Class 11 Chemistry (Chemistry Part-I) Chapter 6 Equilibrium

Key Concept: Chemical Equilibrium

c) The concentrations of A and B remain constant.

[Solution Description] At equilibrium, the forward and reverse reaction rates are equal, leading to constant concentrations of all reactants and products. Thus, the concentrations of A and B remain constant.

Key Concept: Temperature and Catalyst

b) To speed up the reaction rate

[Solution Description] The synthesis of ammonia (NH3) through the Haber process requires a catalyst like iron to achieve satisfactory reaction rates at high temperatures. Without the catalyst, the reaction would proceed too slowly despite the high temperature because the activation energy remains significantly high.

Key Concept: Basic Definition

b) Saturated solution

[Solution Description] A solution in which no more solute can be dissolved at a particular temperature is known as a saturated solution.

Key Concept: Lewis Acid-Base

d) $NH_3$

[Solution Description] A Lewis base is a compound that donates an electron pair. Among the given options, $NH_3$ has a lone pair of electrons which it can donate, making it a Lewis base.

Key Concept: Basic Concept

c) $$[SO_3]^2/[SO_2]^2[O_2]$$

[Solution Description] The equilibrium constant expression $K_c$ is given by the concentration of products raised to their stoichiometric coefficients divided by the concentration of reactants also raised to their stoichiometric coefficients. For this reaction, the expression would be $\frac{[SO_3]^2}{[SO_2]^2[O_2]}$, where $[SO_3]$, $[SO_2]$, and $[O_2]$ are the concentrations of $SO_3$, $SO_2$, and $O_2$, respectively.

Key Concept: Solubility in Buffer

a) 1000 times

[Solution Description] In pure water, the solubility product is expressed by:

$$\text{CaF}_2 \leftrightarrow \text{Ca}^{2+} + 2\text{F}^{-}$$

Given $K_{sp} = [\text{Ca}^{2+}][\text{F}^{-}]^2$ and $K_{sp} = S \times (2S)^2 = 4S^3$.

In a buffer with pH 5.0, we have:

$$\text{HF} \leftrightarrow \text{H}^+ + \text{F}^{-}$$

Since $\text{pH} = 5.0$, $[\text{H}^+] = 10^{-5}$ M.

Using the common ion effect, assume no change in $[\text{H}^+]$, compute the effective concentration of $[\text{F}^{-}]$:

$$[\text{F}^{-}]_{\text{effective}} = \frac{K_w}{[\text{H}^+]} = \frac{10^{-14}}{10^{-5}} = 10^{-9}$$

Let new solubility be $T$. Then,

$$K_{sp} = T \cdot (2T + 10^{-9})^2$$

Assuming $2T >> 10^{-9}$, approximately $T \approx \sqrt{\frac{K_{sp}}{4}}$, thus ratio is $T/S = 1/[(2 \cdot 10^{-9})^{2/3}] \approx 1000$.

Key Concept: Dynamic Equilibrium

c) They remain constant

[Solution Description] At equilibrium, both forward and reverse reactions occur at equal rates, resulting in no net change in the concentration of reactants and products. Therefore, their concentrations remain constant.

Key Concept: Strong vs Weak

b) CH$_3$COOH

[Solution Description] A weak acid does not fully dissociate in solution, donating protons less readily compared to strong acids. Among the options, acetic acid ($CH_3COOH$) is a weak acid because it partially ionizes in solution.

Key Concept: Units of Kc

b) L/mol

[Solution Description] For the given balanced equation $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$, we calculate $$\Delta n = (\text{moles of gaseous products}) – (\text{moles of gaseous reactants}) = 2 – 3 = -1$$. Hence, the units of $K_c$ would be $(mol/L)^{-1} = L/mol$.

Key Concept: Basic Calculation

c) $$K_c = 8.0$$

[Solution Description] Substitute the equilibrium concentrations into the $K_c$ expression:

$$K_c = \frac{[AB]^2}{[A_2][B]^2}$$

Plug in the values:

$$K_c = \frac{(2.0)^2}{(0.5)(1.0)^2} = \frac{4.0}{0.5} = 8.0$$

Key Concept: Equilibrium Conditions

c) Constant temperature and closed system

[Solution Description] Liquid-vapour equilibrium in a closed system is achieved when the rate of evaporation of the liquid equals the rate of condensation of the vapour. This dynamic equilibrium requires a constant temperature and no net change in the amounts of liquid and vapour.

Key Concept: Pressure Effect

b) The equilibrium shifts towards the products.

[Solution Description] Increasing pressure will shift the equilibrium toward the side with fewer gas molecules to counteract the increase in pressure. Thus, it shifts towards the products side.

Key Concept: Dynamic Nature

d) The rates of the forward and reverse reactions are equal.

[Solution Description] Dynamic equilibrium in a chemical system means that the rate of the forward reaction equals the rate of the backward reaction, resulting in no net change in the concentration of reactants and products over time, even though both reactions continue to occur.

Key Concept: Real-World Application, Industrial Applications

a) To increase yield

[Solution Description] Applying Le Chatelier’s Principle, increasing pressure favors the side of the reaction with fewer moles of gas. The forward reaction reduces the number of gas moles from three (1 CO + 2 H$_2$) to one ($\text{CH}_3\text{OH}$). Thus, high pressure increases the yield of methanol.

Key Concept: Inert Gas Addition

c) There is no change in the equilibrium position.

[Solution Description] Adding an inert gas like argon at constant volume does not affect the partial pressures of the reacting gases, as it does not alter the concentration or partial pressures involved in the reaction. Therefore, there is no effect on the equilibrium position.

Key Concept: Basic Concept

c) The system is dynamic

[Solution Description] The dynamic nature of equilibrium implies that the processes in the system are ongoing and occurring at the same rate, even though no net change is observed.

Key Concept: Real-world Application

b) Solubility of CO$_2$ decreases

[Solution Description] When the bottle is sealed, the CO$_2$ above the liquid surface increases the pressure inside the bottle according to Henry’s Law. Upon opening, the pressure drops to atmospheric levels.

According to Henry’s Law, the solubility of a gas in a liquid decreases with decreasing pressure. As the pressure drops, the dissolved CO$_2$ escapes from the liquid, leading to effervescence or fizzing as CO$_2$ gas bubbles out.

Thus, the correct explanation is that the solubility of CO$_2$ decreases when the pressure decreases.

Key Concept: Equilibrium Dynamics

c) Molecules adhere to ice and escape into the liquid phase simultaneously

[Solution Description] At 273 K, dynamic equilibrium is achieved in a closed system where the rate of molecules adhering to ice equals the rate of those escaping into the liquid phase. This results in constant mass despite ongoing exchange.

Key Concept: Attainment of Equilibrium

d) Forward and reverse reactions occur at equal rates.

[Solution Description] At equilibrium in a closed system, the rate of the forward reaction equals the rate of the reverse reaction, maintaining a dynamic state where overall concentrations remain constant.

Key Concept: Gibbs Energy Relation

c) Spontaneity depends on reaction quotient $Q$

[Solution Description] At equilibrium, $\Delta G = 0$. Given $\Delta G^0 = -RT \ln Kc$, substitute:

$$\Delta G^0 = -8.314 \times \ln(0.5).$$

Calculate $\ln(0.5) \approx -0.693$:

$$\Delta G^0 \approx 8.314 \times 0.693 \approx 5.76\, \text{J/mol}.$$

$\Delta G$ becomes negative when $Q < Kc$, indicating spontaneity in the forward direction for $Q < Kc$.

Key Concept: Concept Application

c) It will remain undissolved

[Solution Description] Camphor is known to be only slightly soluble in water due to its non-polar nature. At room temperature, adding camphor to water will result in very little dissolution. Therefore, camphor will remain mostly undissolved in the water.

Key Concept: Equilibrium Identification

b) At 273 K and 1 atm

[Solution Description] Ice and water coexist in equilibrium at 273 K and 1 atm, which are standard conditions for the co-existence of solid and liquid phases of water.

Key Concept: Volatility and Pressure

c) It is more volatile

[Solution Description] A substance with higher vapour pressure at a given temperature tends to be more volatile. Volatility refers to how readily a substance can vaporize. Therefore, if a substance has a high vapour pressure, it indicates that it’s easier for the molecules to escape into the vapour phase, making the substance more volatile.

Key Concept: Role of Water

a) Acid

[Solution Description] In this reaction, water donates a proton to ammonia ($NH_3$), acting as an acid according to the Brønsted-Lowry theory. As a result, it forms $OH^-$.

Key Concept: Real-World Application, Complex Equilibrium Problem

c) System goes to completion as $K_c$ is very large

[Solution Description] – Balanced equation: $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$

– Initial concentrations: [$SO_2$]=0.8 M, [$O_2$]=0.2 M, [$SO_3$]=0 M

– Change in concentrations: [$SO_2$]=$-2x$, [$O_2$]=$-x$, [$SO_3$]=$+2x$

– At equilibrium: [$SO_2$]=0.8-2x, [$O_2$]=0.2-x, [$SO_3$]=2x

– Given [$SO_3$]=0.4 M $\Rightarrow$ 2x=0.4 $\Rightarrow$ x=0.2

– Substitute x into [$SO_2$]=0.8-2(0.2)=0.4 M, [$O_2$]=0.2-0.2=0 M

– $$K_c=\dfrac{[SO_3]^2}{[SO_2]^2[O_2]} = \dfrac{(0.4)^2}{(0.4)^2(0)}$$

– Since [$O_2$] becomes zero, the system reaches completion under these conditions.

Key Concept: pH Meter Precision, Real-world Application

b) 7.00

[Solution Description] A modern digital pH meter measures with high precision. Here, the meter’s error margin is ±0.003. Thus, each reading can fluctuate by this margin:

For each value:

– 3.456 can range between 3.453 and 3.459

– 7.002 can range between 6.999 and 7.005

– 10.523 can range between 10.520 and 10.526

Considering the rounding rules, only the second reading, 7.002, rounded to two decimal places remains stable at 7.00 because its possible values (6.999 to 7.005) round back to 7.00. Hence, it is considered accurate.

Key Concept: Experimental Observation

b) Brown color fades

[Solution Description] When the test tube is placed in cold water, the temperature of the system decreases. According to Le Chatelier’s principle, the equilibrium will shift in the direction that absorbs heat to counteract this change. Since the formation of $N_2O_4$ from $NO_2$ is exothermic (releasing heat), cooling the system causes the equilibrium to shift towards the dimerization of $NO_2$ into $N_2O_4$, leading to a decrease in the brown color intensity as $NO_2$ is consumed.

Key Concept: Equilibrium Constant Change

b) $K_c$ increases

[Solution Description] In endothermic reactions, increasing the temperature increases the value of the equilibrium constant ($K_c$). This happens because the addition of heat favors the forward (endothermic) reaction, increasing the concentration of products at equilibrium.

Key Concept: Basic Concept

c) Equilibrium is achieved

[Solution Description] When the rate of evaporation equals the rate of condensation, the system reaches equilibrium, and the vapor pressure becomes constant.

Key Concept: Real-world Application

c) Equilibrium shifts right

[Solution Description] In the Haber process, continuously removing ammonia causes the equilibrium to shift toward the products’ side, according to Le Chatelier’s principle. By removing $NH_3$, the reaction shifts to the right to produce more ammonia, thereby increasing its yield.

Key Concept: Basic Formula

c) $$Kp = Kc(RT)^{\Delta n}$$

[Solution Description] The relationship between Kp and Kc is given by the formula:

$$Kp = Kc(RT)^{\Delta n}$$

Where $R$ is the gas constant, $T$ is the temperature in Kelvin, and $\Delta n$ is the change in moles of gas (moles of gaseous products minus moles of gaseous reactants).

Key Concept: Reaction Direction Prediction

b) Reverse

[Solution Description] The reaction is:

$$C(s) + CO_2(g) \rightleftharpoons 2CO(g)$$

Since $K_c = 1.7 \times 10^{-4}$ is much smaller than 1, it indicates that at equilibrium, the concentration of products ($CO$) is significantly lower compared to reactants ($CO_2$).

Therefore, the reaction favors the reverse direction, forming more $CO_2$ at equilibrium.

Key Concept: Basic Concept

a) $$K_c = \frac{[CO_2][H_2O]^2}{[CH_4][O_2]^2}$$

[Solution Description] The equilibrium constant $K_c$ is expressed as the ratio of the product of the concentrations of products to the product of the concentrations of reactants, each raised to the power of their coefficients in the balanced chemical equation. For the given reaction:

$$K_c = \frac{[CO_2][H_2O]^2}{[CH_4][O_2]^2}$$

Key Concept: Basic Concept

c) The pressure exerted by vapour at equilibrium

[Solution Description] Equilibrium vapour pressure is the pressure exerted by a vapour in equilibrium with its liquid at a given temperature. This occurs when the rate of evaporation of the liquid equals the rate of condensation.

Key Concept: pH and Hydrogen Ion Concentration

b) $1.0 \times 10^{-4}$ M

[Solution Description] Given the pH, we can find the hydrogen ion concentration using the formula $[H^+] = 10^{-pH}$. Substituting the given pH value:

$$[H^+] = 10^{-4}$$

This results in a hydrogen ion concentration of

$$[H^+] = 1.0 \times 10^{-4} \text{ M}$$

Therefore, the correct hydrogen ion concentration corresponding to pH of 4 is $1.0 \times 10^{-4}$ M.

Key Concept: Phase Transformation

c) It is the temperature where solid and liquid coexist

[Solution Description] The normal melting point is the temperature at which the solid and liquid phases coexist at equilibrium under atmospheric pressure.

Key Concept: Equilibrium Characteristics

c) They remain constant

[Solution Description] At chemical equilibrium, the concentrations of reactants and products remain constant over time. This constancy occurs because the forward and reverse reaction rates are equal, resulting in no overall change in concentration even though reactions continue to occur.

Key Concept: Temperature and Yield

b) The yield of $NO_2(g)$ decreases.

[Solution Description] In this exothermic reaction, a decrease in temperature favors the formation of $N_2O_4(g)$ as per Le Chatelier’s principle. Hence, the yield of $NO_2(g)$ will decrease.

Key Concept: Pressure and Yield

c) The yield of methanol increases.

[Solution Description] Increasing the pressure favors the side of the reaction with fewer moles of gas. In this case, the forward reaction results in the formation of liquid methanol, which effectively removes gaseous reactants from the system, thus increasing the yield of methanol.

Key Concept: Real-world Application

a) To increase solubility

[Solution Description] High pressure increases the solubility of CO$_2$ in the liquid according to Henry’s Law, making it more effective for carbonation.

Key Concept: Temperature and Solubility

b) Solubility decreases

[Solution Description] The solubility of gases in liquids generally decreases with increasing temperature. This is because higher temperatures provide more energy for the gas molecules to escape into the gaseous phase rather than staying dissolved.

Key Concept: Basic Concepts

a) Acetic acid

[Solution Description] A weak electrolyte only partially ionizes in solution. Among the given options, acetic acid ($CH_3COOH$) is known to be a weak electrolyte as it does not dissociate completely in water.

Key Concept: Pressure Effect

d) Shifts equilibrium to the right.

[Solution Description] The equilibrium position for the synthesis of NH$_3$ shifts towards the side with fewer moles of gas. In this reaction, there are 4 moles of reactants (1 mole N$_2$ and 3 moles H$_2$) and 2 moles of product (NH$_3$). According to Le Chatelier’s principle, increasing the pressure will favor the formation of ammonia because it decreases the total number of moles, leading to a shift in equilibrium to the right.

Key Concept: Equilibrium Characteristics

d) Both forward and reverse reactions occur at the same rate

[Solution Description] In an equilibrium state, the rate of the forward reaction is equal to the rate of the reverse reaction. This means that both processes occur at the same rate, leading to no net change in the concentrations of reactants and products.

Key Concept: Common Ion Effect in Solutions

b) Decreases

[Solution Description] When NaF is added to a solution containing CaF$_2$, it introduces additional F$^-$ ions into the solution. According to Le Chatelier’s principle and the common ion effect, adding more F$^-$ ions will shift the equilibrium to the left, reducing the solubility of CaF$_2$. The dissolution equilibrium for CaF$_2$ can be represented as:

$$\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^-(aq)$$

Adding F$^-$ pushes the equilibrium towards solid CaF$_2$, thus decreasing its solubility.

Key Concept: Phase Ignorance

a) Solid

[Solution Description] In heterogeneous equilibria, the concentrations of pure solids and liquids do not appear in the equilibrium constant expression. Thus, the solid phases do not appear in the expression here.

Key Concept: pH of Neutral Salts

b) 7

[Solution Description] NaCl is a neutral salt, which means it is formed from a strong acid (HCl) and a strong base (NaOH). Such salts dissolve in water to produce solutions with a pH of 7.

Key Concept: Equilibrium Shift Prediction

b) The reaction shifts to the right

[Solution Description] By Le Chatelier’s principle, adding more $\text{CH}_3\text{COOH}$ will cause the equilibrium to shift to the right to counteract the change, producing more products.

Key Concept: Solubility Product Calculation

a) 2.14 $\times$ 10$^{-4}$ M

[Solution Description] The dissolution of $CaF_2$ in water can be represented by the equation:

$$CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^{-}(aq)$$.

Therefore, the equilibrium expression for $K_{sp}$ is:

$$K_{sp} = [Ca^{2+}][F^{-}]^2$$.

Let $s$ be the molar solubility of $CaF_2$. Then at equilibrium:

$[Ca^{2+}] = s$ and $[F^{-}] = 2s$.

Substitute these into the $$K_{sp}$$ expression:

$$K_{sp} = s(2s)^2 = 4s^3 = 3.9 \times 10^{-11}.$$

Solving for $s$, we get:

$$s^3 = \frac{3.9 \times 10^{-11}}{4}.$$

Calculate $s^3$:

$$s^3 = 0.975 \times 10^{-11} = 9.75 \times 10^{-12}.$$

Take the cube root of both sides:

$$s = (9.75 \times 10^{-12})^{1/3} \approx 2.14 \times 10^{-4} \text{ M}.$$

Key Concept: Henderson-Hasselbalch Application

a) 3.53

[Solution Description] To find the pH, we will use the Henderson-Hasselbalch equation:

$$\text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)$$

Here, $[\text{A}^-]$ is the concentration of the conjugate base (sodium formate), and $[\text{HA}]$ is the concentration of the weak acid (formic acid).

Given data:

– $[\text{HA}] = 0.25$ M

– $[\text{A}^-] = 0.15$ M

– $\text{p}K_a = 3.75$

Plug these values into the equation:

$$\text{pH} = 3.75 + \log\left(\frac{0.15}{0.25}\right)$$

First calculate the fraction inside the log:

$$\frac{0.15}{0.25} = 0.6$$

Now, calculate the logarithm:

$$\log(0.6) \approx -0.2218$$

Therefore, the pH is:

$$\text{pH} = 3.75 – 0.2218 = 3.5282$$

Hence, the pH of the buffer solution is approximately 3.53.

Key Concept: Catalyst Role

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion states that catalysts are employed in the Haber process to enhance the reaction rate, which is true. The reason given is that catalysts reduce the activation energy needed for a reaction. This statement is also correct and accurately explains why catalysts are used in the process. Thus, both the assertion and reason are true, and the reason correctly explains the assertion. Therefore, the correct answer is option (a).

Key Concept: Basic Concept

c) Yield of products increases

[Solution Description] The increase in pressure will favor the side with fewer moles of gas according to Le Chatelier’s principle. In this reaction, we have 4 moles of reactants and 2 moles of product gases. Increasing the pressure will shift the equilibrium towards the production of $NH_3$, thereby increasing the yield of ammonia.

Key Concept: Solubility and pH

b) Solubility decreases

[Solution Description] The dissolution of Mg(OH)$_2$ can be written as:

$$Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$$

Increasing pH means increasing the OH$^-$ concentration. According to Le Chatelier’s Principle, this will shift the equilibrium to the left, hence reducing the solubility of Mg(OH)$_2$.

Therefore, increasing pH decreases the solubility of Mg(OH)$_2$.

Key Concept: Gibbs Energy Calculation

a) -30.4 kJ/mol

[Solution Description] To find the Gibbs energy change, we use the equation:

$$\Delta G = – RT \ln Kc$$

where $R$ is the universal gas constant ($8.314 \, J/mol\,K$), and $T$ is the temperature in Kelvin. Substituting the values:

$$\Delta G = – (8.314 \, J/mol\,K) \times (500 \, K) \times \ln(1.5 \times 10^3)$$

$$\ln(1.5 \times 10^3) \approx 7.313$$

Therefore,

$$\Delta G = – (8.314) \times (500) \times 7.313 \approx -30.4 \, kJ/mol$$

The negative sign indicates a spontaneous reaction under standard conditions.

Key Concept: Equilibrium Concept

c) The rate of the forward reaction equals the rate of the reverse reaction.

[Solution Description] When a reaction is at equilibrium, the rates of the forward and reverse reactions are equal. This means that the concentrations of reactants and products remain constant over time, even though both reactions continue to occur.

Key Concept: Expression Modification

b) $$\left(\frac{[NO_2]}{[NO][O_2]^{1/2}}\right)$$

[Solution Description] The original expression for $K_c$ is $\frac{[NO_2]^2}{[NO]^2[O_2]}$. When the stoichiometry is halved, the balanced reaction becomes $NO(g) + \frac{1}{2}O_2(g) \rightarrow NO_2(g)$. The new expression for $K_c$ is then $\left(\frac{[NO_2]}{[NO][O_2]^{1/2}}\right)$.

Key Concept: Formula Identification

c) $$K_p = K_c (RT)^{\Delta n}$$

[Solution Description] The correct formula is $K_p = K_c (RT)^{\Delta n}$, where $R$ is the ideal gas constant and $T$ is temperature in Kelvin.

Key Concept: Standard Gibbs Energy

c) $$K_c = 1$$

[Solution Description] The relationship between standard Gibbs energy change and the equilibrium constant is given by $\Delta G^{\circ} = -RT \ln K_c$. If $\Delta G^{\circ}= 0$, then $-RT \ln K_c = 0$, which implies $\ln K_c = 0$. Thus, $K_c = e^0 = 1$.

Key Concept: Basic Concept

d) The forward and reverse reactions continue at the same rate.

[Solution Description] In a closed chemical system, when equilibrium is achieved, it does not mean that reactions stop occurring. Instead, the rates of the forward and reverse reactions become equal, so there is no net change in the concentration of reactants and products over time. This state where both processes occur at the same rate is called dynamic equilibrium.

Key Concept: Temperature Effect

c) Solubility increases

[Solution Description] For most solids, as temperature increases, the solubility also increases due to increased molecular motion leading to greater dissolution.

Key Concept: Equilibrium Expression

b) $$K_a = \frac{[H^+][X^-]}{[HX]}$$

[Solution Description] For a weak acid HX, the equilibrium expression for its ionization is given by:

$$K_a = \frac{[H^+][X^-]}{[HX]}$$

Key Concept: pH Calculation

c) 4.76

[Solution Description] Using the Henderson-Hasselbalch equation:

$$pH = pK_a + \log{\left(\frac{{[\text{Salt}]}}{{[\text{Acid}]}}\right)}$$

Since $[\text{Salt}] = [\text{Acid}]$, the equation simplifies to:

$pH = pK_a = 4.76$

Key Concept: Ionization Equation

b) $$NaOH(aq) \rightarrow Na^+(aq) + OH^-(aq)$$

[Solution Description] The ionization equation for sodium hydroxide ($NaOH$), a strong base, in water is given by:

$$NaOH(aq) \rightarrow Na^+(aq) + OH^-(aq)$$

Sodium hydroxide dissociates completely into sodium ions ($Na^+$) and hydroxide ions ($OH^-$).

Key Concept: Heterogeneous Equilibrium

c) $$K_c = [CO_2]$$

[Solution Description] The equilibrium constant expression for a reaction involving solids and gases only includes gaseous components because the concentration of pure solids is considered constant and does not appear in the expression. Hence, the equilibrium constant expression is $K_c = [CO_2]$.

Key Concept: Basic Concept

a) $$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$$

[Solution Description] A homogeneous equilibrium occurs when all the reactants and products are in the same phase, typically gaseous or aqueous solutions. The reaction in option A, where both reactants and products are gases, fits this description.

Key Concept: Basic Definition

b) Equilibrium in a system with more than one phase

[Solution Description] A heterogeneous equilibrium involves a system where more than one phase is present, such as solid-liquid, liquid-gas, or solid-gas systems.

Key Concept: Saturated Solution

c) A solution where dissolved and undissolved solute are in equilibrium

[Solution Description] A saturated solution is one where no additional solute can dissolve at a given temperature and pressure, meaning it is in dynamic equilibrium with undissolved solute.

Key Concept: Definition Recall

b) Hydroxyl ions ($OH^-$)

[Solution Description] According to the Arrhenius theory, a base is a substance that dissociates in water to yield hydroxyl ions. The presence of $OH^-(aq)$ ions characterizes a solution as basic according to this concept.

Key Concept: Ionization Process

c) It dissociates into hydrogen ions and anions

[Solution Description] During ionization, an Arrhenius acid dissociates to release hydrogen ions $H^+(aq)$ and its corresponding anion. For example, $HNO_3(aq) \rightarrow H^+(aq) + NO_3^-(aq)$ shows the generation of hydrogen ions.

Key Concept: Pressure and Volume

c) The equilibrium will shift to the right, favoring the formation of products.

[Solution Description] According to Le Chatelier’s Principle, reducing the volume increases pressure, causing the system to shift towards the side with fewer moles of gas. Here, reactants have 4 moles, and products have 2 moles. Thus, equilibrium shifts right.

Key Concept: Definition Recall

c) The acid is stronger

[Solution Description] A larger value of $K_a$ indicates that the acid dissociates more in solution, producing more $H^+$ ions, which means the acid is stronger.

Key Concept: Pressure and Equilibrium

b) The concentration decreases

[Solution Description] If the pressure above the liquid is reduced by half, according to Henry’s Law, the solubility of the gas decreases proportionally. Therefore, the equilibrium concentration of carbon dioxide would decrease.

Key Concept: Phase Change

d) 0 K

[Solution Description] When ice and water are in equilibrium, they coexist at 0°C or 273 K at atmospheric pressure. This is the melting/freezing point of water.

Key Concept: Temperature and Yield

c) The yield of $N_2O_4$ decreases.

[Solution Description] The reaction is exothermic, indicated by the negative enthalpy change ($\Delta H = -57.2 \text{ kJ mol}^{-1}$). According to Le Chatelier’s principle, increasing temperature will shift the equilibrium to favor the endothermic direction (producing more reactants). Thus, as temperature increases, the yield of $N_2O_4$ decreases.

Key Concept: Equilibrium Shift Analysis, Equilibrium Constant Calculation

c) The equilibrium shifts towards more reactant formation.

[Solution Description] The equilibrium constant $K_c$ can be calculated as:

$$K_c = \frac{[N_2O_4]}{[NO_2]^2}.$$

Substituting the given concentrations,

$$K_c = \frac{0.5}{(1.0)^2} = 0.5.$$

When the volume is doubled, the concentration of each species is halved. New concentrations:

$$[NO_2] = 0.5 \text{ M}$$, $$[N_2O_4] = 0.25 \text{ M}$$.

We recalculate $Q_c$:

$$Q_c = \frac{0.25}{(0.5)^2} = 1.0.$$

Since $Q_c > K_c$, the reaction will shift to the left, towards forming more $NO_2$.

Key Concept: Dynamic Equilibrium

c) The rates of sublimation and deposition are equal

[Solution Description] When camphor reaches dynamic equilibrium while sublimating in a closed container, the rates of sublimation and deposition become equal. This means there is no net change in the amount of camphor in either the solid or gas phase over time.

Key Concept: Equilibrium Shift Factors

b) Decreasing temperature

[Solution Description] According to Le Chatelier’s principle, an increase in temperature for an endothermic reaction shifts the equilibrium to the right, favoring product formation. Conversely, a decrease in temperature will shift it to the left, favoring reactants. Thus, decreasing temperature will shift the equilibrium left.

Key Concept: Reaction Quotient

c) It remains the same

[Solution Description] From the syllabus, adding an inert gas at constant volume that doesn’t participate in the reaction does not affect the reaction quotient $Q_c$.

Key Concept: Equilibrium State

a) Because the rates of melting and freezing are equal.

[Solution Description] In a dynamic equilibrium state, such as ice-water equilibrium, the rates of melting and freezing are equal, meaning there is no net change in the amount of ice or water, keeping the total mass constant.

Key Concept: Reaction Representation

b) BF3:NH3

[Solution Description] In the reaction between BF3 and NH3, BF3 acts as a Lewis acid, accepting an electron pair from NH3, which acts as a Lewis base. The product formed is a complex where NH3 donates an electron pair to BF3, represented by BF3:NH3.