Prof./ Ayman Mansour — Chemistry Consultant
Chemical Equilibrium
الاتزان الكيميائي
كيمياء ٢٠٢٦ — Chemistry 2026
نماذج تفاعلية · شرح تفصيلي · مراجعة شاملة
The key to these graphs lies in observing where the curves level off (reach equilibrium):
In this system, Ca(OH)2 and CaO are solids, while H2O is a vapor (gas). Solids do not affect the equilibrium position. We only care about the vapor!
This is a two-step calculation: first find the new diluted concentration, then calculate the degree of dissociation (α).
Let's evaluate the chemical additions in each tube based on Le Chatelier's Principle:
Pure distilled water is neutral with a pOH of 7 (and pH of 7).
The total pressure involves only the gases in the system. Solid carbon C(s) is entirely ignored in gas pressure calculations.
This is a reverse-engineering calculation to find Ksp at a higher temperature.
Using Faraday's main formula: Mass = (Current × Time × Equivalent Mass) / 96500
For a system to achieve dynamic equilibrium (Rateforward = Ratebackward), the reaction must be reversible and occur in a closed system where no products escape:
HCl is a strong acid that dissociates completely:
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
This increases the concentration of H3O+ (common ion) in the system. According to Le Chatelier's principle:
The equation is written with "- heat" on the product side, which is equivalent to:
Let's analyze the effects:
The rate of a chemical reaction is influenced by:
Option (d) optimizes all three rate-enhancing factors.
Given [H+] = 1.0 × 10-9 M:
pH = -log[H+] = -log(1.0 × 10-9) = 9
Since pH > 7, the solution is basic (a base) with pH = 9.
The total pressure is the sum of all partial pressures:
Ptotal = PNO2 + PN2 + PO23.2 = 2.0 + PN2 + 1.0PN2 = 3.2 - 3.0 = 0.2 atm
Kp = [PN2 × P2O2] / P2NO2Kp = [0.2 × (1.0)2] / (2.0)2 = 0.2 / 4 = 0.05
Solubility S = Moles / Volume (L) = 2 × 10-4 mol / 0.050 L = 4 × 10-3 M
Mg(OH)2 ⇌ Mg2+ + 2OH-
[Mg2+] = S = 4 × 10-3 M
[OH-] = 2S = 8 × 10-3 M
Ksp = [Mg2+][OH-]2 = (4 × 10-3) × (8 × 10-3)2 = 2.56 × 10-7
From Cell 1 (A/C) and Cell 2 (C/B):
E°ox(A) > E°ox(C) with difference of 1.2 V.
E°ox(C) > E°ox(B) with difference of 0.8 V.
EMF = +2.0 V.EMF = -2.0 V.| Substance | Cl₂ | NO | NOCl |
|---|---|---|---|
| Moles at equilibrium | 3 | 1.5 | 3 |
Kc = [NO]² × [Cl₂] / [NOCl]²
Let V = volume. Concentrations:
Kc = (1.5/V)² × (3/V) / (3/V)² = (2.25/V²)(3/V) / (9/V²)
= (6.75/V³) / (9/V²) = 6.75 / (9V) = 0.75/V
0.25 = 0.75/V → V = 0.75/0.25 = 3 L ✓
Initial moles of NaOH = 0.01 × 0.2 = 0.002 mol
Final moles needed = 0.7 × 0.2 = 0.14 mol (assuming volume stays ≈ 200 mL since solid is added)
Additional moles = 0.14 − 0.002 = 0.138 mol
Mass = 0.138 × 40 = 5.52 g ✓
Initial: H₂ = 1 mol, I₂ = 1 mol, HI = 0
At equilibrium: H₂ = 0.3 mol, I₂ = 0.3 mol
Reacted = 1 − 0.3 = 0.7 mol each → HI formed = 2 × 0.7 = 1.4 mol
Concentrations (V = 2L): [H₂] = 0.3/2 = 0.15 M, [I₂] = 0.15 M, [HI] = 1.4/2 = 0.7 M
Kc = [HI]²/([H₂][I₂]) = (0.7)²/(0.15 × 0.15) = 0.49/0.0225 = 21.78 ✓
| Option | pH | pOH | α |
|---|---|---|---|
| A | 2.87 | 11.13 | 0.0134 |
| B | 11.13 | 2.87 | 0.0134 |
| C | 13 | 1 | 1 |
| D | 9.26 | 4.74 | 1.8×10⁻⁵ |
NH₄OH is a weak base.
α = √(Kb/C) = √(1.8×10⁻⁵/0.1) = √(1.8×10⁻⁴) = 0.0134
[OH⁻] = α × C = 0.0134 × 0.1 = 1.34 × 10⁻³ M
pOH = −log(1.34×10⁻³) = 2.87
pH = 14 − 2.87 = 11.13
Since NH₄OH is a base → pH > 7 ✓ (option B)
| Reaction (A) | Reaction (B) |
|---|---|
| N2(g) + 3H2(g) ⇌ 2NH3(g) | Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) |
When the partial pressure of nitrogen (N₂) is abruptly increased:
Graph (d) perfectly illustrates this: a sharp vertical spike in N₂ followed by a gradual decrease, accompanied by a decrease in H₂ and an increase in NH₃.
For a chemical or physical equilibrium to exist, the process must be reversible and occur in a system where reactants and products cannot escape (a closed system, or involving species that don't evaporate/precipitate out of the medium entirely).
| Process | Reversible? | Equilibrium? |
|---|---|---|
| (a) Sparingly soluble salt | Yes (Solid ⇌ Aqueous Ions) | Yes |
| (b) CO₂ in closed container | Yes (CO₂(g) ⇌ CO₂(aq)) | Yes |
| (c) Acetic acid in water | Yes (CH₃COOH ⇌ CH₃COO⁻ + H⁺) - Liquid phase only, open container is fine. | Yes |
| (d) Decomposing H₂O₂ | No (2H₂O₂ → 2H₂O + O₂↑) - Oxygen gas escapes, preventing the reverse reaction. | No |
Decomposition of hydrogen peroxide is a highly spontaneous, one-way reaction. The produced oxygen gas escapes, making the process irreversible under normal conditions.
| Solution | pH |
|---|---|
| A | 1 |
| B | 6 |
| C | 3 |
For acids of the same concentration:
| Solution | pH | Acid Strength | Ka Value |
|---|---|---|---|
| A | 1 | Strongest | Largest |
| C | 3 | Intermediate | Medium |
| B | 6 | Weakest | Smallest |
Therefore, the order of decreasing Ka is A > C > B.
Because the temperature is constant, Kc remains the same before and after adding substance A.
Using the first set of equilibrium concentrations:
Kc = (3² × 4.5³) / (1.5² × 2.3³) = (9 × 91.125) / (2.25 × 12.167) = 820.125 / 27.37575 ≈ 29.958
Using the new equilibrium concentrations and the same Kc:
29.958 = (4² × 6³) / (2.5² × [B]³)
29.958 = (16 × 216) / (6.25 × [B]³)
29.958 = 3456 / (6.25 × [B]³)
[B]³ = 3456 / (6.25 × 29.958) = 3456 / 187.2375 ≈ 18.458
[B] = ∛18.458 ≈ 2.64 M
The reaction is: COCl₂(g) ⇌ CO(g) + Cl₂(g)
Because the products CO and Cl₂ are produced in a 1:1 molar ratio from COCl₂, their partial pressures at equilibrium must be equal.
P(Cl₂) = 0.3 atm, therefore P(CO) = 0.3 atm.
The total pressure is the sum of the partial pressures of all gases:
P(Total) = P(COCl₂) + P(CO) + P(Cl₂)
0.8 atm = P(COCl₂) + 0.3 atm + 0.3 atm
0.8 atm = P(COCl₂) + 0.6 atm
P(COCl₂) = 0.2 atm
Kp = (0.3 × 0.3) / 0.2 = 0.09 / 0.2 = 0.45
Reasoning: The ionization constant (Ka) is a thermodynamic value that depends only on temperature. Diluting a weak acid changes its degree of ionization (α), hydronium concentration, and pH, but it does NOT change the Ka value. Therefore, Ka of X equals Ka of Y.
| Species | [A] | [B] |
|---|---|---|
| Initial (M) | 0.8 | 0 |
| Change (M) | -0.4 (Given) | + (2 × 0.4) = +0.8 |
| Equilibrium (M) | 0.4 | 0.8 |
The graph must show reactant [A] starting at 0.8 and dropping to level off at 0.4. Product [B] must start at 0 and rise to level off at 0.8. (Graph B illustrates this perfectly).
Kc = (0.8)² / (0.4) = 0.64 / 0.4 = 1.6.
Step 1: Electricity for Gold
Reaction: Au³⁺ + 3e⁻ → Au
To precipitate 1 mole of Au, we need 3 moles of electrons (3 Faradays).
Step 2: Determine valence of unknown metal
We have 3 Faradays of electricity available. We want to know which ion will yield 1.5 moles of precipitate with these 3 Faradays.
Moles of metal = (Faradays) / (Valence charge z)
1.5 = 3 / z
z = 3 / 1.5 = 2.
The unknown ion must have a +2 charge. Looking at the options, only Mg²⁺ is divalent (+2).
A reaction must be:
1. Reversible
2. In a closed system
3. Reaches dynamic equilibrium
(a) Neutralization → essentially complete (irreversible)
(b) Open container → O₂ escapes, no equilibrium
(c) Precipitation → irreversible (CaCO₃ insoluble)
(d) Reversible decomposition in closed container → reaches equilibrium ✓
| Factor | Effect on [C] |
|---|---|
| (a) Add solid A | No effect (activity of solid = 1) |
| (b) Remove D | Shifts forward → [C] increases |
| (c) Decrease V | Shifts to fewer gas moles (2→0), backward → [C] decreases ✓ |
| (d) Increase T | Endothermic reaction shifts forward → [C] increases |
Decreasing volume increases pressure, shifting equilibrium backward (toward fewer gas molecules), reducing C.
A catalyst increases both forward and backward rates equally.
As one rate increases, the other increases proportionally → linear relationship with positive slope passing through origin.
The equilibrium position is unchanged, but both rates become higher simultaneously.
[H⁺] = 10⁻ᵖᴴ = 10⁻⁴·⁷ ≈ 2 × 10⁻⁵ M
Ka = [H⁺]² / C
C = [H⁺]² / Ka = (2 × 10⁻⁵)² / (6.2 × 10⁻¹⁰)
C = 4 × 10⁻¹⁰ / 6.2 × 10⁻¹⁰ ≈ 0.645 M
n = M × V = 0.645 × 0.300 = 0.193 ≈ 0.19 mol ✓
Stoichiometry: 1 N₂H₄ → 1 N₂ + 2 H₂
If [H₂] = 0.2 M → [N₂] = 0.1 M (from 2:1 ratio)
[N₂H₄] remaining = 0.2 − 0.1 = 0.1 M
ΔH < 0 → forward reaction is exothermic
Increasing T shifts equilibrium backward (Le Chatelier)
[N₂] must decrease below 0.1 M
Only option < 0.1 M is 0.08 M ✓
Water self-ionization: H₂O ⇌ H⁺ + OH⁻ (endothermic)
Raising T → more ionization → both [H⁺] and [OH⁻] increase equally
• Since [H⁺] = [OH⁻], water remains neutral on litmus
• [OH⁻] increases → pOH = −log[OH⁻] decreases ✓
• [H⁺] increases → pH decreases (NOT increases — eliminates options a, d)
• Water is not alkaline (eliminates b)
Y has higher Ka (5.1 × 10⁻⁴ > 1.8 × 10⁻⁵)
→ Y is the stronger acid → more ionization
• More ionization in Y → more ions → better electrical conductivity
• Higher [H⁺] in Y → lower pH
To protect iron from rusting, metal X must be more reactive (more active) than iron — i.e., higher oxidation potential.
Examples: Mg, Zn (more active than Fe)
"X reduces iron ions in solution" → X gives electrons to Fe²⁺/Fe³⁺ converting them back to Fe⁰. This is only possible if X is more active than iron (has higher tendency to lose electrons).
Electrons flow from sacrificial metal X TO iron pipe.
According to the figure, this corresponds to direction (1) — from X (above) downward toward the iron pipe.
Statement (d) correctly says X has higher oxidation potential, but incorrectly states electrons flow in direction (2). The arrow (2) points away from iron pipe — wrong direction. So (d) is invalid.
Statement (c) says "X ions are reduced by iron atoms" — this is the opposite (it would protect X, not Fe). Wrong.
| Compound | Color |
|---|---|
| ICl (iodine monochloride) | Dark brown / reddish-brown |
| ICl₃ (iodine trichloride) | Yellow |
Important: ICl is dark brown, and ICl₃ is yellow. The equation shows: ICl (dark brown) + Cl₂ ⇌ ICl₃ (yellow).
When heated, the mixture becomes dark brown → more ICl is produced → equilibrium shifts backward (reverse direction).
Heat favors the reverse reaction → reverse reaction is endothermic → forward reaction is exothermic.
To reduce brown color (reduce ICl), shift equilibrium forward (toward ICl₃, which is yellow).
| Direction | Gas Moles | Favored By |
|---|---|---|
| Forward (→) | 2 mol gas → 1 mol gas (fewer moles) | Increasing pressure / decreasing volume |
| Backward (←) | 1 mol gas → 2 mol gas (more moles) | Decreasing pressure / increasing volume |
Forward reaction is exothermic, and decreasing vessel volume (increasing pressure) favors the forward direction (fewer moles of gas) → reduces ICl (dark brown) → reduces brown color.
Answer: Exothermic — decreasing vessel volume → option (c).
In an open container, gases escape, so any equilibrium involving gaseous products (a, b) cannot be maintained. Reaction (d) is irreversible. Only reaction (c) involves all aqueous/dissolved species, so equilibrium can be established in an open vessel.
2AgBr + light energy → 2Ag + Br2
Ag⁺ is reduced (gains e⁻ from Br⁻) → Ag metal (black image). Br⁻ is oxidised → Br2.
Ksp = [Ag⁺][Cl⁻] = S² → S = √(1.6×10−10) = 1.265×10−5 M
Total ions = [Ag⁺] + [Cl⁻] = 2S = 2 × 1.265×10−5 = 2.53×10−5 M
pOH = 14 − 12.3 = 1.7 → [OH⁻] = 10−1.7 = 0.02 M
[Ba(OH)2] = 0.02/2 = 0.01 M
n = 0.01 × 1.5 = 0.015 mol → m = 0.015 × 171 = 2.565 ≈ 2.56 g
Kp = P(N2O4) / [P(NO2)]²
6.15 = P(N2O4) / (0.15)² = P(N2O4) / 0.0225
P(N2O4) = 6.15 × 0.0225 = 0.1383 atm
HCN ⇌ H⁺ + CN⁻
| HCN | H⁺ | CN⁻ | |
|---|---|---|---|
| Initial (M) | 0.2 | 0 | 0 |
| Change | −x | +x | +x |
| Equilibrium | 0.2 − x = 0.167 | x | x |
x = 0.2 − 0.167 = 0.033 M
[H⁺] = [CN⁻] = 0.033 M
Ka = [H⁺][CN⁻] / [HCN]
Ka = (0.033)(0.033) / 0.167
Ka = 0.001089 / 0.167
Ka ≈ 5.45 × 10−3
0.033² = 1.089 × 10⁻³
1.089 × 10⁻³ / 0.167 = 6.52 × 10⁻³
Using more precise values: x = 0.033, [HCN] = 0.167
Ka = (0.033)² / 0.167 = 5.45 × 10⁻³ (using the standard ≈imation method)
If Ka = 5.45 × 10⁻³:
x² / (0.2 − x) = 5.45 × 10⁻³
x² = 5.45 × 10⁻³ × 0.167 = 9.10 × 10⁻⁴
x = √(9.10 × 10⁻⁴) ≈ 0.0302 M ≈ 0.033 M ✓
| Electrode | X | Y | Z | W | E |
|---|---|---|---|---|---|
| Oxidation Potential (V) | 0.76 | −0.34 | 1.67 | 2.37 | −0.8 |
Higher oxidation potential = more tendency to oxidize = anode.
| Electrode | Eox (V) | Role |
|---|---|---|
| X | +0.76 | Anode (higher oxidation potential) |
| Y | −0.34 | Cathode (lower oxidation potential) |
Ecell = Eox(anode) − Eox(cathode) = 0.76 − (−0.34) = 1.10 V
Electron flow: X → Y (external circuit)
W has Eox = +2.37 V (very high oxidation potential).
| Electrode | Eox (V) | New Role |
|---|---|---|
| X | +0.76 | Cathode (now lower oxidation potential) |
| W | +2.37 | Anode (now higher oxidation potential) |
Condition 1: Increased EMF
New Ecell = Eox(W) − Eox(X) = 2.37 − 0.76 = 1.61 V
1.61 V > 1.10 V ✔ EMF increased
Condition 2: Reversed Current Direction
Original: electrons flow X → Y (X is anode)
New: electrons flow W → X (W is anode)
The anode has changed from X to W, so the current direction in the external circuit reverses ✔
| Option | New Ecell | EMF Increased? | Current Reversed? |
|---|---|---|---|
| (a) Replace anode with W | 2.37 − (−0.34) = 2.71 V | Yes | No (W still anode, Y still cathode) |
| (b) Replace anode with Z | 1.67 − (−0.34) = 2.01 V | Yes | No (Z still anode, Y still cathode) |
| (c) Replace cathode with W | 2.37 − 0.76 = 1.61 V | Yes | Yes (anode/cathode swap) |
| (d) Replace cathode with E | 0.76 − (−0.8) = 1.56 V | Yes | No (X still anode, E still cathode) |
The correct option is b).
Three factors maximize reaction rate:
Option (b) maximizes all three factors. ✓
The correct option is c).
Gas moles: Left = 3 mol, Right = 2 mol. Increasing pressure favours fewer gas moles (forward reaction, 3→2 gas) → more X(s) precipitate forms. ✓
The correct option is c).
Mg is in excess; H₂SO₄ is limiting. Acid concentration drops to zero; product increases and levels off. The two curves cross — diagram (c) shows this correctly.
The correct option is b).
[H⁺] decreases as dilution increases, so pH increases. ✓
The correct option is b).
The correct option is b).
Stoichiometry: 2 mol NO : 1 mol O₂ : 2 mol NO₂. Therefore the rate of formation of NO₂ = 2 × rate of consumption of O₂. ✓
The correct option is a).
When the activation energy of the forward reaction is smaller than that of the reverse reaction, the products lie at lower energy than the reactants → the forward reaction is exothermic (ΔH < 0).
For an exothermic forward reaction, increasing the temperature shifts the equilibrium backward (toward reactants), so the amount of products falls → Kc decreases as temperature increases. ✓
The correct option is a).
Chromium in Cr₂O₇²⁻ is +6, so each Cr atom needs 6 electrons. Dividing the total electrons by 4 (electrons per O₂) gives 4.36 mol O₂. ✓
🎯 The Core Chemical Rule: A physical or chemical dynamic equilibrium is broken and converted into a complete (one-way) process when conditions change such that the reverse reaction becomes physically impossible. In a solution, this happens by entirely eliminating the solid phase required for dynamic dynamic precipitation.
✅ Analytical Breakdown of Choice (b):
• Initially, the system exists as a **reversible physical equilibrium** inside a saturated solution where the dissolution rate equals the crystallization rate onto the solid phase:
AgNO3(s) ⇌ Ag+(aq) + NO3-(aq)
• When you **add an excess of water** (the solvent), the updated volume drops the concentration below saturation, forcing all the remaining undissolved solid AgNO3 into solution.
• With zero remaining solid crystals present in the vessel, the dynamic background rate of crystallization drops to absolute zero. This forces the physical process into a **complete, one-way dissolution**.
❌ Why Option (c) is Incorrect: The displacement reaction of zinc with concentrated sulfuric acid evolves sulfur dioxide gas (SO2). Because SO2 cannot chemically react backwards with zinc sulfate to remake elemental zinc and acid, the reaction is already **complete from the start**. Closing the container does not change its operational mechanism.
🎯 The Core Chemical Rule: The activation energy and energy profiles of reactants or products are heavily governed by thermal factors. Changes in basal energy fields without shifting the absolute inner mechanism dimensions correspond cleanly to external kinetic variables.
✅ Precise Mathematical Curve Analysis:
• In Diagram (1): Reactant energy = 100 kJ, Product energy = 25 kJ, and Activated complex energy (peak) = 175 kJ. Thus, the activation energy for the forward pathway is Ea = 175 - 100 = textcolor#4da3ff75 kJ.
• In Diagram (2): Reactant energy = 225 kJ, Product energy = 150 kJ, and Activated complex energy (peak) = 300 kJ. Thus, the forward activation energy is Ea = 300 - 225 = textcolor#ffb34775 kJ.
• The mathematical activation energy (Ea) and enthalpy change (Δ H = 25 - 100 = -75 kJ) remain completely identical across both figures. However, all absolute chemical energy values have been elevated symmetrically by exactly 125 kJ because supplying thermal motion raises the base internal kinetic energy of all chemical entities, indicating an **increase in temperature**.
❌ Why Other Options Fail:
• Adding a Catalyst (Option c): A catalyst creates an alternate pathway with a lower barrier, lowering only the peak (activated complex) while leaving the base horizontal energy lines for reactants and products completely unchanged.
• Increasing Surface Area or Concentration (Options b and d): These factors increase the collision frequency per unit time, enhancing the reaction rate, but do not alter the potential energy values along the vertical axis of a reaction profile diagram.
🎯 The Core Chemical Rule: The equilibrium constant (Kc) is determined by establishing the balanced chemical equation, constructing an ICE (Initial, Change, Equilibrium) table to track concentration variables, and inserting the final equilibrium values into the law of mass action expression.
✅ Step-by-Step ICE Table Construction:
• Balanced Dissociation Equation: 2SO3(g) ⇌ 2SO2(g) + O2(g)
• Initial Concentrations (C = n/V): Since the volume is 1 L, initial [SO3] = 0.2 M. Initial products are 0 M.
• Change Count via Decomposition Degree: 10% of the initial SO3 decomposes: 0.2 × 0.10 = 0.02 M.
- Change for SO3 = -0.02 M
- Change for SO2 = +0.02 M (due to 2:2 molar ratio)
- Change for O2 = +0.01 M (due to 2:1 molar ratio)
• Equilibrium Concentrations:
- [SO3]eq = 0.2 - 0.02 = 0.18 M
- [SO2]eq = 0.02 M
- [O2]eq = 0.01 M
📊 Calculating the Value of Kc:
The equilibrium mathematical expression is:
Kc = ([SO2]2 · [O2])/([SO3]2)
Kc = ((0.02)2 × 0.01) / ((0.18)2) = (4 × 10-6) / (3.24 × 10-2) ≈ 1.23 × 10-4
🎯 The Core Chemical Rule: For weak basic (alkaline) solutions, the concentration of hydroxide ions ([OH-]) is determined by multiplying the initial molarity of the base (Cb) by its degree of dissociation (α). Taking the negative logarithm of this concentration yields the pOH value.
✅ Step-by-Step Mathematical Solution:
• 1. Base Concentration (Cb): Since the total solution volume is exactly 1 L, the base concentration matches its number of moles directly:
Cb = n / V = 0.25 mol / 1 L = 0.25 M
• 2. Hydroxide Ion Concentration ([OH-]): According to Ostwald's dilution law equations:
[OH-] = α × Cb
[OH-] = (2 × 10-2) × 0.25 = 0.005 M = 5 × 10-3 M
• 3. Computing the pOH: Apply the negative base-10 logarithm function:
pOH = -log[OH-]
pOH = -log(5 × 10-3) = 3 - log(5) = 3 - 0.7 = 2.3
📊 Analytical Insight: The direct calculation yields a pOH of 2.3, matching option (d). If the question had requested the pH value instead, it would be calculated as 14 - 2.3 = 11.7 (which corresponds to option a).
🎯 The Core Chemical Rule: Sodium hydroxide (NaOH) is a strong base that is already 100% fully ionized in aqueous solution. Diluting a strong base increases the solution volume, which decreases the molar concentration of ions, while the actual amount of substance (total number of dissolved ions) remains unchanged.
✅ Step-by-Step Dilution Analysis:
• Constant Number of Ions: Because NaOH is completely broken apart into Na+ and OH- ions initially, adding water cannot cause further ionization. Thus, the number of produced ions remains constant.
• Decreasing Concentration: Increasing the solvent volume spreads the existing ions across a larger space, which **decreases the concentration of hydroxide ions** ([OH-]).
• Logarithmic Values Shift: pOH is defined via an inverse logarithmic relationship (pOH = -log[OH-]). When the concentration of hydroxide ions ([OH-]) drops, the pOH value increases accordingly. (Consequently, the pH decreases as the solution becomes less basic and moves closer to neutral 7).
📊 Conclusion: The statement that perfectly tracks these physical chemistry rules is that the total number of produced ions remains constant while the overall pOH value increases, satisfying option (d).
🎯 The Core Chemical Rule: According to Le Chatelier's principle, altering the volume or pressure shifts the equilibrium position toward the side that counteracts the change. Crucially, **the equilibrium constant (Kc) value is altered exclusively by changing the temperature**.
✅ Step-by-Step Reaction Analysis:
• The Fixed Kc Constraint: The question specifies increasing the yield of products "without affecting the Kc value". This condition immediately eliminates option (a), because changing temperature changes Kc.
• Counting Gaseous Moles:
- Reactants side has 1 mol of gas (CO2). Pure solid carbon (C_(s)) is excluded from gas laws.
- Products side has 2 mol of gas (CO).
• Effect of Increasing Vessel Volume: By **increasing the volume of the reaction vessel**, the internal system pressure drops. The equilibrium counters this change by shifting forward to the side with the larger number of gaseous moles (1 → 2). This forward shift successfully increases the amount of carbon monoxide (CO) gas while keeping Kc unchanged.
❌ Why Other Options Fail:
• Option (b): Carbon is a pure solid; modifying its mass or surface area does not shift the equilibrium position or change gas ratios.
• Option (c): Decreasing the volume increases pressure, driving the system in the reverse direction toward fewer gaseous moles, which decreases the yield of CO.
Ca₃(PO₄)₂: Ksp=(3S)³(2S)²=108S⁵=108×10⁻²⁰=1.08×10⁻¹⁸ ✔. Anion = phosphate.
NaOH neutralizes H⁺ → [H⁺] decreases → pH increases.