIUM CONSTANT AND STANDARD FREE ENERGY
▢ Formula: \(\Delta G^\circ = -2.303RT\log K\)
Interpretation
  • •
    Negative \(\Delta G^\circ\) → forward reaction spontaneous
  • •
    Negative \(\Delta G^\circ\) indicated by high \(K\)
  • •
    Positive \(\Delta G^\circ\) → reverse reaction feasible; \(K < 1\)
  • •
    At equilibrium \((K = 1)\), total change in standard free energy = 0
  • •
    Higher \(K\) → product more thermodynamically stable, reactant more unstable
📚
ENERGY PROFILE DIAGRAM
▢ Exothermic Reaction:
  • •
    \(E_a\) forward < \(E_a\) backward
  • •
    \(\Delta H = E_a\text{ of backward reaction} - E_a\text{ of forward reaction}\)
  • •
    \(\Delta H = -Q\) negative value
▢ Endothermic Reaction:
  • •
    \(E_a\) forward > \(E_a\) backward
  • •
    \(\Delta H = E_a\text{ of forward reaction} - E_a\text{ of backward reaction}\)
  • •
    \(\Delta H = +Q\) positive value
▢ General: \(\Delta H = E_a\text{ forward reaction} - E_a\text{ backward reaction}\)
📚
IMPORTANCE OF EQUILIBRIUM CONSTANT
▢ Uses:
  • •
    Measures tendency of reversible reaction to proceed in forward direction
  • •
    Greater \(K\) → greater forward tendency
  • •
    Indicates thermodynamic stability of products
  • •
    Greater \(K\) → greater product stability + reactant instability
  • •
    Required for calculation of equilibrium concentrations of reactants and products
📚
CALCULATION OF MOLAR CONCENTRATION AT EQUILIBRIUM
▢ Two Reactants Form Product:
❖ Example: \(H_2 + I_2 \rightleftharpoons 2HI\)

Table 1: ICE-type setup

Stage
\(H_2\)
\(I_2\)
\(2HI\)
Initial moles
\(a\)
\(b\)
0
At equilibrium
\(a-x\)
\(b-x\)
\(2x\)
❖ Molar Concentrations:
  • •
    \([H_2] = \dfrac{a-x}{V}\)
  • •
    \([I_2] = \dfrac{b-x}{V}\)
  • •
    \([HI] = \dfrac{2x}{V}\)
  • •
    \(K_c = \dfrac{[HI]^2}{[H_2][I_2]} = \dfrac{4x^2}{(a-x)(b-x)}\)
▢ Single Substance Dissociation:
❖ Example: \(2NH_3 \rightleftharpoons N_2 + 3H_2\)
❖ Per Mole Form: \(NH_3 \rightleftharpoons \dfrac{1}{2}N_2 + \dfrac{3}{2}H_2\)

Table 1: Degree of dissociation

Stage
\(NH_3\)
\(N_2\)
\(H_2\)
Initial moles
1
0
0
Final moles
\(1-x\)
\(\dfrac{x}{2}\)
\(\dfrac{3x}{2}\)
❖ Note: \(x =\) degree of dissociation.
📚
TYPES OF CHEMICAL EQUILIBRIUM

Table 1: Homogeneous vs heterogeneous equilibrium

Type
Meaning
Examples
Homogeneous equilibrium
Reactants and products in same phase
All gaseous / all liquid
Heterogeneous equilibrium
Reactants and products in two or more phases
Solid + gas systems
▢ Homogeneous Gaseous Equilibrium:
  • •
    All reactants and products are gases
  • •
    Two types: \(\Delta n = 0\), \(\Delta n \ne 0\)
▢ Homogeneous Liquid Equilibrium:
  • •
    All reactants and products in liquid state
  • •
    Example: \(CH_3COOH(l)+C_2H_5OH(l) \rightleftharpoons CH_3COOC_2H_5(l)+H_2O(l)\)
  • •
    \(K_c = \dfrac{x^2}{(a-x)(b-x)}\)
📚
HOMOGENEOUS GASEOUS EQUILIBRIUM
▢ Type I: \(\Delta n = 0\):
❖ Example: \(H_2(g) + I_2(g) \rightleftharpoons 2HI(g)\)
❖ Setup:

Table 1: Equilibrium setup

Stage
\(H_2\)
\(I_2\)
\(2HI\)
At start
\(a\)
\(b\)
0
At equilibrium
\(a-x\)
\(b-x\)
\(2x\)
❖ Concentration Constant: \(K_c = \dfrac{4x^2}{(a-x)(b-x)}\)
❖ Pressure Constant:
  • •
    Total moles at equilibrium = \(a+b\)
  • •
    \(P*{H_2}=\dfrac{(a-x)P}{a+b}\)
  • •
    \(P*{I_2}=\dfrac{(b-x)P}{a+b}\)
  • •
    \(P*{HI}=\dfrac{(2x)P}{a+b}\)
  • •
    \(K_p = \dfrac{(P*{HI})^2}{P*{H_2}P*{I_2}} = \dfrac{4x^2}{(a-x)(b-x)}\)
For \(\Delta n=0\)
  • •
    \(K_p\) and \(K_c\) do not involve \(V\) or \(P\)
  • •
    Equilibrium unaffected by change in volume/pressure
  • •
    \(K_p = K_c\)
▢ Type II: \(\Delta n \ne 0\):
❖ Thermal Dissociation of \(PCl_5\):
◉ Reaction: \(PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)\)
◉ Setup:

Table 1: Dissociation of \(PCl_5\)

Stage
\(PCl_5\)
\(PCl_3\)
\(Cl_2\)
At start
\(a\)
0
0
At equilibrium
\(a-x\)
\(x\)
\(x\)
◉ Concentration Constant:
  • •
    \([PCl_3]=\dfrac{x}{V}\), \([Cl_2]=\dfrac{x}{V}\), \([PCl_5]=\dfrac{a-x}{V}\)
  • •
    \(K_c = \dfrac{[PCl_3][Cl_2]}{[PCl_5]} = \dfrac{x^2}{(a-x)V}\)
  • •
    Expression contains \(V\) in denominator → equilibrium affected by volume
◉ Pressure Constant:
  • •
    Total moles at equilibrium = \((a-x)+x+x = a+x\)
  • •
    \(P*{PCl_3}=\dfrac{xP}{a+x}\)
  • •
    \(P*{Cl_2}=\dfrac{xP}{a+x}\)
  • •
    \(P*{PCl_5}=\dfrac{(a-x)P}{a+x}\)
  • •
    \(K_p = \dfrac{P*{PCl_3}P*{Cl_2}}{P*{PCl_5}} = \dfrac{x^2P}{a^2-x^2}\)
◉ Pressure Effect:
  • •
    If \(x\) very small, \(K_p = \dfrac{x^2P}{a^2}\)
  • •
    \(x^2 = \dfrac{K_p a^2}{P}\)
  • •
    \(x \propto \dfrac{1}{\sqrt{P}}\)
  • •
    Degree of dissociation of \(PCl_5\) inversely proportional to square root of pressure
  • •
    Decrease in pressure → dissociation of \(PCl_5\) increases
❖ Synthesis of Ammonia:
◉ Reaction: \(N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\)
◉ Setup:

Table 1: Haber process setup

Stage
\(N_2\)
\(H_2\)
\(2NH_3\)
At start
\(a\)
\(b\)
0
At equilibrium
\(a-x\)
\(b-3x\)
\(2x\)
◉ Constants:
    ■ _*type: bullet
  1. \(K_c = \dfrac{(2x/V)^2}{((a-x)/V)((b-3x)/V)^3} = \dfrac{4x^2V^2}{(a-x)(b-3x)^3}\)
  2. Total moles at equilibrium = \(a+b-2x\)
  3. \(K_p = \dfrac{4x^2(a+b-2x)^2}{(a-x)(b-3x)^3P^2}\)
For \(\Delta n \ne 0\)
  • •
    \(K_p\) and \(K_c\) expressions are different
  • •
    Expression depends on individual reaction
  • •
    \(P\) or \(V\) appears → equilibrium depends on pressure/volume
  • •
    When products have more gaseous moles, \(V\) factor in \(K_c\) denominator → dissociation increases with volume / decreases with pressure
📚
HETEROGENEOUS EQUILIBRIA
▢ Definition: Reactants and products present in two or more phases.
▢ Law of Mass Action: Applicable to heterogeneous equilibria.
▢ Calcium Carbonate Dissociation:
  • •
    \(CaCO_3(s) \xrightarrow{heat,closed\ vessel} CaO(s) + CO_2(g)\)
  • •
    \(K_p = \dfrac{P*{CaO(s)}P*{CO_2(g)}}{P*{CaCO_3(s)}}\)
  • •
    Active mass of solid = 1
  • •
    \(K_p = P*{CO_2}\)
  • •
    Equilibrium constant of \(CaCO_3\) dissociation equals pressure of \(CO_2\) produced
▢ Ammonium Hydrogen Sulphide Dissociation:
  • •
    \(NH_4HS(s) \rightleftharpoons NH_3(g) + H_2S(g)\)
  • •
    \(K_p = \dfrac{P*{NH_3}P*{H_2S}}{P*{NH_4HS(s)}}\)
  • •
    \(K_p = P*{NH_3}P*{H_2S}\)
▢ Ammonium Carbamate Dissociation:
  • •
    \(NH_2COONH_4(s) \rightleftharpoons 2NH_3(g) + CO_2(g)\)
  • •
    \(K_p = (P*{NH_3})^2(P*{CO_2})\)
  • •
    Gases formed in molar ratio \(NH_3:CO_2 = 2:1\)
  • •
    If total pressure = \(P\), \(P*{NH_3}=\dfrac{2P}{3}\), \(P*{CO_2}=\dfrac{P}{3}\)
  • •
    \(K_p = \left(\dfrac{2P}{3}\right)^2\left(\dfrac{P}{3}\right)=\dfrac{4P^3}{27}\)
📚
REACTION QUOTIENT \((Q)\)
▢ Definition: Expression like equilibrium constant, calculated at any time \(t\), not necessarily equilibrium.
▢ For Reaction: \(aA + bB \rightleftharpoons cC + dD\)
▢ Formula: \(Q = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}\)

Table 1: Reaction direction from \(Q\)

Condition
Meaning
\(Q = K_c\)
Reaction at equilibrium
\(Q < K_c\)
Reaction proceeds forward till \(Q = K_c\)
\(Q > K_c\)
Reaction proceeds backward till \(Q = K_c\)
📚
DEGREE OF DISSOCIATION FROM DENSITY
▢ Formula: \(\alpha = \dfrac{D-d}{(n-1)d}\)
▢ Terms:
  • •
    \(d =\) observed density at particular temperature when degree of dissociation is \(\alpha\)
  • •
    \(D =\) vapour density when there is no dissociation
  • •
    \(n =\) number of molecules of products
📚
NUMERICALS

Table 1: Equilibrium constant numericals

No.
Problem
Key working
Answer
1
\(2H_2S(g) \rightleftharpoons 2H_2(g)+S_2(g)\); equilibrium mixture: \(0.5\,mol\ H_2S\), \(0.10\,mol\ H_2\), \(0.4\,mol\ S_2\) in 1 L
\(K = \dfrac{[S_2][H_2]^2}{[H_2S]^2}=\dfrac{0.4\times0.1\times0.1}{0.5\times0.5}\)
\(0.016\,mol\,L^{-1}\)
2
Equimolar alcohol + acetic acid; 66.5% converted to ester
\(CH_3COOH+C_2H_5OH\rightleftharpoons CH_3COOC_2H_5+H_2O\); initial each = 100; ester/water = 66.5; acid/alcohol = 33.5
\(K=\dfrac{66.5\times66.5}{33.5\times33.5}\approx4\)
3
\(AB+CD\rightleftharpoons AD+CB\); 3/4 mole of each reactant converted
\(K=\dfrac{(3/4V)(3/4V)}{(1/4V)(1/4V)}\)
\(K=9\)
4
\(0.80\,mol\ H_2\) + \(0.80\,mol\ I_2\), at equilibrium \(0.60\,mol\ HI\) formed
\(2x=0.60\Rightarrow x=0.30\); \([H_2]=[I_2]=0.50\), \([HI]=0.60\)
\(K_c=\dfrac{0.60^2}{0.50\times0.50}=1.4\)
5
\(2\,mol\ PCl_5\) in 2 L; 40% dissociated
Dissociated = \(0.8\,mol\); concentrations: \(PCl_5=0.6\), \(PCl_3=0.4\), \(Cl_2=0.4\)
\(K_c=\dfrac{0.4\times0.4}{0.6}=0.267\,M\)
6
\(2SO_2+O_2\rightleftharpoons2SO_3\) in 1 L; equilibrium: \(48g\ SO_3\), \(12.8g\ SO_2\), \(9.6g\ O_2\)
\([SO_3]=0.6\), \([SO_2]=0.2\), \([O_2]=0.3\); \(K_c=\dfrac{[SO_3]^2}{[SO_2]^2[O_2]}\)
\(30\,mol\,L^{-1}\)
7
Free energy change for \(K_p=1\times10^{10}\) at 300 K
\(\Delta G^\circ=-2.303RT\log K_p=-2.303\times2\times300\times10\)
\(-13818\,cal\)
8
Relation for \(N_2+3H_2\rightleftharpoons2NH_3+Heat\)
\(\Delta n=2-4=-2\)
\(K_p=K_c(RT)^{-2}\)
9
\(SO_2+NO_2\rightleftharpoons SO_3+NO\); \(K=16\). Find \(K\) for \(2SO_2+2NO_2\rightleftharpoons2SO_3+2NO\)
Equation multiplied by 2
\(K=16^2=256\)
10
Same parent reaction; find \(K\) for \(\frac{1}{2}SO_2+\frac{1}{2}NO_2\rightleftharpoons\frac{1}{2}SO_3+\frac{1}{2}NO\)
Equation multiplied by \(1/2\)
\(K=16^{1/2}=4\)
11
Forward rate constant \(=11.25\times10^{-4}\), equilibrium constant \(=1.5\)
\(K=\dfrac{K_f}{K_b}\)
\(K_b=7.5\times10^{-4}\)
12
\(PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)\); \(K=24\times10^{-3}\). Backward constant?
\(K' = \dfrac{1}{K}\)
\(4.2\times10^1\)
📚
EXOTHERMIC AND ENDOTHERMIC REACTIONS
▢ Definitions:
  • •
    Exothermic reaction: heat evolved
  • •
    Endothermic reaction: heat absorbed
▢ Examples:
  • •
    \(N_2+3H_2\rightleftharpoons2NH_3+23\,kcal\); exothermic; \(\Delta H=-23\,kcal\)
  • •
    \(N_2+O_2\rightleftharpoons2NO-43.2\,kcal\); endothermic; \(\Delta H=+43.2\,kcal\)
📚
LE CHATELIER'S PRINCIPLE
▢ Statement: If a system at equilibrium is subjected to change in temperature, pressure or concentration, equilibrium shifts to nullify the effect of change.
▢ Temperature:
  • •
    Temperature raised → reaction proceeds in direction absorbing heat
  • •
    Temperature raised → endothermic direction favoured
  • •
    Temperature lowered → exothermic direction favoured
  • •
    \(N_2+O_2+43.2\,kcal\rightleftharpoons2NO\): forward favoured by high temperature
  • •
    \(N_2+3H_2\rightleftharpoons2NH_3+23\,kcal\): backward favoured by high temperature
▢ Pressure:
  • •
    Pressure increased → equilibrium shifts to direction with fewer gaseous moles
  • •
    Pressure decreased → equilibrium shifts to direction with more gaseous moles
  • •
    \(PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)\): backward favoured by increased pressure
  • •
    \(N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)\): forward favoured by increased pressure
  • •
    If total gaseous moles same on both sides → pressure has no effect
  • •
    Examples unaffected by pressure: \(CO+H_2O\rightleftharpoons CO_2+H_2\), \(2HI\rightleftharpoons H_2+I_2\)
  • •
    Solid/liquid systems: pressure change negligible effect
▢ Concentration:
  • •
    Reactant concentration increased → forward shift
  • •
    Product removed → forward shift
  • •
    Reactant removed → backward shift
  • •
    Product added → backward shift
▢ Catalyst:
  • •
    Speeds up forward and backward reactions equally
  • •
    No effect on equilibrium position
  • •
    Equilibrium achieved quickly
▢ Inert Gas:
  • •
    Constant volume with same total moles → no effect
  • •
    Constant pressure → shifts toward side with increased number of moles
  • •
    At constant pressure: \(\Delta n=0\) no effect
  • •
    At constant pressure: \(\Delta n>0\) forward reaction favoured
  • •
    At constant pressure: \(\Delta n<0\) backward reaction favoured

Table 1: Concentration change summary

Change at equilibrium
Shift
Addition of reactant
Forward direction
Removal of product
Forward direction
Removal of reactant
Backward direction
Addition of product
Backward direction
📚
APPLICATIONS OF LE CHATELIER'S PRINCIPLE
▢ Synthesis of Ammonia / Haber Process:
❖ Reaction: \(N_2 + 3H_2 \rightleftharpoons 2NH_3 + 23\,kcal\)
❖ Moles: 1 vol + 3 vol ⇌ 2 vol
❖ Favourable Conditions:
  • •
    High pressure \((\Delta n<0)\)
  • •
    Low temperature
  • •
    Excess \(N_2\) and \(H_2\)
  • •
    Removal of \(NH_3\)
▢ Contact Process:
❖ Reaction: \(2SO_2 + O_2 \rightleftharpoons 2SO_3 + 45\,kcal\)
❖ Moles: 2 vol + 1 vol ⇌ 2 vol
❖ Favourable Conditions:
  • •
    High pressure \((\Delta n<0)\)
  • •
    Low temperature
  • •
    Excess \(SO_2\) and \(O_2\)
▢ Birkeland-Eyde Process:
❖ Reaction: \(N_2 + O_2 \rightleftharpoons 2NO - 43.2\,kcal\)
❖ Moles: 1 vol + 1 vol ⇌ 2 vol
❖ Favourable Conditions:
  • •
    High temperature
  • •
    Excess \(N_2\) and \(O_2\)
  • •
    Pressure has no effect because \(\Delta n=0\)
▢ Formation of Nitrogen Dioxide:
❖ Reaction: \(2NO + O_2 \rightleftharpoons 2NO_2 + 27.8\,kcal\)
❖ Moles: 2 vol + 1 vol ⇌ 2 vol
❖ Favourable Conditions:
  • •
    High pressure
  • •
    Low temperature
  • •
    Excess NO and \(O_2\)
▢ Bosch Process:
❖ Reaction: \(CO + H_2O \rightleftharpoons CO_2 + H_2 + x\,kcal\)
❖ Moles: 1 vol + 1 vol ⇌ 1 vol + 1 vol
❖ Favourable Conditions:
  • •
    Low temperature
  • •
    Large excess of steam and CO
  • •
    Pressure has no effect
▢ Dissociation of \(PCl_5\):
❖ Reaction: \(PCl_5 \rightleftharpoons PCl_3 + Cl_2 - 15\,kcal\)
❖ Moles: 1 vol ⇌ 1 vol + 1 vol
❖ Favourable Conditions:
  • •
    Low pressure / high volume; \(\Delta n>0\)
  • •
    High temperature
  • •
    Excess \(PCl_5\)
▢ Dissociation of Nitrogen Tetroxide:
❖ Reaction: \(N_2O_4 \rightleftharpoons 2NO_2 - 14\,kcal\)
❖ Moles: 1 vol ⇌ 2 vol
❖ Favourable Conditions:
  • •
    Low pressure
  • •
    High temperature
  • •
    Excess \(N_2O_4\)
▢ Melting of Ice:
❖ Reaction: Ice \(\rightleftharpoons\) Water \(-x\,kcal\)
❖ Volume: Ice greater volume; water lesser volume
❖ Effects:
  • •
    High temperature → more water formed
  • •
    High pressure → more water formed due to volume decrease
  • •
    Higher pressure → melting point of ice lowered
  • •
    Higher pressure → boiling point of water increased
▢ Melting of Sulphur:
❖ Reaction: \(S(s) \rightleftharpoons S(l)-x\,kcal\)
❖ Volume: Melting accompanied by volume increase
❖ Effects:
  • •
    High temperature → more liquid sulphur formed
  • •
    High pressure → less sulphur melts
  • •
    Higher pressure → melting point of sulphur increased
▢ Boiling of Water:
❖ Reaction: Water \(\rightleftharpoons\) water vapour \(-x\,kcal\)
❖ Volume: Water low volume; vapour high volume
❖ Effects:
  • •
    High temperature → more vapour formed
  • •
    High pressure → vapour converted to liquid
  • •
    Higher pressure → boiling point of water increased; principle of pressure cooker
▢ Solubility of Salts:
❖ Heat Absorption:
  • •
    Solubility increases with rise in temperature
  • •
    Examples: \(NH_4Cl\), \(K_2SO_4\), \(KNO_3\)
  • •
    \(KNO_3(s)+aq \rightleftharpoons KNO_3(aq)-Q\,kcal\)
❖ Heat Evolution:
  • •
    Solubility decreases with rise in temperature
  • •
    Examples: \(CaCl_2\), \(Ca(OH)_2\), \(NaOH\), \(KOH\)
  • •
    \(Ca(OH)_2(s)+aq \rightleftharpoons Ca(OH)_2(aq)+Q\,kcal\)
▢ Pressure on Solubility of Gases:
❖ Principle: When gas dissolves in liquid, volume decreases; increasing pressure increases solubility.
❖ Henry's Law:
  • •
    Mass of gas dissolved in given mass of solvent at any temperature is directly proportional to pressure of gas above solvent
  • •
    \(m \propto p\)
  • •
    \(m = Kp\)
  • •
    \(K =\) Henry's constant; depends on nature of gas, liquid and temperature
▢ Transport of Oxygen by Haemoglobin:
❖ Reaction: \(Hb(s)+O_2(g)\rightleftharpoons HbO_2(s)\)
❖ Explanation:
  • •
    In tissues: partial pressure of \(O_2\) low → equilibrium shifts left → oxygen released
  • •
    In lungs: partial pressure of \(O_2\) high → equilibrium shifts right → oxyhaemoglobin formed
📚
READ & DIGEST
▢ Key Points:
    ◉ _*type: bullet
  1. Rate of reaction \(\propto\) active masses
  2. Law of mass action by Guldberg and Waage cannot be applied to thermal decomposition of \(KClO_3\): \(2KClO_3\rightarrow2KCl+3O_2\) because oxygen escapes
  3. For \(N_2O_4\rightleftharpoons2NO_2\), if \(\alpha\) is degree of dissociation, total moles at equilibrium = \(1+\alpha\)
  4. Active mass means \(\dfrac{moles}{litre}\)
  5. Opening soda bottle → gas comes out with fizz due to decrease in pressure
  6. Liquid ⇌ gas equilibrium: vapour pressure constant
  7. Liquid in equilibrium with vapour at boiling point: molecules in both phases have equal kinetic energy
  8. If equilibrium constant of \(N_2+O_2\rightleftharpoons2NO\) is \(k_1\), then for \(\frac{1}{2}N_2+\frac{1}{2}O_2\rightleftharpoons NO\), \(k_2=\sqrt{k_1}\)
  9. \(K_p=K_c\) for \(N_2+O_2\rightleftharpoons2NO\) [IOM 2004]
  10. For \(SO_2+\frac{1}{2}O_2\rightleftharpoons SO_3\) and \(2SO_3\rightleftharpoons2SO_2+O_2\), relation: \(K_1^2=\dfrac{1}{K_2}\)
  11. For \(MgCO_3(s)\rightleftharpoons MgO(s)+CO_2(g)\), \(K_p=P*{CO_2}\)
  12. Inert gas added to \(N_2+3H_2\rightleftharpoons2NH_3\) at constant volume → equilibrium unaffected
  13. For gaseous homogeneous reaction: active mass \(=\dfrac{P}{RT}\)
  14. For reversible type \(A+B\rightleftharpoons AB\), combination is exothermic and dissociation is endothermic
  15. \(2SO_2+O_2\rightleftharpoons2SO_3\) is exothermic; increased pressure and decreased temperature favours forward reaction [MOE 2061]
  16. \(N_2+3H_2\rightleftharpoons2NH_3\): ammonia formation favoured by increased pressure and decreased temperature [MOE 2060, IOM 2003]
▢ Effect of Inert Gas:

Table 1: Inert gas addition

Condition
\(\Delta n\)
Effect
\(\Delta V=0\), \(V=\) constant
\(0,+ve,-ve\)
No effect
\(\Delta V\ne0\), \(V\ne\) constant
\(0\)
No effect
\(\Delta V\ne0\), \(V\ne\) constant
\(>0\)
Forward shift
\(\Delta V\ne0\), \(V\ne\) constant
\(<0\)
Backward shift
▢ Temperature and Pressure Summary:

Table 1: Effect of increased temperature and pressure

Nature of reaction
Effect of increased temperature
\(\Delta n\)
Effect of increased pressure
Exothermic
Backward shift
0
No shift
Exothermic
Backward shift
-ve
Forward shift
Endothermic
Forward shift
-ve
Forward shift
Endothermic
Forward shift
+ve
Backward shift
Exothermic
Backward shift
+ve
Backward shift
Exothermic
Backward shift
-ve
Forward shift
Endothermic
Forward shift
+ve
Backward shift
Endothermic
Forward shift
+ve
Backward shift
📚
CHEMICAL EQUILIBRIUM
▢ Basic Concept:
❖ Chemical Equilibrium: Dynamic state in reversible reaction where rate of forward reaction = rate of backward reaction
❖ Dynamic Nature: Reaction does not stop; forward and backward reactions continue at equal rate
❖ Condition: Closed system required
❖ Macroscopic Properties: Concentration, pressure, colour and density remain constant at equilibrium
▢ Types of Equilibrium:

Table 1: Types of Equilibrium

Type
Meaning
Example
Homogeneous equilibrium
Reactants and products in same phase
N2(g) + 3H2(g) ⇌ 2NH3(g)
Heterogeneous equilibrium
Reactants and products in different phases
CaCO3(s) ⇌ CaO(s) + CO2(g)
▢ Law of Mass Action:
❖ Statement: At constant temperature, rate of chemical reaction is directly proportional to product of active masses of reacting substances
❖ For Reaction: aA + bB ⇌ cC + dD
❖ Equilibrium Constant: Kc = [C]^c[D]^d / [A]^a[B]^b
❖ Important Point: Pure solids and pure liquids are not written in equilibrium constant expression
▢ Active Mass:
❖ Meaning: Effective concentration of reacting species
❖ For Solution: Active mass = molar concentration
❖ For Gas: Active mass may be expressed by partial pressure
❖ For Pure Solid / Pure Liquid: Taken as constant
▢ Equilibrium Constants:

Table 1: Important Equilibrium Constants

Constant
Expression / Meaning
Kc
Equilibrium constant in terms of molar concentration
Kp
Equilibrium constant in terms of partial pressure
Kx
Equilibrium constant in terms of mole fraction
Ksp
Solubility product
❖ Relation: Kp = Kc(RT)^Δn
❖ Delta n: Δn = gaseous moles of products − gaseous moles of reactants
❖ Special Cases:
  • •
    If Δn = 0, Kp = Kc
  • •
    If Δn > 0, Kp > Kc
  • •
    If Δn < 0, Kp < Kc
▢ Equilibrium Constant Meaning:

Table 1: Value of K and Extent of Reaction

Value of K
Meaning
K very large
Products predominate; reaction nearly complete
K very small
Reactants predominate; reaction proceeds very little
K ≈ 1
Comparable amounts of reactants and products
▢ Reaction Quotient:
❖ Symbol: Q
❖ Meaning: Ratio of product concentration to reactant concentration at any instant

Table 1: Q and Direction of Reaction

Condition
Direction
Q < K
Forward reaction favoured
Q > K
Backward reaction favoured
Q = K
System at equilibrium
▢ Le Chatelier Principle:
❖ Statement: When a system at equilibrium is disturbed by changing concentration, pressure or temperature, the equilibrium shifts in a direction that opposes the change
❖ Concentration Effect:
  • •
    Adding reactant → shifts forward
  • •
    Adding product → shifts backward
  • •
    Removing product → shifts forward
  • •
    Removing reactant → shifts backward
❖ Pressure Effect:
  • •
    Increase in pressure favours side having fewer gaseous moles
  • •
    Decrease in pressure favours side having more gaseous moles
  • •
    No pressure effect if gaseous moles are equal on both sides
❖ Temperature Effect:
  • •
    Increase in temperature favours endothermic direction
  • •
    Decrease in temperature favours exothermic direction
❖ Catalyst Effect: Catalyst does not change equilibrium constant or position of equilibrium; it only helps equilibrium attain faster
▢ Important Industrial Equilibria:

Table 1: Industrial Reactions

Reaction
Favourable Conditions
N2 + 3H2 ⇌ 2NH3 + heat
Low temperature, high pressure, catalyst
2SO2 + O2 ⇌ 2SO3 + heat
Low temperature, high pressure, catalyst
N2 + O2 ⇌ 2NO − heat
High temperature
▢ Solubility Product:
❖ Definition: Product of ionic concentrations of a sparingly soluble salt in saturated solution at constant temperature
❖ For PbCl2: PbCl2 ⇌ Pb2+ + 2Cl−; Ksp = [Pb2+][Cl−]^2
❖ For AgCl: AgCl ⇌ Ag+ + Cl−; Ksp = [Ag+][Cl−]
❖ Precipitation Rule:
  • •
    Ionic product < Ksp → no precipitation
  • •
    Ionic product = Ksp → saturated solution
  • •
    Ionic product > Ksp → precipitation occurs
▢ Degree of Dissociation:
❖ Symbol: α
❖ Meaning: Fraction of initial molecules dissociated at equilibrium
❖ Formula: α = number of moles dissociated / initial number of moles
❖ For N2O4 ⇌ 2NO2: If α is degree of dissociation, total moles = 1 + α
▢ High-Yield Points:
  • •
    Equilibrium constant depends only on temperature
  • •
    Catalyst does not change Kc or Kp
  • •
    For exothermic reaction, increase in temperature decreases K
  • •
    For endothermic reaction, increase in temperature increases K
  • •
    Increase in pressure favours ammonia formation in Haber process
  • •
    Increase in pressure favours SO3 formation from SO2 and O2
  • •
    In heterogeneous equilibrium, solids and pure liquids are omitted from K expression
  • •
    Kp = Kc when gaseous moles are equal on both sides
  • •
    Greater K means greater product formation
  • •
    At equilibrium, rate of forward reaction equals rate of backward reaction
Q1.
A chemical equilibrium is dynamic because
Q2.
For a reversible reaction at equilibrium
Q3.
Theory of active mass states that rate of reaction is directly proportional to
Q4.
For the reaction aA + bB ⇌ cC + dD, Kc is
Q5.
The equilibrium constant of a reaction is affected by change in
📅MOE Model
Q6.
At equilibrium, a catalyst
Q7.
For N2(g) + O2(g) ⇌ 2NO(g), Kp changes only with change in
📅MOE Model
Q8.
For 2SO2(g) + O2(g) ⇌ 2SO3(g), ΔH is negative. Product formation is favoured by
📅MOE
Q9.
For N2(g) + 3H2(g) ⇌ 2NH3(g), increase in pressure favours
Q10.
In Haber process, formation of ammonia is favoured by
Q11.
According to Le Chatelier principle, if concentration of H2 is increased in N2 + 3H2 ⇌ 2NH3, equilibrium shifts
Q12.
For an exothermic equilibrium reaction, increase in temperature shifts equilibrium
Q13.
For an endothermic equilibrium reaction, increase in temperature shifts equilibrium
Q14.
The relation between Kp and Kc is
Q15.
For H2(g) + I2(g) ⇌ 2HI(g), relation between Kp and Kc is
Q16.
For PCl5(g) ⇌ PCl3(g) + Cl2(g), Δn is
Q17.
For PCl5(g) ⇌ PCl3(g) + Cl2(g), if Kp = 26 at 523 K, approximate Kc is
📅IOM 2005
Q18.
For CaCO3(s) ⇌ CaO(s) + CO2(g), Kp is expressed as
Q19.
In heterogeneous equilibrium, pure solids are not included in K expression because their
Q20.
For the reaction 2HI(g) ⇌ H2(g) + I2(g), Kc is
Q21.
For the reaction N2(g) + 3H2(g) ⇌ 2NH3(g), Kc is
Q22.
If K for N2 + 3H2 ⇌ 2NH3 is K, then K for 2NH3 ⇌ N2 + 3H2 is
Q23.
If K for N2 + 3H2 ⇌ 2NH3 is K, then K for 1/2N2 + 3/2H2 ⇌ NH3 is
Q24.
If an equilibrium reaction is multiplied by 2, its new equilibrium constant becomes
Q25.
If Q < K for a reaction mixture, the reaction proceeds
Q26.
If Q > K for a reaction mixture, the reaction proceeds
Q27.
At equilibrium, reaction quotient Q is
Q28.
A large value of K indicates that
Q29.
If equilibrium constant is 1, it generally means
Q30.
For PbCl2(s) ⇌ Pb2+(aq) + 2Cl−(aq), Ksp is
📅MOE 2061
Q31.
For AgCl(s) ⇌ Ag+(aq) + Cl−(aq), Ksp is
Q32.
Precipitation occurs when ionic product is
Q33.
When ionic product is equal to Ksp, the solution is
Q34.
For 2SO2(g) + O2(g) ⇌ 2SO3(g) + heat, backward reaction is favoured by
Q35.
For N2(g) + O2(g) ⇌ 2NO(g), formation of NO is favoured by
Q36.
In the reaction CaCO3(s) ⇌ CaO(s) + CO2(g), increasing pressure shifts equilibrium
Q37.
In CaCO3(s) ⇌ CaO(s) + CO2(g), removal of CO2 shifts equilibrium
Q38.
Which change favours the reverse reaction in chemical equilibrium?
Q39.
For a reversible reaction, if concentration of reactants is doubled, equilibrium constant will be
Q40.
For a gaseous homogeneous reaction, active mass of a reactant may be obtained by
Q41.
The number of gram molecules of a substance present in unit volume is called
Q42.
At equilibrium, 10 g of CaCO3 corresponds to how many moles?
Q43.
In N2O4(g) ⇌ 2NO2(g), if α is degree of dissociation of N2O4, total moles at equilibrium starting from 1 mole is
Q44.
For N2O4(g) ⇌ 2NO2(g), if α = 0.5 starting from 1 mole, total moles at equilibrium will be
Q45.
For H2(g) + I2(g) ⇌ 2HI(g), if 0.45 mol each of H2 and I2 are present and 3 mol HI is formed at equilibrium in 10 L vessel, Kc is
Q46.
At a certain temperature, if 50% HI is dissociated in 2HI ⇌ H2 + I2, Kc is
Q47.
If equal moles of ethanol and acetic acid are mixed and 2/3 of each reacts, Kc for esterification is
📅IOM 2006
Q48.
In N2(g) + 3H2(g) ⇌ 2NH3(g), addition of inert gas at constant volume
Q49.
In N2(g) + 3H2(g) ⇌ 2NH3(g), addition of inert gas at constant pressure shifts equilibrium
Q50.
For synthesis of ammonia, if temperature is increased, equilibrium yield of ammonia
Q51.
For manufacture of ammonia, catalyst is used mainly to
Q52.
For the reaction N2 + 3H2 ⇌ 2NH3 + heat, optimum industrial conditions are
Q53.
For 2SO3(g) ⇌ 2SO2(g) + O2(g), increase in volume shifts equilibrium
Q54.
If heat is absorbed in the forward reaction, then increasing temperature will
Q55.
If heat is evolved in the forward reaction, then increasing temperature will
Q56.
In an equilibrium mixture, maximum yield of product can often be obtained by
Q57.
For the reaction A + B ⇌ C + D, if one mole each of A and B gives 0.6 mole each of C and D at equilibrium, Kc is
Q58.
For the reaction A + 2B ⇌ C + D, the unit of Kc is
Q59.
For the reaction 2NO2(g) ⇌ N2O4(g), increasing pressure favours
Q60.
For 2NO2(g) ⇌ N2O4(g), brown colour decreases when
Q61.
If Kc for a reaction is less than Kp, then Δn is
Q62.
For a reaction where Δn = −2, Kp and Kc are related as
Q63.
For N2 + O2 ⇌ 2NO, pressure has little effect because
Q64.
At constant temperature, changing concentration of reactants
Q65.
The equilibrium constant of a reaction with products written as reactants is
Q66.
For 4NH3(g) + 5O2(g) ⇌ 4NO(g) + 6H2O(g), Δn is
Q67.
Which of the following is an example of heterogeneous equilibrium?
Q68.
Which equilibrium is homogeneous?