Equilibrium

This question is based on the concept of dynamic equilibrium between a liquid and its vapour. At a fixed temperature, molecules continuously evaporate from the liquid surface while vapour molecules simultaneously condense back into the liquid. At equilibrium:

\[ \text{Rate of Evaporation}=\text{Rate of Condensation} \]

The pressure exerted by vapour molecules under this equilibrium condition is called the equilibrium vapour pressure or saturated vapour pressure.

An important principle is that the vapour pressure of a pure liquid depends only on temperature and is independent of the volume of the container, provided some liquid remains present.

Important Theory

• Evaporation occurs only from the liquid surface.
• Condensation occurs when vapour molecules strike the liquid surface.
• Increasing container volume decreases the concentration of vapour molecules.
• Lower vapour concentration reduces the frequency of collisions with the liquid surface.
• As a result, condensation decreases immediately while evaporation remains almost unchanged.
• More liquid evaporates until the equilibrium vapour pressure is attained again.

The equilibrium constant in terms of concentration, denoted by \(K_c\), expresses the ratio of the concentration of products to the concentration of reactants at equilibrium.

Each concentration is raised to the power of its stoichiometric coefficient in the balanced chemical equation.

A larger value of \(K_c\) indicates that, at equilibrium, products are present in greater amounts than reactants.

Important Theory

• Only equilibrium concentrations are substituted into the equilibrium constant expression.
• Pure solids and pure liquids are not included in the expression, but all gaseous species are.
• The coefficients in the balanced chemical equation become the exponents in the equilibrium constant expression.
• The value of \(K_c\) depends only on temperature.

Formula Used

For the reaction

\[ 2\mathrm{SO_2}(g)+\mathrm{O_2}(g)\rightleftharpoons2\mathrm{SO_3}(g) \]

the equilibrium constant is

\[ K_c=\frac{[\mathrm{SO_3}]^2}{[\mathrm{SO_2}]^2[\mathrm{O_2}]} \]

This question is based on the relationship between partial pressure and the equilibrium constant expressed in terms of pressure (\(K_p\)).

For gaseous equilibria, the equilibrium constant is calculated using the partial pressures of gaseous reactants and products.

The percentage by volume of a gas is numerically equal to its mole fraction, which helps determine the partial pressure of each gas.

Important Theory

• The partial pressure of a gas is given by \[ P_i=x_iP_{\text{total}} \] where \(x_i\) is the mole fraction.
• Percentage by volume of a gas is equal to its mole percentage.
• The equilibrium constant in terms of pressure for the reaction \[ \mathrm{I_2(g)\rightleftharpoons2I(g)} \] is \[ K_p=\frac{(P_I)^2}{P_{I_2}} \]
• The value of \(K_p\) depends only on temperature.

Formula Used

For the reaction

\[ \mathrm{I_2(g)\rightleftharpoons2I(g)} \]

the equilibrium constant is

\[ K_p=\frac{(P_I)^2}{P_{I_2}} \]

The equilibrium constant in terms of concentration, \(K_c\), is written by taking the ratio of the equilibrium concentrations of products to reactants, with each concentration raised to the power of its stoichiometric coefficient.

While writing the equilibrium constant expression, it is important to remember that:

• Pure solids are not included.
• Pure liquids are not included.
• Only gases and aqueous species are included.
• The coefficients in the balanced equation become exponents.

Important Theory

The equilibrium constants \(K_p\) and \(K_c\) are related for gaseous reactions by the equation

\[ K_p=K_c(RT)^{\Delta n} \]

where

• \(R=0.0831\ \text{L bar mol}^{-1}\text{K}^{-1}\) (or an equivalent gas constant depending on units)
• \(T\) is the absolute temperature (K).
• \(\Delta n=\) (moles of gaseous products) − (moles of gaseous reactants).

Solids and pure liquids are not considered while calculating \(\Delta n\).

Important Theory

• If \(\Delta n>0\), then \(K_p>K_c\).
• If \(\Delta n<0\), then \(K_c>K_p\).
• If \(\Delta n=0\), then \(K_p=K_c\).
• Only gaseous species are counted while calculating \(\Delta n\).

Formula Used

\[ K_p=K_c(RT)^{\Delta n} \]

Therefore,

\[ K_c=\frac{K_p}{(RT)^{\Delta n}} \]

The equilibrium constant of a reaction depends on the direction in which the reaction is written. If a chemical equation is reversed, the equilibrium constant becomes the reciprocal of the original equilibrium constant.

Thus, if

\[ \mathrm{A}\rightleftharpoons\mathrm{B} \]

has equilibrium constant

\[ K_c, \]

then for the reverse reaction

\[ \mathrm{B}\rightleftharpoons\mathrm{A}, \]

the equilibrium constant is

\[ K_c'=\frac{1}{K_c} \]

Important Theory

• Reversing a chemical equation changes the equilibrium constant to its reciprocal.
• Multiplying a balanced equation by a number raises the equilibrium constant to the same power.
• The equilibrium constant depends only on temperature.
• A very large value of \(K_c\) indicates that the forward reaction is strongly favoured.

Formula Used

For the reverse reaction,

\[K_c(\text{reverse}) = \frac{1}{K_c(\text{forward})}\]

The equilibrium constant is written in terms of the active masses (activities) of the reactants and products. For pure solids and pure liquids, the activity remains constant throughout the reaction. Therefore, they do not affect the numerical value of the equilibrium constant.

Since their concentration does not change appreciably during the reaction, they are incorporated into the value of the equilibrium constant itself and are omitted from the equilibrium constant expression.

Important Theory

• The equilibrium constant depends on the activities of the reacting species.
• For a pure solid or pure liquid, the activity is always equal to 1.
• The concentration (or density) of a pure solid or liquid remains constant at a fixed temperature.
• Only gases and dissolved (aqueous) species have concentrations that change significantly during a reaction.
• Hence,only gases and aqueous species are included in the expressions of \(K_c\) and \(K_p\).

Theory Explanation

Consider the equilibrium:

\[ \mathrm{CaCO_3}(s)\rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO_2}(g) \]

If all species were included, the equilibrium constant would be written as

\[ K_c= \frac{ [\mathrm{CaO}] [\mathrm{CO_2}] } { [\mathrm{CaCO_3}] } \]

However, the concentrations of pure calcium carbonate and calcium oxide remain constant throughout the reaction because they are solids.

Therefore,

\[ [\mathrm{CaCO_3}] =\text{constant} \]

and

\[ [\mathrm{CaO}] =\text{constant} \]

These constant values are absorbed into the equilibrium constant.

Hence,

\[ K_c= [\mathrm{CO_2}] \]

Thus, the concentrations of pure solids are omitted from the equilibrium constant expression.

Similarly, consider the reaction

\[ \mathrm{CH_3COOC_2H_5}(aq)+\mathrm{H_2O}(l) \rightleftharpoons \mathrm{CH_3COOH}(aq)+\mathrm{C_2H_5OH}(aq) \]

Water is a pure liquid. Its concentration remains practically constant throughout the reaction.

Therefore, water is omitted while writing the equilibrium constant expression.

This problem involves calculating the equilibrium composition using an ICE (Initial–Change–Equilibrium) table. Since the equilibrium constant is extremely small, the forward reaction proceeds only to a very small extent. Consequently, only a negligible amount of product is formed, and the reactant concentrations remain almost unchanged.

Such problems test the understanding of equilibrium constants, stoichiometry and approximation methods used in chemical equilibrium.

Important Theory

• The equilibrium constant is given by \[ K_c=\frac{\text{Product Concentrations}}{\text{Reactant Concentrations}} \]
• A very small value of \(K_c\) indicates that the equilibrium lies predominantly towards the reactants.
• An ICE table helps determine the equilibrium concentrations systematically.
• If \(K_c\ll1\), the change in reactant concentration is usually negligible.

Formula Used

For the reaction

\[ 2\mathrm{N_2}(g)+\mathrm{O_2}(g)\rightleftharpoons2\mathrm{N_2O}(g) \]

the equilibrium constant is

\[ K_c= \frac{[\mathrm{N_2O}]^2} {[\mathrm{N_2}]^2[\mathrm{O_2}]} \]

This problem is based on the stoichiometry of a reversible reaction. The equilibrium amount of one substance is given, and the amounts of the remaining substances are calculated using the balanced chemical equation.

The balanced equation shows that:

\[ 2\text{ mol NO}+1\text{ mol Br}_2 \longrightarrow 2\text{ mol NOBr} \]

Thus, the number of moles of NO consumed is equal to the number of moles of NOBr formed, while the number of moles of Br₂ consumed is one-half the number of moles of NOBr formed.

Important Theory

• Balanced chemical equations provide the mole ratios between reactants and products.
• The change in the number of moles of each substance follows these stoichiometric ratios.
• Equilibrium amount = Initial amount − Amount consumed (for reactants).
• Equilibrium amount = Initial amount + Amount formed (for products).

For gaseous equilibria, the equilibrium constants expressed in terms of pressure (\(K_p\)) and concentration (\(K_c\)) are related by the equation

\[ K_p=K_c(RT)^{\Delta n} \]

where

• \(R\) is the universal gas constant.
• \(T\) is the absolute temperature.
• \(\Delta n\) is the difference between the total moles of gaseous products and gaseous reactants.

The first step in every such problem is to calculate the value of \(\Delta n\).

Important Theory

• Only gaseous species are considered while calculating \(\Delta n\).
• If \(\Delta n<0\), then \[ K_c=K_p(RT) \] when \(\Delta n=-1\).
• The values of \(K_p\) and \(K_c\) depend only on temperature.
• The balanced chemical equation must always be written before calculating \(\Delta n\).

Formula Used

\[ K_p=K_c(RT)^{\Delta n}\]

Rearranging,

\[ K_c=\frac{K_p}{(RT)^{\Delta n}} \]

This problem is based on the calculation of the equilibrium constant using partial pressures. Since only HI is initially present, the partial pressures of H₂ and I₂ at equilibrium can be determined using the stoichiometry of the balanced chemical equation.

An ICE (Initial–Change–Equilibrium) table provides a systematic method for solving such gaseous equilibrium problems.

Important Theory

• The equilibrium constant expressed in terms of pressure is \[ K_p= \frac{\text{Partial pressures of products}} {\text{Partial pressures of reactants}} \]
• For a balanced equation, the stoichiometric coefficients become the exponents in the equilibrium expression.
• Partial pressures change according to the stoichiometric coefficients of the reaction.
• An ICE table helps determine equilibrium partial pressures accurately.

Formula Used

For the reaction

\[ 2\mathrm{HI}(g) \rightleftharpoons \mathrm{H_2}(g)+\mathrm{I_2}(g) \]

the equilibrium constant is

\[ K_p= \frac{P_{\mathrm{H_2}}\,P_{\mathrm{I_2}}} {(P_{\mathrm{HI}})^2} \]

To determine whether a reaction mixture is at equilibrium, we calculate the reaction quotient (\(Q_c\)) and compare it with the equilibrium constant (\(K_c\)).

The reaction quotient has the same mathematical form as the equilibrium constant but is calculated using the initial concentrations instead of the equilibrium concentrations.

The comparison is interpreted as follows:

• If \[ Q_c = K_c, \] the system is already at equilibrium.
• If \[ Q_c < K_c, \] the forward reaction proceeds to form more products.
• If \[ Q_c > K_c, \] the reverse reaction proceeds to form more reactants.

Important Theory

• The reaction quotient predicts the direction in which the reaction will proceed.
• The expression for \(Q_c\) is identical to that of \(K_c\).
• Only concentrations at a particular instant are substituted while calculating \(Q_c\).
• The reaction continues until \(Q_c\) becomes equal to \(K_c\).

Formula Used

For the reaction

\[ \mathrm{N_2}(g)+3\mathrm{H_2}(g)\rightleftharpoons2\mathrm{NH_3}(g) \]

the reaction quotient is

\[ Q_c= \frac{[\mathrm{NH_3}]^2} {[\mathrm{N_2}][\mathrm{H_2}]^3} \]

The equilibrium constant expression is directly related to the balanced chemical equation.

The species appearing in the numerator are the products, while those in the denominator are the reactants.

The exponents of the concentration terms become the stoichiometric coefficients of the corresponding substances in the balanced chemical equation.

Important Theory

• The general form of an equilibrium constant is \[ K_c=\frac{\text{Products}}{\text{Reactants}} \]
• Each exponent in the equilibrium expression is equal to the stoichiometric coefficient of that species.
• Only gaseous and aqueous species appear in the equilibrium constant expression.
• Pure solids and pure liquids are omitted because their activities remain constant.

This question is solved using an ICE (Initial–Change–Equilibrium) table. The percentage of water reacted is first converted into the number of moles consumed. Using the stoichiometric coefficients of the balanced reaction, the equilibrium concentrations of all species are calculated. These concentrations are then substituted into the equilibrium constant expression.

Important Theory

• The equilibrium constant is calculated using equilibrium concentrations only.
• Since all stoichiometric coefficients are equal to one, every reactant consumed produces an equal amount of each product.
• For this reaction, \[ \Delta n=0, \] therefore, \[ K_p=K_c. \]
• An ICE table provides a systematic way to determine equilibrium concentrations.

Formula Used

For the reaction

\[ \mathrm{H_2O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H_2}(g)+\mathrm{CO_2}(g) \]

the equilibrium constant is

\[ K_c= \frac{[\mathrm{H_2}][\mathrm{CO_2}]} {[\mathrm{H_2O}][\mathrm{CO}]} \]

Since the reaction mixture initially contains only hydrogen iodide (HI), the equilibrium is established by the decomposition of HI into hydrogen and iodine.

An ICE (Initial–Change–Equilibrium) table helps determine the equilibrium concentrations of the reactants and products.

Because H₂ and I₂ are produced in equal amounts, their equilibrium concentrations are equal.

Important Theory

• The equilibrium constant is calculated using equilibrium concentrations.
• For the reaction \[\mathrm{H_2}(g)+\mathrm{I_2}(g)\rightleftharpoons2\mathrm{HI}(g),\] the equilibrium constant is \[K_c=\frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]}\]
• When only HI is present initially, equal amounts of H₂ and I₂ are formed because of the stoichiometric ratio.
• An ICE table provides a systematic method for solving equilibrium problems.

Formula Used

\[K_c=\frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]}\]

This problem is solved using an ICE (Initial–Change–Equilibrium) table. Initially, only iodine monochloride (ICl) is present. As equilibrium is established, ICl decomposes to form iodine (I₂) and chlorine (Cl₂).

The equilibrium concentrations are substituted into the equilibrium constant expression, resulting in a quadratic equation. Solving this equation gives the equilibrium composition of the reaction mixture.

Important Theory

• The equilibrium constant is calculated using only equilibrium concentrations.
• For every 2 moles of ICl consumed, 1 mole each of I₂ and Cl₂ is produced.
• An ICE table is the most systematic approach for equilibrium calculations.
• The physically meaningful solution of the quadratic equation is always chosen.

Formula Used

For the reaction

\[ 2\mathrm{ICl}(g) \rightleftharpoons \mathrm{I_2}(g)+\mathrm{Cl_2}(g) \]

the equilibrium constant is

\[ K_c= \frac{[\mathrm{I_2}] [\mathrm{Cl_2}]} {[\mathrm{ICl}]^2} \]

This problem involves gaseous equilibrium expressed in terms of partial pressures. Since only ethane (C₂H₆) is initially present, part of it dissociates into equal amounts of ethene (C₂H₄) and hydrogen (H₂).

An ICE (Initial–Change–Equilibrium) table is used to determine the equilibrium partial pressures. These values are substituted into the equilibrium constant expression to calculate the equilibrium pressure of ethane.

Important Theory

• The equilibrium constant in terms of pressure is calculated using equilibrium partial pressures.
• For every mole of C₂H₆ decomposed, one mole each of C₂H₄ and H₂ is formed.
• Partial pressures follow the stoichiometric coefficients of the balanced equation.
• The equilibrium pressure of a gas is its partial pressure in the equilibrium mixture.

Formula Used

For the reaction

\[ \mathrm{C_2H_6}(g) \rightleftharpoons \mathrm{C_2H_4}(g)+\mathrm{H_2}(g) \]

the equilibrium constant is

\[ K_p= \frac{P_{\mathrm{C_2H_4}} P_{\mathrm{H_2}}} {P_{\mathrm{C_2H_6}}} \]

This question illustrates the application of the reaction quotient (\(Q_c\)) and the equilibrium constant (\(K_c\)).

Since water is not the solvent and is produced during the reaction, it must be included in both the reaction quotient and the equilibrium constant expressions.

The comparison between \(Q_c\) and \(K_c\) indicates whether equilibrium has been established and predicts the direction in which the reaction will proceed.

Important Theory

• If water acts as the solvent, it is omitted from the equilibrium expression.
• If water is a reactant or product and is not the solvent, it must be included in the equilibrium expression.
• If \[ Q_c=K_c, \] the system is at equilibrium.
• If \[ Q_c < K_c, \] the forward reaction proceeds.
• If \[ Q_c > K_c, \] the reverse reaction proceeds.

Initially, the vessel contains only phosphorus pentachloride (PCl₅). During the reaction, it partially dissociates into phosphorus trichloride (PCl₃) and chlorine (Cl₂).

Since both products are formed in equal amounts according to the balanced chemical equation, their equilibrium concentrations are identical.

The equilibrium constant expression is then used to calculate the unknown concentration.

Important Theory

• Only equilibrium concentrations are substituted into the equilibrium constant expression.
• According to the balanced equation, one mole of PCl₅ produces one mole each of PCl₃ and Cl₂.
• When products are formed in equal amounts, both concentrations can be represented by the same variable.
• An ICE table provides a systematic method for solving equilibrium problems.

Formula Used

For the reaction

\[ \mathrm{PCl_5}(g) \rightleftharpoons \mathrm{PCl_3}(g) +\mathrm{Cl_2}(g) \]

the equilibrium constant is

\[ K_c= \frac{ [\mathrm{PCl_3}] [\mathrm{Cl_2}] } { [\mathrm{PCl_5}] } \]

This is a gaseous equilibrium problem involving partial pressures. Since FeO and Fe are pure solids, they are omitted from the equilibrium constant expression because their activities remain constant.

An ICE (Initial–Change–Equilibrium) table is used to determine the equilibrium partial pressures.

Important Theory

• Pure solids do not appear in the equilibrium constant expression.
• Only gaseous species are included while writing \(K_p\).
• The equilibrium constant depends only on equilibrium partial pressures.
• An ICE table simplifies equilibrium calculations.

Formula Used

For the reaction

\[ \mathrm{FeO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Fe}(s)+\mathrm{CO_2}(g) \]

the equilibrium constant is

\[ K_p= \frac{P_{\mathrm{CO_2}}} {P_{\mathrm{CO}}} \]

To determine whether a reaction mixture is at equilibrium, we calculate the reaction quotient (\(Q_c\)) and compare it with the equilibrium constant (\(K_c\)).

The reaction quotient has the same mathematical expression as the equilibrium constant but is calculated using the concentrations at any instant rather than the equilibrium concentrations.

The comparison is interpreted as follows:

• If \[ Q_c=K_c, \] the reaction is at equilibrium.
• If \[ Q_c < K_c, \] the forward reaction is favoured.
• If \[ Q_c > K_c, \] the reverse reaction is favoured.

Important Theory

• The expression for \(Q_c\) is identical to that of \(K_c\).
• Only the concentrations at the given instant are substituted while calculating \(Q_c\).
• The reaction proceeds in the direction that makes \(Q_c\) equal to \(K_c\).
• The comparison of \(Q_c\) and \(K_c\) predicts the direction of the net reaction.

Formula Used

For the reaction

\[ \mathrm{N_2}(g)+3\mathrm{H_2}(g)\rightleftharpoons2\mathrm{NH_3}(g) \]

The reaction quotient is

\[ Q_c= \frac{[\mathrm{NH_3}]^2} {[\mathrm{N_2}][\mathrm{H_2}]^3} \]

Initially, only bromine monochloride (BrCl) is present. As the reaction proceeds, BrCl decomposes to produce bromine (Br₂) and chlorine (Cl₂).

An ICE (Initial–Change–Equilibrium) table is used to express the equilibrium concentrations in terms of a single variable. These concentrations are substituted into the equilibrium constant expression to determine the equilibrium concentration of BrCl.

Important Theory

• The equilibrium constant is calculated using equilibrium concentrations.
• For every 2 moles of BrCl decomposed, 1 mole each of Br₂ and Cl₂ is produced.
• Because Br₂ and Cl₂ are formed in equal amounts, their equilibrium concentrations are equal.
• An ICE table provides a systematic method for solving equilibrium problems.

Formula Used

For the reaction

\[ 2\mathrm{BrCl}(g) \rightleftharpoons \mathrm{Br_2}(g)+\mathrm{Cl_2}(g) \]

the equilibrium constant is

\[ K_c= \frac{ [\mathrm{Br_2}] [\mathrm{Cl_2}] } {[\mathrm{BrCl}]^2} \]

This problem combines concepts of mass percentage, mole fraction, partial pressure and the relationship between \(K_p\) and \(K_c\).

Since carbon is a pure solid, it is omitted from the equilibrium constant expression. The percentage by mass is first converted into the mole ratio of CO and CO₂. From the mole fractions and the total pressure, the equilibrium partial pressures are calculated. Finally, \(K_p\) is determined and converted into \(K_c\).

Important Theory

• Pure solids do not appear in equilibrium constant expressions.
• Mass percentages must first be converted into moles.
• Partial pressure is calculated using \[ P_i=X_iP_{\text{total}} \] where \(X_i\) is the mole fraction.
• The relationship between \(K_p\) and \(K_c\) is \[ K_p=K_c(RT)^{\Delta n}. \]

Formula Used

For the reaction

\[ \mathrm{C}(s)+\mathrm{CO_2}(g)\rightleftharpoons2\mathrm{CO}(g) \]

\[ K_p= \frac{(P_{\mathrm{CO}})^2} {P_{\mathrm{CO_2}}} \]

Since

\[ \Delta n=2-1=1, \]

\[ K_p=K_c(RT) \]

Therefore,

\[ K_c=\frac{K_p}{RT} \]

The standard Gibbs free energy change of a reaction is obtained from the standard Gibbs free energies of formation of the reactants and products.

Once \[ \Delta G^\Theta \] is known, the equilibrium constant can be calculated using the thermodynamic relationship between Gibbs free energy and the equilibrium constant.

Important Theory

• The standard Gibbs free energy of reaction is given by

\[ \Delta G^\Theta = \sum \Delta_fG^\Theta(\text{Products}) - \sum \Delta_fG^\Theta(\text{Reactants}) \]

• The relationship between Gibbs free energy and equilibrium constant is

\[ \Delta G^\Theta=-RT\ln K \]

• If \[ \Delta G^\Theta<0, \] the reaction is spontaneous under standard conditions and \[ K>1. \]
• If \[ \Delta G^\Theta>0, \] the reaction is non-spontaneous under standard conditions and \[ K<1. \]

Formula Used

\[ \Delta G^\Theta = \sum \Delta_fG^\Theta(\text{Products}) - \sum \Delta_fG^\Theta(\text{Reactants}) \]

\[ \Delta G^\Theta=-RT\ln K \]

This question is based on Le Chatelier's Principle, which states that when an equilibrium system is disturbed, it shifts in a direction that opposes the disturbance.

When the pressure decreases (or the volume increases), the equilibrium shifts towards the side having a greater number of gaseous moles. Solid substances do not affect the pressure because their concentration remains constant.

Important Theory

• Only gaseous species are considered while comparing the number of moles.
• Pure solids and liquids do not influence pressure changes.
• Decrease in pressure favours the side containing more gaseous molecules.
• If both sides contain an equal number of gaseous moles, no shift occurs.

According to Le Chatelier's Principle, increasing the pressure shifts the equilibrium towards the side having the smaller number of gaseous molecules.

Only gaseous species are considered while comparing the number of moles because solids and liquids do not contribute to pressure.

Important Theory

• If the reactant side has more gaseous moles, increasing pressure favours the forward reaction.
• If the product side has more gaseous moles, increasing pressure favours the backward reaction.
• If both sides contain the same number of gaseous moles, pressure has no effect.
• Pure solids and liquids are ignored while counting gaseous moles.

Initially, only hydrogen bromide (HBr) is present. Since the reaction is reversible, a small amount of HBr dissociates into hydrogen (H₂) and bromine (Br₂) until equilibrium is established.

Because the reaction starts with only HBr, an ICE (Initial–Change–Equilibrium) table is the most systematic approach. The equilibrium constant expression is then used to calculate the unknown equilibrium pressures.

Important Theory

• For gaseous equilibria, equilibrium constants may be expressed in terms of partial pressures.
• Since \[ \Delta n=2-(1+1)=0, \] we have \[ K_p=K_c=1.6\times10^5. \]
• A very large equilibrium constant indicates that the products of the forward reaction (HBr) are highly favoured.
• Only a very small fraction of HBr dissociates at equilibrium.

Formula Used

For the reaction

\[ \mathrm{H_2}(g)+\mathrm{Br_2}(g) \rightleftharpoons 2\mathrm{HBr}(g) \]

\[ K_p= \frac{(P_{\mathrm{HBr}})^2} {P_{\mathrm{H_2}}P_{\mathrm{Br_2}}} \]

This question is based on Le Chatelier's Principle and the dependence of the equilibrium constant on temperature. The equilibrium constant depends only on temperature, whereas pressure and catalysts affect the position of equilibrium or the rate at which equilibrium is reached.

The reaction is endothermic, so heat behaves as a reactant. Therefore, increasing the temperature favours the forward reaction, producing more hydrogen gas.

Important Theory

• The equilibrium constant in terms of pressure is written using the partial pressures of gaseous reactants and products.
• Increasing pressure shifts equilibrium towards the side having fewer gaseous molecules.
• For an endothermic reaction, increasing temperature shifts equilibrium towards the products.
• A catalyst speeds up both forward and reverse reactions equally and therefore does not alter the equilibrium constant or equilibrium composition.

Formula Used

For the reaction

\[ \mathrm{CH_4}(g)+\mathrm{H_2O}(g) \rightleftharpoons \mathrm{CO}(g)+3\mathrm{H_2}(g) \]

The equilibrium constant is

\[ \boxed{ K_p= \frac{ P_{\mathrm{CO}} \left(P_{\mathrm{H_2}}\right)^3 } { P_{\mathrm{CH_4}} P_{\mathrm{H_2O}} } } \]

This question is based on Le Chatelier's Principle, which states that when a system at equilibrium is disturbed, it shifts in a direction that minimizes the effect of the disturbance.

When the concentration of a reactant or product changes, the equilibrium shifts to oppose that change by consuming the added substance or replacing the removed substance.

Important Theory

• Adding a reactant shifts the equilibrium towards the products.
• Adding a product shifts the equilibrium towards the reactants.
• Removing a reactant shifts the equilibrium towards the reactants.
• Removing a product shifts the equilibrium towards the products.
• The value of the equilibrium constant remains unchanged because the temperature is not changed.

This question examines the concepts of equilibrium constant, reversible reactions, and the effect of concentration, pressure and temperature on chemical equilibrium.

The value of the equilibrium constant depends only on temperature. Concentration and pressure may shift the position of equilibrium but do not change the value of the equilibrium constant.

Since the decomposition of PCl₅ is an endothermic reaction, increasing the temperature favours the forward reaction and increases the equilibrium constant.

Important Theory

• The equilibrium constant is written using only gaseous reactants and products.
• The equilibrium constant of the reverse reaction is the reciprocal of the forward equilibrium constant.
• Changing concentration or pressure changes the equilibrium position but not the value of \(K_c\).
• Only temperature changes the numerical value of the equilibrium constant.

Initially, only carbon monoxide and steam are present. As the reaction proceeds, carbon dioxide and hydrogen are formed in equal amounts according to the stoichiometry of the balanced equation.

Since the stoichiometric coefficients of all gaseous species are equal, an ICE (Initial–Change–Equilibrium) table provides a straightforward method for calculating the equilibrium composition.

The equilibrium constant is then used to determine the unknown equilibrium pressure.

Important Theory

• For gaseous equilibria, the equilibrium constant is expressed using partial pressures.
• Because the stoichiometric coefficients are equal, all pressure changes are represented by the same variable.
• The equilibrium constant is calculated using equilibrium partial pressures only.
• Since \[ \Delta n=0, \] the total pressure remains unchanged during the reaction.

Formula Used

For the reaction

\[\mathrm{CO}(g)+\mathrm{H_2O}(g)\rightleftharpoons\mathrm{CO_2}(g)+\mathrm{H_2}(g)\]

The equilibrium constant is

\[ K_p = \frac{P_{\mathrm{CO_2}}P_{\mathrm{H_2}}}{P_{\mathrm{CO}}P_{\mathrm{H_2O}}}\]

The magnitude of the equilibrium constant indicates the relative amounts of reactants and products present at equilibrium.

• If \[ K_c\gg1, \] products are strongly favoured.
• If \[ K_c\ll1, \] reactants are strongly favoured.
• If \[ K_c\approx1, \] both reactants and products are present in appreciable amounts.

Thus, by comparing the numerical values of the equilibrium constants, we can predict the composition of the equilibrium mixture without performing any calculations.

Important Theory

• A very large equilibrium constant indicates that the forward reaction is nearly complete.
• A very small equilibrium constant indicates that very little product is formed.
• An equilibrium constant close to unity indicates that neither reactants nor products dominate the equilibrium mixture.

The equilibrium constant relates the equilibrium concentrations of reactants and products according to the balanced chemical equation.

Since the equilibrium concentration of oxygen is known and the equilibrium constant is given, the concentration of ozone can be calculated directly by substituting the values into the equilibrium constant expression.

Important Theory

• Equilibrium concentrations must be used while evaluating the equilibrium constant.
• The exponents in the equilibrium constant expression are the same as the stoichiometric coefficients in the balanced chemical equation.
• A very small value of \[ K_c \] indicates that the equilibrium strongly favours the reactants.

Formula Used

For the reaction

\[ 3\mathrm{O_2}(g)\rightleftharpoons2\mathrm{O_3}(g) \]

the equilibrium constant is

\[ K_c= \frac{[\mathrm{O_3}]^2} {[\mathrm{O_2}]^3} \]

Since the system is already at equilibrium, the given concentrations can be directly substituted into the equilibrium constant expression. The only unknown quantity is the concentration of methane, which can be calculated by simple algebra.

Because the volume of the flask is

\[ 1\ \text{L}, \]

the numerical values of moles and molar concentrations are identical.

Important Theory

• At equilibrium, the concentrations satisfy the equilibrium constant expression.
• For a 1 L vessel, \[ \text{Concentration}=\text{Number of moles}. \]
• The powers in the equilibrium constant expression are equal to the stoichiometric coefficients.

Formula Used

For the reaction

\[\mathrm{CO}(g)+3\mathrm{H_2}(g)\rightleftharpoons\mathrm{CH_4}(g)+\mathrm{H_2O}(g)\]

the equilibrium constant is

\[K_c=\frac{[\mathrm{CH_4}][\mathrm{H_2O}]}{[\mathrm{CO}][\mathrm{H_2}]^3}\]

This question is based on the Brönsted-Lowry theory of acids and bases. According to this theory:

• An acid donates a proton (\(\mathrm{H^+}\)).
• A base accepts a proton (\(\mathrm{H^+}\)).

Two species that differ by exactly one proton (\(\mathrm{H^+}\)) form a conjugate acid-base pair.

Important Theory

• The conjugate acid is obtained by adding one proton (\(\mathrm{H^+}\)) to a species.
• The conjugate base is obtained by removing one proton (\(\mathrm{H^+}\)) from a species.
• The conjugate acid-base pair differs only by one hydrogen ion.

Definition of Conjugate Acid-Base Pair

A conjugate acid-base pair consists of two chemical species that differ by one proton (\(\mathrm{H^+}\)). When an acid loses a proton, it forms its conjugate base, and when a base gains a proton, it forms its conjugate acid.

General representation:

\[\boxed{\mathrm{HA}\rightleftharpoons\mathrm{H^+}+\mathrm{A^-}}\]

Here,

• \(\mathrm{HA}\) is the acid.
• \(\mathrm{A^-}\) is its conjugate base.

This question is based on the Lewis concept of acids and bases. Unlike the Brönsted-Lowry concept, the Lewis concept does not involve proton transfer. Instead, it is based on the transfer of an electron pair.

Important Theory

• A Lewis acid is an electron-pair acceptor.
• A Lewis base is an electron-pair donor.
• A species having an incomplete octet, a vacant orbital, or a positive charge generally behaves as a Lewis acid.
• A species possessing one or more lone pairs generally behaves as a Lewis base.

Definition

Lewis Acid: A chemical species that accepts a pair of electrons to form a coordinate (dative) covalent bond.

Lewis Base: A chemical species that donates a lone pair of electrons.

According to the Brönsted-Lowry theory, an acid is a substance that donates a proton (\(\mathrm{H^+}\)). When an acid loses one proton, it forms its conjugate base.

Thus, the conjugate base of an acid is obtained by removing one hydrogen ion (\(\mathrm{H^+}\)).

Important Theory

• A Brönsted acid donates a proton (\(\mathrm{H^+}\)).
• The species formed after the loss of one proton is called its conjugate base.
• A conjugate acid-base pair differs by exactly one proton.

General Representation

\[\boxed{\mathrm{HA}\rightleftharpoons\mathrm{H^+}+\mathrm{A^-}}\]

Here,

• \(\mathrm{HA}\) is the acid.
• \(\mathrm{A^-}\) is the conjugate base.

According to the Brönsted-Lowry theory, a base is a substance that accepts a proton (\(\mathrm{H^+}\)). When a base gains one proton, it forms its conjugate acid.

Thus, the conjugate acid of a base is obtained by adding one hydrogen ion (\(\mathrm{H^+}\)).

Important Theory

• A Brönsted base accepts a proton (\(\mathrm{H^+}\)).
• The species formed after accepting one proton is called its conjugate acid.
• A conjugate acid-base pair differs by exactly one proton.

General Representation

\[\boxed{\mathrm{B}+\mathrm{H^+}\rightleftharpoons\mathrm{BH^+}}\]

Here,

• \(\mathrm{B}\) is the base.
• \(\mathrm{BH^+}\) is its conjugate acid.

Some substances can act both as proton donors and proton acceptors. Such species are called amphiprotic (or amphoteric in the Brönsted-Lowry sense).

When an amphiprotic species:

• Accepts one proton (\(\mathrm{H^+}\)), it forms its conjugate acid.
• Donates one proton (\(\mathrm{H^+}\)), it forms its conjugate base.

Important Theory

• Conjugate acid = Base + \(\mathrm{H^+}\)
• Conjugate base = Acid − \(\mathrm{H^+}\)
• Members of a conjugate acid-base pair differ by exactly one proton.
• Species capable of both donating and accepting protons are called amphiprotic species.

According to the Lewis theory, an acid-base reaction involves the transfer of an electron pair rather than the transfer of a proton.

• A Lewis acid accepts an electron pair.
• A Lewis base donates an electron pair.

Lewis acids generally possess an incomplete octet, an empty orbital, or a positive charge, whereas Lewis bases contain one or more lone pairs of electrons available for donation.

Important Theory

• Species with lone pairs usually behave as Lewis bases.
• Electron-deficient molecules such as \(\mathrm{BCl_3}\) are Lewis acids.
• A proton (\(\mathrm{H^+}\)) is a Lewis acid because it accepts an electron pair to form a coordinate covalent bond.

The pH of a solution is a measure of its acidity or basicity. It is defined as the negative logarithm (base 10) of the hydrogen ion concentration.

A higher hydrogen ion concentration corresponds to a lower pH, indicating a more acidic solution.

Important Theory

• pH is used to express the acidity of aqueous solutions.
• The greater the value of \([\mathrm{H^+}]\), the lower the pH.
• Acidic solutions have \[ \mathrm{pH}<7, \] neutral solutions have \[ \mathrm{pH}=7, \] and basic solutions have \[ \mathrm{pH}>7. \]

Formula Used

\[ \boxed{\mathrm{pH}=-\log[\mathrm{H^+}]} \]

The pH scale is a logarithmic measure of the hydrogen ion concentration in a solution. If the pH of a solution is known, the hydrogen ion concentration can be obtained using the inverse logarithmic relationship.

Vinegar contains acetic acid, making it an acidic solution with a pH less than 7.

Important Theory

• The pH of a solution is defined as: \[ \mathrm{pH}=-\log[\mathrm{H^+}] \]
• Rearranging the equation gives: \[ [\mathrm{H^+}]=10^{-\mathrm{pH}} \]
• Lower pH values correspond to higher hydrogen ion concentrations.

Formula Used

\[\boxed{[\mathrm{H^+}]=10^{-\mathrm{pH}}}\]

The strength of a weak acid and its conjugate base are related through the ionic product of water.

For every conjugate acid-base pair,

\[ K_a\times K_b=K_w \]

At

\[ 298\ \text{K}, \]

\[ K_w=1.0\times10^{-14} \]

Therefore, if the acid ionization constant is known, the base ionization constant can be calculated using the above relation.

Important Theory

• For a conjugate acid-base pair, \[ K_aK_b=K_w. \]
• At \[ 298\ \text{K}, \] \[ K_w=1.0\times10^{-14}. \]
• A stronger acid has a weaker conjugate base and vice versa.

Formula Used

\[ \boxed{ K_b=\frac{K_w}{K_a} } \]

Phenol is a weak acid and ionizes only slightly in water.

\[ \mathrm{C_6H_5OH} \rightleftharpoons \mathrm{H^+} + \mathrm{C_6H_5O^-} \]

The ionization constant relates the equilibrium concentrations of phenol, hydrogen ions and phenolate ions.

In the second part, sodium phenolate provides the common ion \(\mathrm{C_6H_5O^-}\), suppressing the ionization of phenol due to the common ion effect.

Important Theory

• For a weak acid, \[ K_a=\frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]} \]
• Since phenol is weak, \[ x\ll C, \] so \[ C-x\approx C. \]
• The common ion effect decreases the ionization of a weak electrolyte.

Formula Used

\[ K_a=\frac{x^2}{C} \]

where

• \(C\) = initial concentration of phenol
• \(x\) = concentration of ionized phenol

Hydrogen sulphide is a diprotic weak acid and ionizes in two successive steps.

First ionization:

\[ \mathrm{H_2S} \rightleftharpoons \mathrm{H^+} + \mathrm{HS^-} \]

Second ionization:

\[ \mathrm{HS^-} \rightleftharpoons \mathrm{H^+} + \mathrm{S^{2-}} \]

The addition of HCl increases the hydrogen ion concentration, causing the equilibrium to shift towards the left due to the common ion effect. Consequently, the concentrations of both \(\mathrm{HS^-}\) and \(\mathrm{S^{2-}}\) decrease.

Important Theory

• For a weak acid, \[ K_a=\frac{x^2}{C} \] when \[ x\ll C. \]
• The common ion effect suppresses the ionization of weak acids.
• The second ionization of a diprotic acid is usually much smaller than the first.

Acetic acid is a weak acid and undergoes only partial ionization in water. Its ionization equilibrium is

\[ \mathrm{CH_3COOH} \rightleftharpoons \mathrm{H^+} + \mathrm{CH_3COO^-} \]

Since the ionization constant is very small, only a small fraction of the acid molecules dissociate. Therefore, the approximation

\[ C-x\approx C \]

can be safely used to simplify the calculations.

Important Theory

• The ionization constant of a weak acid is \[ K_a=\frac{[\mathrm{H^+}][\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]} \]
• For a weak acid, \[ K_a=\frac{x^2}{C} \] where \(x\) is the concentration of hydrogen ions produced.
• The degree of dissociation is \[ \alpha=\frac{x}{C} \]
• The pH is calculated using \[ \mathrm{pH}=-\log[\mathrm{H^+}] \]

An organic acid is a weak acid that undergoes only partial ionization in water.

\[ \mathrm{HA} \rightleftharpoons \mathrm{H^+} + \mathrm{A^-} \]

The given pH allows us to calculate the hydrogen ion concentration. Since one mole of acid produces one mole of hydrogen ion and one mole of anion, the concentration of the anion is equal to the hydrogen ion concentration.

Using the equilibrium expression, the ionization constant and the corresponding \(pK_a\) can then be determined.

Important Theory

• The hydrogen ion concentration is related to pH by \[ \mathrm{pH} = -\log[\mathrm{H^+}] \]
• For a weak acid, \[ K_a = \frac{[\mathrm{H^+}][\mathrm{A^-}]} {[\mathrm{HA}]} \]
• The acid dissociation constant is related to \(pK_a\) by \[ pK_a=-\log K_a \]