Law of Mass Action — K p & K c Dynamic equilibrium: balance of opposing rates Why this matters: Many real reactions do not go to 100% products. Instead, they reach a steady state where reactant and product concentrations stop changing, yet both forward and reverse reactions still run. This is dynamic equilibrium. At equilibrium, the forward rate equals the reverse rate, so macroscopic concentrations stay constant even though molecules keep reacting microscopically. E1 E2 E3 Dynamic equilibrium condition: forward and reverse rates become equal. E4 Mass-action view of aA + bB ⇌ cC + dD: forward and backward rates depend on the concentrations raised to stoichiometric powers. 2026-05-26T17:04:31.847Z Approach to equilibrium: concentrations level off as forward and reverse rates converge. gpt-image-2 Concentration vs time and rate vs time for a reversible reaction. Plot [A], [B] decreasing, [C], [D] increasing; separate panel showing Rate forward and Rate reverse curves crossing and then equal. Clean 2D axes, labels, neutral textbook palette, arrows in red. No embedded text aside from axis labels. Law of Mass Action → K c Guldberg and Waage (1864) observed: at a fixed temperature, any given reaction at equilibrium has a constant ratio of product concentrations to reactant concentrations, each raised to their stoichiometric coefficients. From the rate equality and dividing both sides by [A] a[B] b , we get the equilibrium constant in terms of concentration (mol L -1 ), denoted K c . Law of Mass Action (concentration form). E5 Stoichiometric coefficients from the balanced equation become exponents in K c . K c depends only on temperature for a given reaction. remember K p and the K p – K c relation For gas-phase equilibria, it is often convenient to write the constant using partial pressures in atm or bar. This is K p . Law of Mass Action (pressure form for gases). E6 Relates K p and K c for gas reactions, where n g = moles of gaseous products − moles of gaseous reactants. Use R = 0.0821 L atm mol -1 K -1 with P in atm and T in K. Infographic linking K p and K c : highlight Δ n g cases (positive, zero, negative). 2026-05-26T17:04:32.304Z Infographic with three panels for Δ n g > 0, = 0, < 0 showing Kp = Kc(RT) Δ n g . Include typical gas reactions examples (symbolic), arrows indicating increase/decrease with T factor, and units cues. Vector style, clean icons, no text beyond symbols. gpt-image-2 Aspect K c (concentration) K p (partial pressure) K c vs K p at a glance Feature Definition [Products] stoich /[Reactants] stoich ( P products ) stoich /( P reactants ) stoich When used Any equilibrium; especially solutions Gas-phase equilibria Relation K p = K c (RT) n g Units (typical NEET) (mol L -1 ) n (pressure) n g Special case If Δn = 0 → unitless (NEET view) If Δ n g = 0 → K p = K c Use consistent units: if you plug P in atm, use R = 0.0821 L atm mol -1 K -1 . If you plug P in bar, use R = 0.08314 L bar mol -1 K -1 . neet-alert Units of K c and K p For a reaction aA + bB cC + dD, define n = (c + d) - (a + b) for concentration form and n g for gas moles in pressure form. In NEET convention (using concentrations/pressures rather than activities): - Units of K c : (mol L -1 ) n - Units of K p : (pressure) n g If n = 0 or n g = 0 , the constant is dimensionless in this convention. Remember: in rigorous thermodynamics, K is defined with activities and is dimensionless; NEET uses the above practical unit treatment. Homogeneous vs heterogeneous equilibria Homogeneous: all reactants/products in the same phase (often all gases or all solutes). Heterogeneous: more than one phase (solids, liquids, gases present together). In equilibrium expressions, pure solids and pure liquids are omitted (their effective concentration/activity is constant and folded into K). Only gaseous species and solutes appear. Reaction (balanced) Type K expression (NEET form) K expressions: homogeneous vs heterogeneous Example N2(g) + 3 H2(g) 2 NH3(g) Homogeneous (all gases) K p = ( P NH3 ) 2 / ( P N2 ( P H2 ) 3) CaCO3(s) CaO(s) + CO2(g) Heterogeneous K c = [CO2] (or K p = P CO2 ) H2O(l) H + (aq) + OH - (aq) Heterogeneous K c = [ H + ][ OH - ] (omit pure liquid H2O) Quick check: If it’s a pure solid or a pure liquid with constant density, don’t include it in K. tip Reaction quotient Q vs K: predicting direction Q has the same algebraic form as K, but uses the current (not necessarily equilibrium) concentrations or pressures. Compare Q and K to see the spontaneous direction of change at that moment (at constant T). Comparison Consequence Q vs K: what happens next? Q < K System moves forward (right) to form more products Q = K System is already at equilibrium Q > K System moves backward (left) to form more reactants Horizontal number line labeled with K at center. Three sample Q markers: left of K (arrow → right), at K (no arrow), right of K (arrow → left). Minimalist vector graphic, clean arrows in red/blue, no extra text. gpt-image-2 Q vs K number line: show Q points relative to K and arrow indicating direction to reach K. 2026-05-26T17:04:32.206Z Dynamic equilibrium as a balanced seesaw: a change in the 'mass' (concentration/pressure) pushes the system to re-balance according to Q vs K. Thermodynamic link (at standard state): large K implies negative G . Cross-link to Thermodynamics. Industrial connections (why K guides operating conditions) Haber–Bosch: N2(g) + 3 H2(g) 2 NH3(g). Fe–Mo catalyst, ~200 atm, ~700 K. K p decreases with higher T for this exothermic synthesis, so industry uses a compromise T for reasonable rate and yield. Contact process: 2 SO2(g) + O2(g) 2 SO3(g), V2O5 catalyst, ~720 K, 1–2 atm. Steam–methane reforming: CH4(g) + H2O(g) CO(g) + 3 H2(g) (syngas). Here n g = +2 , so K p grows with the (RT) factor when comparing to K c ; pressure and temperature choices balance rate and yield. Worked numericals (NEET-style) 1) ICE-table with K c Reaction: H2(g) + I2(g) 2 HI(g). At a certain T, K c = 50.0. Initial: [H2] = 1.00 M, [I2] = 1.00 M, [HI] = 0.00 M. Find equilibrium concentrations. Initial (I) 1.00 1.00 0.00 Change (C) -x -x +2x Equilibrium (E) 1.00 - x 1.00 - x 2x ICE table: H2 + I2 2 HI [H2] (M) [I2] (M) [HI] (M) Line K c = [HI] 2/([H2][I2]) = (2x) 2/((1 - x)(1 - x)) = 50.0 → 4x 2/(1 - x) 2 = 50. Taking square root: 2x/(1 - x) = 50 = 7.071. Solve: 2x = 7.071(1 - x) → 9.071x = 7.071 → x = 0.780. Therefore [HI] = 1.560 M; [H2] = [I2] = 0.220 M. 2) Converting K c to K p Reaction: A(g) 2 B(g) at 400 K. Given K c = 4.00. Here n g = 1 . Using R = 0.0821 L atm mol -1 K -1 : K p = K c (RT) n g = 4.00 × (0.0821 × 400) = 4.00 × 32.84 = 131.36. 3) Direction using Q p vs K p For N2(g) + 3 H2(g) 2 NH3(g) at a certain T, suppose K p = 5.0 × 10 -3 . A mixture has P N2 = 0.80 atm, P H2 = 0.90 atm, P NH3 = 0.10 atm. Compute Q p = ( P NH3 ) 2/( P N2 ( P H2 ) 3) = (0.10) 2/(0.80 × 0.90 3 ) = 0.010/(0.80 × 0.729) ≈ 0.010/0.5832 ≈ 1.71 × 10 -2 . Since Q p > K p , the system will move left (consume NH3) to reach equilibrium. tip Always check stoichiometric powers in Q or K expressions. For N2 + 3 H2 ⇌ 2 NH3, the cube on H2 is a common place for mistakes. Real-life: hemoglobin–oxygen equilibrium Hb(aq) + O2(g) HbO2(aq). In lungs (high O2 partial pressure), Q < K so equilibrium shifts right → O2 loads. In tissues (lower O2 partial pressure, higher CO2 affecting pH), Q > K for binding so O2 is released. remember Hemoglobin–oxygen equilibrium shifts with O2 partial pressure: right in lungs, left in tissues. What changes K — and what does not? Temperature-dependent only (for a given reaction). Independent of initial concentrations/pressures. Independent of catalyst: catalysts speed up reaching equilibrium; they do not change K. For the reverse reaction, the equilibrium constant is 1/K. Characteristics of the equilibrium constant Do not mix K c and K p in the same expression. Convert using K p = K c (RT) n g first, then compare or calculate. neet-alert Common traps (and fixes) Equilibrium means the reaction has stopped. Equilibrium is dynamic: forward and reverse reactions continue at equal rates; concentrations are constant but molecules still react. The exponents in the equilibrium constant expression are the reaction orders from the rate law. In K expressions, exponents are always the stoichiometric coefficients from the balanced equation, regardless of the mechanism. Do not. Their effective concentrations are constant and absorbed into K. Only gaseous species and solutes appear. Include concentrations of pure solids and pure liquids in K. Changing initial amounts or adding a catalyst changes K. At a fixed temperature, K is unchanged. Initial amounts or catalysts only change how fast (or how far) the system moves to reach the same K. Omit S and L from K: "S and L are Set and Locked" (pure Solids and pure Liquids don’t enter K). State where forward and reverse reaction rates are equal, so concentrations remain constant while reactions continue microscopically. Dynamic equilibrium At constant temperature, the equilibrium ratio of concentrations (or pressures) raised to stoichiometric powers is constant for a given reaction. Law of Mass Action Equilibrium constant in terms of molar concentrations (mol L -1 ). K c Equilibrium constant in terms of partial pressures of gases (atm or bar). K p Expression with the same form as K but using current, not equilibrium, concentrations or pressures; predicts direction to reach equilibrium. Reaction quotient (Q) Change in moles of gaseous species: moles of gaseous products minus moles of gaseous reactants in the balanced equation. Δ n g Tabular method to track Initial, Change, and Equilibrium amounts or concentrations to solve for equilibrium compositions. ICE table All reactants and products in the same phase (e.g., all gases, or all solutes). Homogeneous equilibrium Reactants and products in different phases; pure solids and liquids are omitted in K. Heterogeneous equilibrium Temperature-dependent constant describing the position of equilibrium for a given reaction; relates to thermodynamics via G = -RT K . Equilibrium constant Key terms Conjugate Acid Conjugate Base Strength Relation Species REACTION VARIANTS Strong parents have weak children: the stronger the Bronsted acid, the weaker its conjugate base. Mastery of Bronsted-Lowry theory application. Inorganic Chemistry Equilibrium Bronsted-Lowry Theory NEET High Yield Conjugate Acid-Base Pairs HClO 4 ClO 4 - Strongest mineral acid yields the weakest conjugate base. HI I - Very strong acid yields a negligible conjugate base. HCl Cl - Strong acid yields a weak, stable conjugate base. H 2SO 4 HSO 4 - Strong first dissociation yields a very weak conjugate base. HNO 3 NO 3 - Strong acid yields a negligible conjugate base. H 3O + H 2O Strongest acid that can exist in water; water is its weak base. HSO 4 - SO 4 2- Moderately weak acid yields a weak conjugate base. H 3PO 4 H 2PO 4 - Weak acid yields a weak conjugate base. HF F - Weak acid yields a relatively strong weak conjugate base. CH 3COOH CH 3COO - Weak acid yields a weak conjugate base. H 2CO 3 HCO 3 - Weak acid yields a weak conjugate base. H 2S HS - Weak acid yields a weak conjugate base. NH 4 + NH 3 Weak acid yields a weak conjugate base. HCN CN - Very weak acid yields a stronger weak conjugate base. HCO 3 - CO 3 2- Very weak acid yields a moderately strong conjugate base. H 2O OH - Extremely weak acid yields a strong conjugate base. C 2H 5OH C 2H 5O - Negligible acid yields an extremely strong conjugate base. NH 3 NH 2 - Negligible acid yields a very strong conjugate base (amide ion). H 2 H - Negligible acid yields a powerful conjugate base (hydride ion). CH 4 CH 3 - Negligible acid yields one of the strongest possible conjugate bases. Increase Temperature (Endothermic, H > 0 ) Forward Shift Increases Absorption of heat is favored; K eq is temperature dependent Increase Temperature (Exothermic, H < 0 ) Backward Shift Decreases Release of heat is opposed; K eq is temperature dependent Decrease Temperature (Endothermic, H > 0 ) Backward Shift Decreases System shifts to generate heat; K eq is temperature dependent Decrease Temperature (Exothermic, H < 0 ) Forward Shift Increases System shifts to generate heat; K eq is temperature dependent Increase Concentration of Reactants Forward Shift No Change Q c < K c ; system consumes extra reactants Increase Concentration of Products Backward Shift No Change Q c > K c ; system consumes extra products Increase Pressure (where n g > 0 ) Backward Shift No Change System shifts toward the side with fewer moles of gas Increase Pressure (where n g < 0 ) Forward Shift No Change System shifts toward the side with fewer moles of gas Increase Volume (where n g > 0 ) Forward Shift No Change System shifts toward the side with more moles of gas Increase Volume (where n g < 0 ) Backward Shift No Change System shifts toward the side with more moles of gas Inert Gas Addition (Constant Volume) No Shift No Change Partial pressures of reacting species remain constant Inert Gas Addition (Constant Pressure, n g > 0 ) Forward Shift No Change Total volume increases; system shifts toward more moles Inert Gas Addition (Constant Pressure, n g < 0 ) Backward Shift No Change Total volume increases; system shifts toward more moles Addition of a Catalyst No Shift No Change Equilibrium attained faster; E a lowered equally for both directions Pressure Change (where n g = 0 ) No Shift No Change Moles of gas are equal on both sides Removal of Product Forward Shift No Change Q c < K c ; system replaces the removed species Shift Direction Effect on K eq Reason Condition Change TREND Equilibrium Le Chateliers Principle Chemical Kinetics Physical Chemistry Predicting system response to external stress. Heat the endo, squeeze the gas to the small side, and remember only Temperature can change the K constant! Le Chatelier's Principle Effects Here are a few prompt variations optimized for high-quality AI generation (Midjourney, DALL-E 3, or Stable Diffusion), specifically designed for a NEET chemistry study table. Option 1: The Sequential "Trend" View (Best for showing Cause & Effect) This prompt creates a composite image showing the "Before," "During," and "After" states, which is ideal for a "Trend" table. > Prompt: > A professional educational vector illustration divided into three vertical panels on a pure white background. Panel 1 (Top): A perfectly horizontal, balanced seesaw. The left side is labeled "Reactants" (blue spheres), the right side "Products" (red spheres). Panel 2 (Middle): The seesaw tilts heavily down to the left. Extra blue spheres are added, labeled "Stress: Added Reactant." Panel 3 (Bottom): A large directional arrow points from left to right, indicating "Forward Shift." The spheres are redistributing to level the beam. Style: High contrast, 2D flat textbook vector, clear sans-serif typography, scientific diagram aesthetic, distinct distinct outlines. Option 2: The Single Focus Diagram (Best for a small table cell) This prompt focuses on the moment of imbalance and the direction of the shift, suitable for a smaller image inside a table. > Prompt: > A single, high-contrast scientific vector diagram of a seesaw analogy for Le Chatelier's Principle. The seesaw is tilted downwards on the left side, which is overloaded with blue beaker icons labeled "High Concentration." The right side is raised high with fewer red beaker icons. A bold, thick green arrow curves from the left side toward the right side, labeled "Equilibrium Shift." Style: Clean lines, educational textbook illustration, flat colors, white background, minimalist but accurate, clear labels in bold black text. Option 3: Abstract/Icon Style (Best for "Quick Revision" cards) > Prompt: > A stylized, minimal vector icon representing chemical equilibrium shift. A balance scale (fulcrum) tipped to one side. On the heavier side, a "+" symbol indicates addition. A motion arrow indicates flow toward the lighter side. Style: Infographic vector art, crisp edges, deep blue and bright orange accent colors, pure white background, no shading, high visibility. Tips for Best Results: Aspect Ratio: If using Midjourney, add --ar 3:2 for Option 1 (vertical stack) or --ar 1:1 for Option 2 (square). Text Handling: AI struggles with specific text. If the generated text is gibberish, use a photo editor to replace the AI text with "Reactants," "Products," and "Shift" using a clean font like Arial or Helvetica. Color Coding: The prompt specifies Blue (Reactants) and Red (Products) to maintain visual logic for students.