Chemical Reactions

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Last Updated: Thursday, 01 October 2026 03:02

Published: Thursday, 16 April 2026 03:28

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Chemical Reactions:

Khan Academy => https://www.khanacademy.org/science/ap-chemistry-beta/x2eef969c74e0d802:chemical-reactions

Chemical reactions cause chemical changes, while non-chemical reactions cause only physical change. Ex: ice melting to water is physical change, while propane burning is a chemical change as it results in formation of new compounds. Chemical change results in making and breaking of bonds.

Consider the reaction between NaCl (sodium Chloride) and AgNO₃ (silver Nitrate). They are both solids, but they can be dissolved in a solution (usually water), and then they are denoted by "aq" to indicate they are in aqueous solution. This changes how the 2 compounds react.

Salt (NaCl) dissolves in water. We think of crystalline salt as solid NaCl, and these indeed exist as molecule of NaCl. But when dissolved n water, Na and Cl actually dissociate to form Na⁺ ions and Cl^- ions, surrounded by water molecules, with Na+ surrounded by partially -ve Oxygen atoms of water, while Cl- surrounded by partially +ve Hydrogen atoms of water. Similarly Silver Nitrate dissociates into Ag⁺ and NO₃^-_.

We write eq as follows in Molecular Eq as follows: NaCl (aq) + AgNO₃ (aq) => NaNO₃ (aq) + AgCl (s) [NOTE: AgCl solid is formed from the above reaction, as it precipitates out of the solution, it doesn't have high solubility]

To better represent that the aq solution actually consists of dissociated ions, we can write eq in complete Ionic Eq as follows:  Na⁺ (aq) + Cl^- (aq) +  Ag⁺ (aq) + NO₃^- (aq) => Na⁺ (aq)  + NO₃^- (aq) + AgCl (s) 

We can write above eqn in Net ionic eqn as follows: Cl^- (aq) +  Ag⁺ (aq) => AgCl (s) [By cancelling out spectator ions which are the ions which are just watching the reaction, In this case Na⁺ (aq) and NO₃^- (aq) are spectator ions]

+ve ions are called cations, as they get attracted to cathode (-ve terminal), while -ve ions are called anions, as they get attracted to anode (+ve terminal),

Balancing number of atoms of both sides of eqn is needed, as no atoms can disappear or appear out of no where.

ex: CH₄ (g) + H₂O (g) => CO (g) + 3H₂ (g) [Here we have to have 3 molecules of H₂ gas on RHS to have same number of H atoms on LHS (total of 6 H atoms on both sides of the eqn]

Stoichiometry: Here we solve chemical eqn to find how many grams of compounds react to produce how many grams of final product. This can be done by first balancing the eqn, then finding out molar mass of molecules on the 2 sides, and then covert it to grams. This is pure Maths addition/subtraction.

Oxidation Reduction (redox reaction):  An oxidation–reduction is a reaction that involves the transfer of electrons between chemical species (the atoms, ions, or molecules involved in the reaction). During a redox reaction, some species undergo oxidation, or the loss of electrons, while others undergo reduction, or the gain of electrons.

Oxidation is process of something combining with oxygen. Since oxygen has 6 electrons in outermost shell (n=2), it needs 2 more electrons to complete it's shell, which requires those electrons coming from the other compound, so oxidation is defined as loss of electrons (more +ve charge, gain of charge). Oxygen itself gains electrons, and that is defined as reduction (more -ve charge, loss of charge).

For example, consider the reaction between iron and oxygen to form rust:

4Fe(s) + 3O₂ (g) => 2Fe₂O₃ (s) (rusting of iron)

Fe has 2 electrons in outermost 4s shell, so it loses 2 electrons to for Fe²⁺. But more stable form is Fe³⁺ where it loses an extra electron from 3d shell to form half filled 3d shell (from 3d⁶ to 3d⁵ ). So Fe₂O₃ is a chain like structure with 2 oxygen on the end attaching to 2 Fe atoms, while oxygen in the center forms 1 bond with each Fe atom, forming O^2- for all 3 O atoms, and Fe loses 3 electrons for each Fe atom to form Fe³⁺. Fe is oxidized, while O is reduced.

An atom’s oxidation number (or oxidation state) is the imaginary charge that the atom would have if all of the bonds to the atom were completely ionic. Oxidation involves an increase in oxidation number, while reduction involves a decrease in oxidation number.

The sum of the oxidation numbers for all atoms in a neutral compound is equal to zero, while the sum for all atoms in a polyatomic ion is equal to the charge on the ion. Look in KA for examples on Oxidation number.

Half reaction method: To balance a redox equation using the half-reaction method, the equation is first divided into two half-reactions, one representing oxidation and one representing reduction, and then both are balanced independently and then combined in the end.

Equilibrium: In chemical eqn, we have reactants and products. At time=0, we only have reactants, and as time progresses, products are formed, and finally both reactants and products are in equilibrium at some time=T. We denote this by writing eqn with arrow pointing in both forward and backward dir. At equilbrium, the conc of reactants and products remain constant. Reactants are still changing into products and products are still changing into reactants, but it happens at the same rate, so their conc don't change at equilibrium. 

Let's say reaction is aA + bB ⇔ cC + dD where A and B are reactants and C and D are products. a, b, c, d refer to number of molecules of each of these needed to form balanced eqn.

Molar conc: [A] (within square brackets) refers to molar conc of chemical A in moles/liter, written as M. So, 0.5M implies 0.5moles per liter. 1 liter of water weighs 1000g, 1 mole of water molecules weigh 18g, so molar conc of water is = 1000/18=55.5moles/liter = 55.5M.

Forward rate of reaction (R) = Kf * [A]^a * [B]^b

Forward rate of reaction (R) = Kr * [C]^c * [D]^d

We can plot graph of rate vs time,

• At start of reaction, forward reaction rate R keeps decreasing with time as conc of reactants dec, while backward reaction rate R keeps increasing with time as conc of products inc.In this scenario, reactants get converted into products.
• At equilibrium, these 2 rates become equal. 
• If it so happens that rate of backward reaction is greater than rate of forward reaction, then products get converted back into reactants. 

Equilibrium Constant (K_c): We add subscript c to indicate it's constant for concentration. 

K_c = [A]^a * [B]^b / ([C]^c * [D]^d) => Here [A], [B], [C], [D] refer to equilibrium conc in M. The unit of K_c is either dimensionless or has M raised to the power of something. For only gases involved in reactions, it's more convenient to measure their partial pressure to determine their conc, so, we define K_c = [P_c]^c * [P_d]^d / ( [P_a]^a * [P_b]^b) where P reps partial pressure of each gas.

K_c is usually specified at 25C (room temp), as K_c  varies with temp (K_c generally dec with inc temp for exothermic reaction, and inc with inc temp for endothermic reaction). K_c  is defined for a given reaction written that particular way. If we change how we write the reaction by doubling the number of molecules on each side of eqn, then K_c for that new reaction will be sqaure of the original one. If we reverse the eqn, then new K_c  will be reciprocal of original value. If we get a reaction by adding up 2 reactions, then new K_c  will be product of the individual K_c  of ach reaction.

ex: HF (aq) ⇔ H⁺ (aq) + F^- (aq)  For this reaction, K_c = 6.8*10^-4. If we multiply the reaction by 2, i.e 2HF (aq) ⇔ 2H⁺ (aq) + 2F^- (aq) , then new K_c = (6.8*10^-4)^2 as all products and reactants get squared now. If we reverse the reaction to H⁺ (aq) + F^- (aq) ⇔ HF (aq), then for this reaction, new K_c will be inverse of original one, i.e new K_c = 1/(6.8*10^-4).

However, if reaction involves a mix of solid and gas, then it's a heterogeneous reaction (in contrast to homogeneous reaction where all products and reactants are in same phase). We define this as heterogeneous Equilibrium Constant, and in calculating this, we leave out pure solids and pure liquid in heterogeneous equilibrium constant since their conc don't change, so it doesn't help having this in the equilibrium constant defn.

Magnitude of K_c says how many products and reactants we have at equilibrium.

• K_c < 1 => There are more reactants than products at equilibrium. If K_c << 1, this reaction barely proceeds in forward dir.
• K_c > 1 => There are more products than reactants at equilibrium. If K_c >> 1, this reaction proceeds to completion.

Reaction Constant (Q_c): Another term Q_c is defined which is the Reaction Constant at any time, where the conc are instantaneous conc of reactants and products at that time. Term Q_p is similarly defined for gases when rep in terms of partial pressures. 

• If Q_c < K_c  , that implies that there are more reactants than products, hence reaction is proceeding forward (to the right)
• If Q_c = K_c  , that implies that reaction is at equilibrium
• If Q_c > K_c  , that implies that there are more products than reactants, hence reaction is proceeding backward (to the left)

ICE table: An ICE table is made for any reaction, where I is initial partial pressure of all reactants/products, C=change in partial pressure and E is the equilibrium partial pressure. This ICE table can also be made for conc (instead of pressures).

To find E, we put I=initial Pressure, C=x for reactant (and -x for products, assuming coeff of 1 for both products and reactants), and then take E=I+C, and then solve for x by putting in K_c  to find final equilibrium pressures.

Le Chatelier's Principle: When stress is applied to a reaction at equilibrium, net reaction moves in a direction that relieves the stress.

• Concentration: Changing conc of reactants and products is one way to change stress at equilibrium. If we have a reaction where gas A is changing into gas B, then inc conc of Gas A, will cause the reac to go in dir where gas A will dec, meaning reaction moves forward. This can also be found out by calc Q_c which is less than K_c , hence reaction moves forward
• Volume: Changing vol of reactants and products is another way to change stress at equilibrium. Let's say we have a reaction of solid turning into another solid and gas in a container, and we dec the vol of container. Then pressure of gas will inc, and the reaction will try to dec pressure of gas, making reaction go to the left to dec gas pressure and inc amount of reactant.
  • NOTE: adding any inert gas at equil doesn't change the reaction at equil, since inert gas is not participating in the reaction, so Q_c remains the same and equal to K_c 
• Temperature: Changing Temp is another stress. This is the only one where we K_c for the reaction changes.
  • For an exothermic reaction (ΔH < 0), heat is released. Let's say we inc the temp, which inc heat. If heat is treated as a product, then reaction will try to go in a dir that moves it to the left so that heat is decreased. If we dec the temp, then reverse happens (reaction moves to right). We can also evaluate this by calc K_c , which inc as temp dec, so Q_c < K_c , implying reaction will move to right.
  • For an endothermic reaction (ΔH > 0), heat is consumed. Reverse happens over here, as compared to exothermic reaction. Let's say we inc the temp, which inc heat. If heat is treated as a reactant, then reaction will try to go in a dir that moves it to the right so that heat is decreased. If we dec the temp, then reverse happens (reaction moves to left). We can also evaluate this by calc K_c , which inc as temp inc, so Q_c < K_c , implying reaction will move to right.
  • Catalyst: One way to inc rate of reaction for exothermic reaction is to increase temp. However that changes K_c which changes the amount of products. If we don't want to change the amount of products, we can add a catalyst, which allows the reaction to reach equilibrium faster. Catalyst speeds up both the forward and backward reaction by same amount, hence K_c remains the same.

Dissociation of Water molecules: 

When we talk of water, we treat it as having all it's molecules present as H₂O. In reality, some of it's molecules dissociate into H⁺ and OH^-. The H+ combines with a lone H₂O molecule to form hydronium ion, H₃O⁺.  About 6 in every 100 million (6 in 10⁸) water molecules undergo the following reaction: 

  H₂O (l) +  H₂O (l) <=>  H₃O⁺ (aq) + OH^- (aq) Here l refers to liquid while aq refers to aqueous soln. The extra proton forms a covalent bond with 1 of the lone pair electron in oxygen. The reaction is bidir.

Dissociation of water => https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/06%3A_Acids_and_bases/6.05%3A_Dissociation_of_water

This process is called the autoionization of water and occurs in every sample of water, whether it is pure or part of a solution. This ionization of water gives rise to conductivity of water, albeit very low.

We can find the equilibrium.constant 

Conductivity of water: 

Let's take water as a solution in our standard cell to measure it's conductivity: electrodes of area A=1 cm2A=1 cm^2, gap d=1 cmd=1 cm, applied voltage V=1 VV=1 V, measured current I=5.5×10−8 AI=5.5×10^-8A (the 55 nA). It's extremely low current. Using R=V/I, we get R=1.8*10^7 ohms. So, resistivity ρ = R*L/A = 1.8*10^7 Ω-cm. So, resistivity is very high (almost insulator) as metals have conductivity in range of 10^-7 Ω-cm (10^14 times higher conductivity in metals).

The ionic conductivity of water arises from conductivity of  H⁺ and OH^-. We saw in "passive elements" section, that for any conducting medium, σ = -q*n*µ = n*q²*t/m, where n=number of electrons crossing per unit area, q=charge and µ= electron mobility. If there are other carriers besides electrons, then they all have to be added, i.e in semiC, we have holes (p) too, so σ = - (q*n*µ_n + q*n*µ_p) . For solutions, we have similar mobility concept for ions, as the free ions in a solution are what become carriers (as the ions are the ones that move in response to electric field. it's not electrons or holes). 

Electrolyte (ion) conductivity in a solution (κ similar to σ in metals/SemiC) = ∑ (z_i * F) * u_i * C_i = ∑  λ_i * C_i , where λ_i= (z_i * F) * u_i = molar conductivity of ions per mole (charge_per_ion*Faraday*mobility_of_ion) and C_i= ion concentration (mole/cm^3). 

Faraday is defined as total charge (in Coulombs) on 1 mole of electrons = 1.6*10^-19 C * 6.02*10^23 e/mole = 96,500 C. So, (z_i*F)= total charge on 1 mole of ions, which is equiv to "q" in our above classic resistivity eqn. This multiplied with mobility gives conductivity per mole of ions. When we multiply it with conc (mol/cm^3), we get total conductivity κ (in units 1/(Ω-cm) = Siemens/cm (S/cm) or in per volume unit of solution) = ∑  q*n*µ, where q=z_i*F, n=C_i , µ=u_i. 

However, in solutions, talking about conductivity in units of Ω-cm doesn't make much sense, because it will vary based on concentration of the solution. In metals, that was not a problem, since conc of electrons is fixed in metals. However for electrolytes, the same salt solution will have different conductivity based on conc of salt in water (i.e how many moles of salt are present per cm^3 of solution). So, we use Molar conductivity per mole λ_i as a more common unit to define conductivity of solutions, and then multiply it by conc (moles per unit volume) to get ionic conductivity. λ_i = (z_i*F)*u_i = charge*mobility = C*cm^2/(V-s)/mole = S.cm^2/mol (units obtained by dividing σ (S/cm) by conc (mol/cm^3)).

For most questions in electrolytic conductivity, you will see molar conductivity λ_i used for conductivity of solutions. We just need to multiply it by conc to get real conductivity, which is in same units as metal conductivity (S/cm). 

Acid and Base: 

Dissociation constant:

Definition of Acid/Base:

Bronsted and Lowry defined acid and base based on this. They defined acid as anything that is a proton/H⁺ donor. Coversely base was defined as anything that is a proton/H⁺ acceptor. 

ex: HCl (aq) + H₂O (l) → Cl^- + H₃O⁺ Here HCl donated a proton (H⁺), so it's a acid, and H₂O accepted a proton, so it's a base. The product formed from the acid after donating a proton is called "conjugate base" of the acid. Similarly the product formed from the base after accepting a proton is called "conjugate acid" of the base. 

The above eqn are generally rep by arrow in both direction, indicating the forward and backward reaction are both taking place at the same time, and it's a reaction that is in equilibrium.

Acid: donates H⁺. Product after donating becomes a conjugate base. It's called conjugate base, as it can accept H⁺. 

Base: accepts H⁺ . Product after accepting becomes a conjugate acid. It's called conjugate acid, as it can donate H+. 

Titration: It's a process for determining the conc of a acid/base in a solution. We find the unknown molar conc of a solution (i.e acid) by mixing it with known molar conc of another slution (i.e base) and note where does the resulting solution go neutral.