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Published: Sunday, 06 September 2026 21:29

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scalars and vectors:

This topic is covered in Physics as well as in maths (in calculus). I'm covering in Physics as most of it's application is in Physics.

Scalars: Things which don't have any sense of direction, are called scalars. As an example of scalar, mass of a thing, is a scalar quantity. It doesn't have any sense of direction, whether it's going north, south, etc, the mass is the same. Similarly for time, temp, volume, etc.

Vectors: Things which have direction along with a magnitude, are called vectors. As an example of vector, force is a vector quantity. It depends on which direction you apply force, so it has a sense of direction. If the force is applied in same direction as the movement of the body, then the body will accelerate, while if it's applied in opposite dir, then the body slows down.So, direction of force matters. Similarly for velocity, acceleration, etc.

Link => https://math.libretexts.org/Bookshelves/Calculus/Supplemental_Modules_(Calculus)/Vector_Calculus/1%3A_Vector_Basics

Vector: A vector is rep by an arrow, whose length is the magnitude of the vector, and whose dir is the angle that that vector makes wrt x and y axis (for 2D vector), or wrt x, y and z axis (for 3D vector). Vectors can be moved anywhere in space and will remain same, as long as mag and dir remain same (i.e starting point and ending pt of vector can be anywhere).

Vector is rep by i, j, k for x,y and z dir or by coords <x,y,z>, (x,y,z) or in matrix form as x, y, z written vertically in 1st column. An arrow is put on top of the line segment to indicate it's vector. Magnitude or norm of vector is denoted by ||PQ||, while dirn is denoted by an angle (tan(θ)=slope). If ||v||=1, or v/||v|| it's called a unit vector. i, j, k are called std unit vectors in dir of x, y and z axis. 

For simplicity, initial point of vector is usually chosen as origin, so only the end pt need to be specified in such a case. This rep of vector is called std position. or component form of vector. End point Coords <v1,v2> are called components of vector v. 

Direction cosine: The 3 angles (α, β, γ) that a vector makes with x, y and z axis are called direction angles of v, and the 3 cosine values are called Direction cosine. For v=<v1,v2,v3>, v1=||v||*cos(α),  v2=||v||*cos(β),  v3=||v||*cos(γ). This yields cos²(α) +  cos²(β)  + cos²(γ) = 1 

NOTE: differentiation/integration of vectors is discussed in Calculus section.

 We'll study some operations (algebra) on vector quantities.

• Addition/subtraction: Addition/subtraction of vectors is done by placing end point of a vector to the start point of next vector and so on, until all vectors are connected. Then you connect the start point of 1st vector to end point of last vector, and that's the sum of all the vectors. For subtraction, you still treat it as addition, except that the direction of vector is reversed for the subtracting vectors. You still place them the same way as in addition, except that -ve vector is pointing in other dir.
  • For vector <x1,y1> and <x2,y2>, the sum is <x1+x2,y1+y2>. The magnitude of vector is sq rt of [(x1+x2)^2 + (y1+y2)^2]. Mag may also be calc by using cosine thm, , if you know the mag of 2 vectors and angle b/w them.
  • Many properties as Commutative, associative, identity, distributive properties hold true for vector and can be proved easily by using above formula. Any set of vectors that satisfy these 8 properties is called "vector space"
• Multiplication: There's no multiplication of vectors. We call it the product of vectors. Product of vectors may give a scalar or a vector depending on how we define it. There are 2 kinds of product. Link => https://math.libretexts.org/Bookshelves/Calculus/Supplemental_Modules_(Calculus)/Vector_Calculus/1%3A_Vector_Basics/1.5%3A_The_Dot_and_Cross_Product 
  • Dot product (scalar product): It's a scalar, and is determined by multiplying coefficients as shown in link above.
    • Dot product is simply multiplying the length of one vector, with the length of the shadow (or projection) of the other vector on the 1st vector. So, u.v = ||u||.||v cos θ||  =  ||u||.||v|| cos θ. It can also be calculated by using formula u.v=u1.v1 + u2.v2 + u3.v3.  
      • Cosine Thm: By using cosine thm, we may write  ||u - v||^2 = ||u||^2 + ||v||^2 - 2 ||u||.||v||. cos θ => (u-v).(u-v) =   ||u||^2 + ||v||^2 - 2 ||u||.||v||. cos θ . Cancelling terms, we get u.v = ||u||*||v|| * cos(θ). This is how we get this formula for u.v. This implies that dot product of orthogonal vectors is 0. 
      • Projection of vectors: Many times, we decompose a vector into orthogonal components, once pointing towards X axis and other towards Y axis, which is called the projection of vector on x and y axis. These are the i,j coeff of vector. Some times, we want to find component of a vector in the dir of another vector (not necessarily in dir of i,j), then we need to find projection onto that vector.
        • Proj of u vector in dir of v vector is given by w1 = proj_v(u) = (u.v)/[||v||^2] * v. This is easy to prove, as dot product gives the product of projected vector component with the other vector. But then we need to divide by ||v|| since it includes that in dot product. So, (u.v)/||v|| gives the mag of such projection. We need to put dir in it (in dir of v vector), so we take unit vector in that dir, which is v/||v||. That gives above formula. ||u||cos(θ) is the component of u in dir of v. The component perpendicular to w1 is w2=u-w1.
    • Angle θ b/w the 2 vectors is taken to be b/w 0 to π. So, cos θ value goes from +1 to -1. Since ||u|| and ||v|| are +ve, u.v and cos θ will have same sign.
    • Dot product has many of the same properties as  Commutative, distributive properties and can be proved easily by using above formula.
  • Cross product (vector product): It's a vector, and is determined by multiplying coefficients as shown in link above. It also has dir (as the product is a vector) which is orthogonal to both the vectors whose cross product is being determined.
    • Cross product is found by getting determinant of the matrix with alternating +ve and -ve sign. It also has dir which is perpendicular to the plane of the 2 vectors u, v (i.e u X v is perpendicular to both u and v). 
      • Mag of u X v = ||u X v|| = |u|| * ||v|| * Sin(θ) (Note dot product was similar except that dot product had cos(θ)). This can be proved by using dot product. 
        • |u|| * ||v|| * Sin(θ) = |u|| * ||v|| * √ (1-Cos²(θ)) =  |u|| * ||v|| * √ (1 - [u.v/(||u||*||v||)]^2) =  √ (||u||^2*||v||^2 - [u1v1 + u2v2 + u3v3]^2) =   √ [(u1^2 + u2^2 + u3^2)*(v1^2+v2^2+v3^2) -  (u1v1 + u2v2 + u3v3)^2] which gives the reqd answer
        • ||u X v|| gives area of parallelogram with sides u and v. Easy to see as ||v|| sin(θ) is the height of the parallelogram.
      • u X v forma a right handed system.  curl your fingers in dir from u to v, then the thumb points in dir of u X v. k = i X j.
    • Cross product has properties as distributive and few others.
      • u X v = -(v  X u). 
      • u X (v+w) =  u X v + u X w.  
      • Triple scalar product: u.(v X w) = v.(w X u) = w.(u X v) => dot here reps the dot product
        • Triple Scalar product gives the vol of parallelopiped (a polyhedron, whose all sides are parallelogram) only if u, v, w don't lie in same plane. 
        • 3 vectors u, v, w which have same starting pt are coplanar if and only if their triple scalar product is 0.
      • u X v =0 if u and v are scalar multiples of each other (i.e u and v have same dir, i.e parallel to each other)
  • Scalar multiplication of vector v by a constant c, yields cv, i. mag of new vector is multiplied by c, with no change in dir (for c=-ve, dir is reversed). Parallel vectors are such that u=cv, where c is a scalar and u.v are 2 vectors. 

 Matrix: Matrices are useful when evaluating dot and cross product of matrices. Matrices are studied in Maths and provided in Advanced Maths section

3D geometry as vectors:

3D figures in space can be expressed as vectors, which sometimes yields eaiser way to manipulate them.

Lines in space:  A line in space can be rep as vector. Consider a line "L" which is parallel to vector v=<a,b,c>, and passes thru pt P (x1,y1,z1). Vector v is dir vector for line L, and a,b,c are dir numbers. We are looking for eqn of vector PQ, where Q is any pt <x,y,z>. Since PQ is parallel to v, we get PQ=t.v, where t is a scaling factor. Vec PQ =  (x-x1)i + (y-y1)j + (z-z1)k = t(ai + bj = ck) => x-x1=ta, y-y1=tb, z-z1=tc. This is called parametric eqn where "t" is an independent parameter which can take any real value, x=x1+at, y=y1+bt, z=z1+ct. This parametric eqn can also be written as symmetric eqn by eliminating t => (x-x1)/a = (y-y1)/b = (z-z1)/c

NOTE: neither parametric eqn nor  symmetric eqn are unique. By choosing some t, we can get other point on that line, and use that pt to get parametric and symmetric eqn which would be different. 

Planes in space:  A plane in space can be rep as an eqn by having a pt that lies on that plane, and having vector "n" which is perpendicular to that plane. Let's consider pt (x1,y1,z1) which lies on a plane and a normal vector n=<a,b,c>. Then plane consists of all points Q=(x,y,z) such that vec PQ is perpendicular to n. i.e PQ.n=0 => [(x-x1)i + (y-y1)j + (z-z1)k].<a.b.c>=0 => (x-x1)a + (y-y1)b +(z-z1)c = 0 => This is called the std form of plane. We can rearrange it to write general form of plane which is ax + by +cz + d =0. Given the general form of plane, it's easy to find a vector normal to plane as it's coeff a,b,c, so n=<a,b,c>

Find eqn of plane given 3 points in plane: First we form 2 vec from the 3 points, and find their cross product to get a vector normal to plane. Then we use above eqn to plug in a.b.c and then find d by using any of the points.

Find angle b/w 2 planes:  The angle b/w 2 planes can be found by finding normal vector of the 2 planes, and then finding angle b/w the 2 normal vectors using dot product.

Finding line of intersecton of 2 planes: Given 2 planes, they may either be parallel or intersect in a line. A line will only involve 2 var (x,y or y,z or x,z), so solve the 2 simultaneous eqn for 2 planes to eliminate a var, and we get the eqn for line intersecting them. We can also write it as parametric eqn by introducig var "t". Other way to get line of intersection, is to get normal vec n1=<a1,b1,c1> and n2=<a2,b2,c2> and then find cross product n1 X n2, The vec obtained is perpendicular to both vec, and hence has to be a intersecting line, since it's common to both of the planes. 

Trace of the plane: Plane in space intersects the coord plane (xy, yz, xz planes). The line of intersection is called the trace of the given plane in the coord plane. To find the xy trace, we set z=0, and so on for other planes. Any eqn of a straight line in 2 var is also a plane in 3D space, as it extends indefinitely in the plane of missing var (i.e 2x+y=5 is a line in 2D space, but a plane in 3D space). 

• If 1 var is missing in eqn of plane, then it's parallel to the x axis. i.e 2x+y=5 is a plane parallel to the z axis.
• If 2 var are missing in eqn of plane, then it's parallel to the coord plane. ex: x=2 is a plane parallel to y-z coord plane

Distance b/w point and plane: To find the distance of a plane to a pt Q, we take a arbitrary pt P (choose x=0, y=0 to get a simple pt z) on plane, and form a vector PQ, then project this vector onto normal vector. So D = ||proj_n(PQ)|| = (PQ.n)/||n||. The choice of pt P on plane doesn't change the answer.

• For plane ax+by+cz+d=0, pt P on plane <x1,y1,z1> and pt Q = (x0,y0,z0), then D = PQ.n/||n|| = [(x0-x1)i + (y0-y1)j + (z0-z1)k].<a,b,c> / ||n|| = a(x0-x1) + b(y0-y1) + c(z0-z1) / √ (a^2+b^2+c^2) = | a*x0+b*y0+c*z0+d | / √ (a^2+b^2+c^2) ( we use the fact that a*x1+b*y1+c*z1+d=0, since pt (x1,y1,z1) lies on plane)

Distance b/w 2 parallel planes: To find the distance b/w 2 parallel planes, use a pt (x0,y0,z0) on 1 plane and then use the normal vector of other plane to find distance as shown above. So, formula is same as above

Distance b/w point and a line in space: To find the distance b/w pt and a line, we just do cross product instead of dot product. So, D= ||PQ X u|| / ||u||, where P is the pt o line, and Q is the other point, and u is direction vector of the line. Proof is simple PQ*Sin(θ) is the distance, which is what the cross product gives.

Spheres in space: The eqn for sphere is just an extension of circles in 2D, (x-x0)^2 + (y-y0)^2+ (z-z0)^2 = r^2 

Cylinders in space: Cylinders are any general curve, which is extended by copying and placing that curve on top of each other indefinitely in the 3D space. Such a curve is called the generating curve, since the 3D figure is obtained by repeating this curve by parallel lines, which are called rulings.

Simplest ex of cylinder is a cylinder obtained by repeating a circle: x^2+y^2=r^2. This is a circle in 2D, but in3D it becomes a cyclinder since z can take on any value.

Other ex: z = y^2, Z=sin(x), etc

Quadric Surfaces: These are 3D analog of conic sections. It's 2nd degree eqn of form ax^2 + by^2 + cz^2 + Dxy + Exz + Fyz + Gx + Hy + Iz + J = 0

There are 6 types of Quadric surfaces:

1. Ellipsoid: 3D extension of ellipse in 2D: x^2/a^2 + y^2/b^2 + z^2/c^2 = 1. The surface is a sphere if a=b=c≠0.
2. Hyperboloid: Here all var are raised to 2nd power, and one or two of the var have -ve coeff. 
   1. Hyperboloid of 1 sheet (only one coeff is -ve): x^2/a^2 + y^2/b^2 - z^2/c^2 = 1. It's hyperbola rotated around one of the axis. The axis of the Hyperboloid corresponds to the var whose coeff is -ve (in his case, that axis is "z" axis)
   2. Hyperboloid of 2 sheets (two coeff are -ve): x^2/a^2 - y^2/b^2 - z^2/c^2 = 1. It's hyperbola rotated and separated in 2 parts. The axis of the Hyperboloid corresponds to the var whose coeff is +ve (in his case, that axis is "x" axis)
3. Elliptic cone: It's a cone shape opening in top and bottom. Eqn is x^2/a^2 + y^2/b^2 - z^2/c^2 = 0 (NOT 1 as in Ellipsoid).  The axis of the cone corresponds to the var whose coeff is -ve (in his case, that axis is "z" axis)
4. Paraboloid: One of the var is raised to only 1st power.
   1. Elliptic paraboloid: It's a parabola in shape of ellipse (similar to a cone pointing upwards, but with gradual edges, i.e parabola rotated around). Eqn is z= x^2/a^2 + y^2/b^2. The axis of the paraboloid corresponds to the var raised to 1st power (in his case, that axis is "z" axis)
   2. Hyperbolic paraboloid: It's a parabola in shape of hyperbola. Eqn is z= y^2/b^2 - x^2/a^2(NOTE the -ve sign). The axis of the paraboloid corresponds to the var raised to 1st power (in his case, that axis is "z" axis)

 NOTE: some 2nd degree eqn in x,y,z don't represent any quadric surface. Ex: ^2=y^2+z^2 is a single point (0,0,0). x^2+y^2=1 is a right circular cylinder.

The general eqn ax^2 + by^2 + cz^2 + Dxy + Exz + Fyz + Gx + Hy + Iz + J = 0 shows a quadric surface not centered at origin. 

Surface of revolution: This is 5th special kind of surface that employs calculus to find surface area, vol, etc. 

• Consider eqn y=r(z) in yz plane.
  • If we revolve this around z axis, we'll get a surface of revolution around z axis. The trace will be a circle, so x^2+y^2=[r(z)]^2.
  • If we revolve this around y axis, we'll get a surface of revolution around z axis. The trace will be a circle, so x^2+z^2=[r(y)]^2.
  • If we revolve this around x axis, we'll get a surface of revolution around z axis. The trace will be a circle, so y^2+z^2=[r(x)]^2.

NOTE: The generating curve for a surface of revolution is not unique.

Cylindrical and spherical coordinates:

Representing coords in (x,y,z) in rectangular coord is how we represent surfaces in space. However, sometimes, some other coord system are easier to represent some particular surfaces.

Cylindrical Coord => This is an extension of polar coord in a plane to 3D space. It's rep by (r, θ, z), where (r, θ) is the polar rep of projection of pt P onto planar xy plane, and z is the distance from  (r, θ) to P. Since rep in polar coord is not unique, so rep in cylindrical coord is also not unique.

• Cylindrical to rectangular coord: x=r*Cos(θ),  y=r*Sin(θ), z=z
• Rectangular to Cylindrical coord: r= √ (x^2+y^2), tan(θ) =y/x, z=z
• Vertical plane rep: θ=c, Horizontal plane rep: z=c

Spherical Coord => This is rep by (ρ, θ, φ), where ρ = distance of pt P from origin, θ is the same angle used in cyclindrical coord (ngle formed by projectionn of Pt P onto xy plane), and φ is the angle b/w +ve z axis and line from origin to point P.  So, spherical coords are what is used to rep points on earth, where θ=longitude and φ=latitude. Sphericl coord is useful for surfaces that have a point or center of symmetry.

ρ and  φ are non -ve (i.e always +ve)

• Spherical to rectangular coord: x=ρSin(φ)Cos(θ), y=ρSin(φ)Sin(θ), z=ρCos(φ)
• Rectangular to Spherical coord: ρ= √(x^2+y^2+z^2), tan(θ) =y/x, Cos(φ)=z/√(x^2+y^2+z^2)
• Spherical to Cylindrical coord (r ≥ 0): r^2=ρ^2Sin²(φ), θ=θ, z=ρCos(φ)
• Cylindrical to Spherical coord (r ≥ 0): ρ= √(r^2+z^2), θ=θ, Cos(φ)=z/√(r^2+z^2)

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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.

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Published: Sunday, 29 March 2026 14:01

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Mag 7 (Magnificent 7)

This is a term referred to top 7 companies in USA, which are in tech and highly successful. They have dominated the total stock market cap, and have provided higher return than S&P500. These are the 7 stocks => Alphabet, Amazon, Apple, Meta, Microsoft, Nvidia, and Tesla. 

ETF: MAGS (Roundhill Magnificent Seven ETF) is the top ETF with 0.29% expense ratio and $3B in AUM to track Mag 7. It was launched on 04/11/23 and initially traded as ticker "BIGT". It has provided 100% return in 3 yrs since it's launch, excluding dividend.

We discuss the 7 stocks to see how worthy of investment are they:

Apple: 

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Published: Tuesday, 24 March 2026 17:44

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IT BPO Companies

BPO is Business Process Outsourcing. It involves outsourcing services such as customer service, technical support, and back-office operations to other countries where cost of operation is lower. IT BPO refers specifically to IT service that is outsourced to other companies. These companies are more efficient in providing entire IT services for other companies. IT works out well for other companies too, as they don't want to have their own IT department to take care of billing, paychecks, etc. Total Market size of IT BPO is $450B as of 2026, and growing at 5%-8% a yr (estimates vary widely). I assume that all markets grow at same rate as what US GDP grows by, which is 5%/yr. USA is the largest market by revenue accounting for 30%-40%. India is the largest provider of IT BPO services, as is evident by the fact that most IT companies are Indian Co, or are situated in India. Besides India, other countries as Philippines, Vietnam, Indonesia are also emerging players. 

Global IT spending is $5.5T as of 2025, of which $1.7T is spent on IT services. BPO part of these IT services seems to be only $0.5T.