Chemical Reactions

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Last Updated: Sunday, 01 November 2026 22:40

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.

Rate of reaction: defined as change in conc of reactant or product vs change in time => R = ΔP/Δt  (where P is reactant or product, t=time). Units are M/s (molar/sec). For reaction A -> B , we write R = -ΔA/Δt = ΔB/Δt (Assuming same number of moles of products and reactants, so rates remain same for both). We put -ve sign for reactant side, so that final rate is +ve for both reactant and product (As we don't want -ve rates, that's just the defn thing).

ex: 2N₂O₅ (g) -> 4NO₂ (g) + O₂ (g) => If Rate of appearance of O₂ = 9*10^-6 M/s, then rate of appearance of NO₂ = 4*rate of O₂ = 4*9*10^-6 M/s. Rate of disappearance of N₂O₅ = -2*9*10^-6 M/s. Since we want the rate of reaction to be the same irrespective of the product or reactant, we divide each of them by their coeff, i.e rate of reaction = ΔO₂/Δt = 1/4 * ΔNO₂/Δt = -1/2 * ΔN₂O₅/Δt =  9*10^-6 M/s (we put -ve sign for ΔN₂O₅/Δt since ΔN₂O₅/Δt  by itself is -ve, so putting -ve sign keeps the rate +ve)

For any general reaction aA + bB ⇔ cC + dD, R = -1/a * ΔA/Δt = -1/b *  ΔB/Δt = + 1/c *  ΔC/Δt = 1/d *  ΔD/Δt

Factors affecting rate of reaction:

• Reactant conc: More the cons of reactions, faster the reaction
• Surface are of solid reactant: More the surface area of the solid reactant by breaking it into smaller pieces, inc the rate
• Temperature: Inc the temp of the reaction, more the rate (higher temp => more collisions, and hence higher rate)
• Catalyst: We can speed up the reaction by adding catalyst (explained later too)

Let's consider a reaction aA + bB=> Products, we want to know how does R depend on conc of A and B. Experimentally, we determine how does R depend on conc of A an B. Let's say R ∝[A] and R ∝[B]², i.e when conc of A (noted as [A] which is M=moles/liter) changes from 1M to 2M, rate changes from 0.01M/sec to 0.02M/sec (linear dependence), but when conc of B (noted as [B]) changes from 1M to 2M, rate changes from 0.01M/sec to 0.04M/sec (square dependence).

So, R = K * [A] * [B]² where K= Reaction constant. Units of K are different for different reaction as units of R is in M/s, and those of A,B are in M. Reaction is called 1st order in A, and 2nd order in B, with net order of the reaction being the sum of the two, which is 3. If R is indep of [A], then it's called 0 order reaction. i.e R=K[A]⁰ => R=K.

Zero Order Reaction: For reaction A -> Product where R ∝[A]⁰ ,R = K[A]⁰ =>  -d[A]/dt = K =>  [A]_t = -K*t + [A]₀ . If plotted [A] vs t, this is a straight line wit slope =-K (-ve slope, conc dec with time)

1st Order Reaction: For reaction A -> Product where R ∝[A], , R = K [A] => -d[A]/dt = K[A] => ∫ dA/A = ∫ Kdt => ln [A] = -Kt => A_t / A₀ = -K(t-0) => [A]_t = [A]₀ e^-Kt (exponentially decaying func)

If we write eqn in log form, than it's a straight line with A on Y axis and t on x axis: ln [A]_t = -Kt + ln [A]₀ (-ve slope, conc dec with time)

If reaction was aA -> Product, then R = 1/a * -d[A]/dt = K [A] / a. So, eqn changes to  ln [A]_t = -a*K*t + ln [A]₀ . (slope changes from K to K*a)

Half Life: The time it takes for reactants to get to 1/2 of it's initial value =>  [A]_t =  [A]₀ / 2.  t_1/2 = ln(2)/K. Radiaactive decay is also a first order process, so we use same eqn there too.

2nd Order Reaction: For reaction A -> Product where R ∝[A]² , R = K [A]² => -d[A]/dt = K[A]² => - ∫ dA/A² = ∫ Kdt => 1/[A]_t - 1/ [A]₀  = Kt =>1/ [A]_t = Kt + 1/ [A]₀ . Here the plot of 1/[A] vs t will be a straight line (+ve slope since we are plotting 1/[A], conc still dec with time)

NOTE:

• For gases, instead of using molar conc, we can use partial pressure too. PV=nRT => P =n/V*RT = MRT, so P ∝M.
• Rate calc above is a func of time, and is changing at every point in time. Rate constant is fixed, but rate at which products and reactants are changing is dependent on conc at that time. 

Elementary Reactions:

So far, we said that coeff of reactants may not affect the rate of reaction. However for some reactions which are 1 step reactions, the reacton equation itself along with. coef used may directly affect the rate (i.e we'll have to rewrite the eqn with correct coeff o reflect that). 

Unimolecular reaction: Only 1 molecule of a reactant produces products. A -> product. Rate R = K[A]¹

Bimolecular reaction: 2 molecules of reactants (1 of each) produces products. A + B  -> product. Rate R = K[A]¹[B]¹. This is determined by experiments if this is what the rate law follows. In such reactions, rate is proportional to conc of both molecules, as by inc conc of any of them them, the chance of collision inc. IF we have A + A -> product, then Rate =  K[A]¹[B]¹ = K[A]².

Termolecular reaction: 3 molecules of reactants (1 of each) produces products. A + B +C  -> product. Rate R = K[A]¹[B]¹[C]¹ . This is determined by experiments. In such reactions, rate is proportional to conc of all 3 molecules, so if we have  A + A + A -> product, then Rate =  K[A]¹[B]¹[C]¹ = K[A]³.

Reaction mechanism: Many reactions are composed of multiple elementary steps (each step being a elementary reaction). For these reactions, we don't know the exact relationship of rate vs conc of reactants. But by looking at each elementary step, we can determine the dependency. We verify it experimentally. The reaction mechanism is the seq of elementary steps by which the reaction proceeds

ex 1: Find K of a reaction comprised of multiple elementary steps:

NO₂ + CO -> NO + CO₂ . For this reaction, experimentally it's determined that R = K [NO₂]² (i.e the rate doesn't even depend on conc of CO, and in fact has sq dependence on NO₂. This is what is going on => The 2 reactants don't go into products in 1 step. Instead, there are 2 elementary steps that the reaction is composed of:

• Elementary Step 1 (Slow, takes 1 sec): NO₂ + NO₂ -> NO + NO₃ . Since this elementary step, R = K₁ * [NO₂] [NO₂] = K₁ * [NO₂]². This slowest step of the reaction is called the rate determining step (RDS) as most of the time of the reaction takes place during this step.
• Elementary Step 2 (Fast, takes 10-8 sec): NO₃ + CO -> NO₂ + CO₂ . Since this step is too fast, it doesn't contribute to the rate of reaction. Even if we put more or talke out CO, the change in this fast rate is very small compared to the slow rate.
• Net reaction (by adding the 2 steps above): NO₂ + NO₂ +  NO₃ + CO -> NO + NO₃ +  NO₂ + CO₂ . NO₃ is an intermediate product as it gets produced and consumed, and hence cancels out
• => NO₂ + CO -> NO + CO₂ (we get the final eqn after we cancel out the same compounds on LHS and RHS, which gives our original eqn). So rate of overall reaction R = R_RDS =  K₁ * [NO₂]²

ex 2: Find K for a reaction with fast equilibrium step:

2 NO + Br₂ -> 2 NOBr => It's comprised of 2 elementary steps

• Elementary Step 1 (fast, forward rate=K₁ , reverse rate=K_-1 ): NO + Br₂ ⇔ NOBr₂ (We are considering equil condition here so arrow should be bidir)
• Elementary Step 2 (slow, rate= K₂ ):  NOBr₂ + NO -> 2 NOBr 
• Adding these 2, we get the original eqn. NOBr₂ is an intermediate reactant.
• From slow step, R = K₂ * [NOBr₂] [NO] . However, this has intermediate reactant, whose conc we don't know, so we need to get rid of it. We can do that by assuming that 1st reaction comes to a fast equilibrium. We assume pre equilibrium condition to be approximated as equilibrium condition. Why can we do that? Because even if the pre-equil value is different, when taken over a longer time of slow reaction, the conc of fast step will always get close to the equil vale (or we can take avg value, which wil be very close to equil value). 
• R of 1st reaction = K₁ , [NO] [Br₂] = K_-1 [NOBr₂] =>  [NOBr₂] =  K₁ / K_-1 , [NO] [Br₂]  => This gives us the conc of NOBr₂ at equil (which is assumed to be true for any point in time). Conc of NOBr₂ remains constant as any equil condition will have constant amount of reactants and products.
• Substituting in 1st eqn, we get R = K₂ * [NOBr₂] [NO] =  K₂ *K₁ / K_-1 , [NO] [Br₂] . [NO]  = K * [NO]² [Br₂], where overall Rate K =  K₂ *K₁ / K_-1 . It's tempting to think that we could have gotten this eqn by just raising the conc to power of coeff of each reactant. But that's not how we got this. We got this rate eqn via experimentation, and we proved it via elementary reaction

NOTE: For elementary reactions ONLY, coeff of reactants is used in power of the reactant. For all other reactions, they have to be broken down into elementary reaction, so that powers of coeff may be used.

Conditions for a reaction to occur:

3 conditions for a reaction A + B-C -> A-B +C, where bond b/w B-C is broken and new compound with bond A-B is formed.

• molecules must collide to react
• collisions must have correct orientation, else bonds may not be formed.
• collisions must have enough energy to break existing bond (by overcoming the activation energy)

We studied Bond Energy in "periodic Table" section. If we make a plot of Energy of the compound (A, B-C) Vs time (reaction progress), we start with BE level of the reactants. The final energy is the BE of the products. Every reaction has an intermediate hill to climb (peak of the hill is the energy of the transition state), which is the activation energy, which allows the bonds to beak, before it gets to the final energy of the products. Energy diff is the energy released if reaction is exothermic, else energy is absorbed if reaction is endothermic.

As per Maxwell- Boltzmann distribution (MBD in "Phases of matter" section), we know that not all molecules have same energy. They have a distribution, so not all reactants are at BE energy level shown. Some of them are higher, some lower. Higher ones may be high enough to cross the hump and actually form product. So, depending on level of E_a and MBD, we may get a fraction of reactants getting converted into products.

Ex: Consider a reaction B-C + D -> B + C-D. Let's say it's a 2 step reactions. A is the catalyst (as it's not there in original eqn at all) and A-C is an intermediate product (as it's cancelled out)

step 1: A + B-C -> A-C + B 

step 2: A-C + D -> A + C-D

Here the graph of Potential Energy (BE) Vs reaction time shows 2 humps for the 2 reactions. 1st reaction is done at the valley at the end of 1st hill. 2nd reaction is done at the valley of 2nd hill. The height of 2 hills reflect E_a of the 2 reactions.  E_a of the 2 reaction is the height from he valley of 1st reaction to the top of the 2nd hump. In the ex on KA, E_a1 has higher height, so 1st step is slower and hence RDS.  E_a2 has lower height, so 2nd step is faster. This is an endothermic reaction, since Final BE > Initial BE, so energy has to be supplied for this reaction to happen.

Arrhenius Equation: Gives us a formula for rate constant K = Ae^(-E_a/RT),where R=gas constant, T=Temp and E_a=Activation energy, A=frequency factor. The term f=e^(-E_a/RT) gives us the fraction of collisions that are above the activation energy, and hence can react. As E_a goes down, f goes up.As T goes up, f goes up. At E_a=40KJ/mole, T=373K, we get f=2.5*10^-6 implying only 2.5 collisions out of 1M collisions are able to react and give products. If  E_a goes down to 10KJ/mole, f = 0.04 implying every 4 collisions out of 100 are able to react. If T gies up to 473K, then f=0.08, implying more reactive collisions are happening.

Catalyst: Catalyst is defined as a substance that inc rate f reaction, w/o being consumed in the process. It lowers the E_a of the reaction, allowing the reaction to go faster. 

ex: Hydrogen Peroxide transforming into water and oxygen. This reaction is very slow, but on adding Iodide ions, which acts as catalyst, it moves much faster.

2H₂O₂ -> 2H₂O + O₂ , When we add Iodide ions (I^-) which acts as catalyst, we can see the reaction as 2 elementary eqn

• Elementary step 1 (slow, so RDS): H₂O₂ + I^- -> H₂O + IO^- (IO^- is an intermediate product). Rate R = K [H₂O₂] [I^-] 
• Elementary step 2 (fast): H₂O₂ + IO^- -> H₂O + O₂ + I^- 

First let's calc BE of reactants and products.

• Reactant: H₂O₂ has 2 O-H bonds and 1 O-O bond. BE of 2 moles of H₂O₂ = 4 moles of O-H bond +  2 moles of O-O bond = 4*464+2*146=2148 kJ. Tis is total energy needed to break the bonds.
• Product: H₂O has 2 O-H bonds and O₂ one double O=O bond. BE of 2 moles of H₂O = 4 moles of O-H bond = 4*464=1856 kJ. BE of 1 mole of O=O = 1*498KJ. So, total energy released during bond formation = 2354 KJ.
• So, net energy released during this reaction = 2354 - 2148 = 206 KJ.

However, the BE graph for this reaction is not a straight line with Final BE lower than initial BE. Instead, the energy first goes up and then comes down. This hump is the activation energy E_a, which is presence of catalyst is lowered making the reaction faster. We have 2 lower humps now, each accounting for the 2 elementary reaction. E_a is higher for step 1 than for step2 since step 1 is RDS. See in Khan Academy vid. 

Different kinds of Catalysts:

• Enzymes: These are biological catalyst, which have active sites. The substrate binds to the Enzyme at the active site, and breaks the bond
• Homogeneous catalyst: A catalyst present in same phase as the reactants. Hydronium ion acts as catalyst in hydrolysis of sucrose to speed up dissociation of sucrose into glucose and fructose
• Heterogeneous catalyst: A catalyst present in different phase than the reactants. Ethene and Hydrogen react to form Ethane, which are all gases. If reaction is allowed to take place on the surface of solid platinum, then platinum acts as a catalyst and it being is a diff phase than reactants, it's called Heterogeneous catalyst. 

If reaction is elementary, then rates of reactions are:

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

Reverse 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 =  Kf / Kr = [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.