Chemistry from First Principles: Atoms, Bonding, Energy & Reactions
A first-principles journey through chemistry, starting with atoms and electrons and building toward chemical bonding, energy, Gibbs free energy, equilibrium, reaction rates, activation energy, and stability.
Chemistry
Chemistry: Chemistry is matter or anything that has mass & occupies space.
Why exists
- atoms are not stable alone. They combine to reach lower energy states.
Everything depends on
- Electrons
- Energy
- forces
- They interact because of electromagnetic force.
Atoms
Atom = smallest unit of an element that retains its properties.
Structure
- Nucleus (center)
- protons (+ charge)
- Neutrons (no charge)
- Electrons orbit around nucleus.
Element
- defined by:
- No. of protons in nucleus. called as Atomic number (z)
exm
Hydrogen -> 1 proton -> z = 1
change of proton -> change element.
Atomic Number = z = no. of protons
Mass number = A = protons + neutrons
electron = protons
charge = 0
Real core -> why atoms react ?
Valence electrons -> electrons in outermost layer.
Atoms wants
- full outer shell
- lower energy
- stability
Octet rule
- Most atoms wants 8 e⁻ in outer shell.
- why ? bcz noble gases (like Neon) are extremely stable.
- exm:
Sodium
1 valence e⁻ -> loses 1 -> stable
chlorine (Cl)
- 7 valence electron -> need 1 -> stable.
So:
$$
Na^+ + Cl^- \rightarrow NaCl
$$
Types
1. Ionic
Metal + Non metal -> NaCl
2. Covalent
Non-metal + Non-metal
-> sharing -> H₂O
3. Metallic
sea of free e⁻, Metals.
Everything else is variation.
Chemical reaction
- rearrangement of atoms
Law of conservation of mass
Mass of reactants = Mass of products
$$
2H_2 + O_2 \rightarrow 2H_2O
$$
Energies
- Ionization energy (IE) -> energy needed to remove an e⁻ from an atom in the gas phase.
- Electron affinity (EA) -> energy released when an atom gains an e⁻.
So
- if an atom loses an e⁻ -> you pay energy -> IE.
- If an atom gains an e⁻ -> you get energy = EA.
Reaction happen spontaneously only if the overall energy change favours lower energy.
eg
- Remove e⁻ from Na -> cost energy -> IE
- Cl gains e⁻ -> releases energy -> EA.
- Energy released > energy spent -> stable product forms.
How ?
energy = ability to do work or cause change.
- e⁻ held by electrostatic force.
- Nucleus : positive charge
- Electron : Negative charge,
Unstable Atom -> high potential energy -> far from nucleus -> less strong forces -> unstable
Stable Atom -> all gaps are filled -> have potential energy -> e⁻ are tightly bounded -> overall systems energy decreases.
e⁻ held in atoms by electrostatic force
$$
F = \frac{k |q_1 q_2|}{r^2}
$$
k = coulomb's constant = $8.99 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$
$q_1 / q_2$ = charge of each sphere in coulomb
r = distance between two charge bodies
Energy released = potential energy drop of system - initial kinetic energy of e⁻
$$
\Delta G = \Delta H - T\Delta S
$$
$\Delta G$ = Gibbs free energy change
$\Delta H$ = Enthalpy change (heat energy)
$T$ = Temperature (Kelvin)
$\Delta S$ = entropy change (disorder change)
What is $\Delta H$ ?
= heat change of reaction
$\Delta H < 0$ -> heat released -> exothermic
$\Delta H > 0$ -> heat absorbed -> endothermic
$\Delta S$
-> $\Delta S$ = Entropy = measure of disorder or number of possible arrangements.
More randomness = higher entropy.
Gas > liquid > solid
More particles > fewer particles.
$\Delta G$ ?
tells you if reaction is spontaneous.
-> spontaneous = done or happening suddenly
-> without external energy ex-reaction
- $\Delta G < 0 \implies$ spontaneous
- $\Delta G > 0 \implies$ non-spontaneous
- $\Delta G = 0 \implies$ equilibrium
exm
Ice -> water
- It absorbs heat ($\Delta H > 0$)
- But entropy increases ($\Delta S > 0$)
at higher temperature
$$
T\Delta S > \Delta H
$$
So $\Delta G$ becomes negative.
That's why ice melts above 0°C.
Now understand
$\Delta H$ = negative or $< 0$ exm :
$\Delta H = -12$ to compensate -> heat released eg : 12 heat units so -> exothermic
same vice versa :
$\Delta H = 12$ to compensate -> heat absorbed eg : -12 heat units so -> endothermic,
$\Delta G$ = Gibbs free energy is amount of energy in a system that is available to do useful work at constant temperature & pressure.
Deep by physics view.
spontaneous processes increase entropy of the universe.
real law
$$
\Delta S_{\text{universe}} > 0
$$
-> Now
$$
\Delta G = -T\Delta S_{\text{universe}}
$$
if spontaneous increase entropy increases with Temperature.
So if:
- $\Delta S_{\text{universe}} > 0$
- Then $\Delta G < 0$
That means
when universe becomes more disordered, $\Delta G$ becomes negative.
Why Nature prefer lower G ?
lower G means:
- less usable energy trapped
- More energy dispersed into surroundings
- Greater entropy.
- Nature loves spreading energy out.
Energy concentrated = unstable. Energy dispersed = stable.
Examples
Explosion -> energy spreads -> stable
Hot object -> cools -> energy spreads -> stable.
Equilibrium
Most reactions don't go "100 % complete"
They reach a balance point called equilibrium.
$$
\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}
$$
For reaction
$$
a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}
$$
Equilibrium constant
$$
K = \frac{[\text{C}]^c [\text{D}]^d}{[\text{A}]^a [\text{B}]^b}
$$
example
reaction
$$
\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}
$$
equilibrium expression
$$
K = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}
$$
if at equilibrium
- $[\text{HI}] = 2\text{ mol/L}$
- $[\text{H}_2] = 1\text{ mol/L}$
- $[\text{I}_2] = 1\text{ mol/L}$
Then:
$$
K = \frac{(2)^2}{1 \times 1} = 4
$$
So $K > 1 \implies$ product dominate, more products
$K < 1 \implies$ reactants dominate, more reactants
$K = 1 \implies$ both present in comparable amounts
So $K = 4 \implies$ More products than reactants.
Not extreme, But product-favored.
$K = 4$ is ratio
$$
\frac{\text{products (at equilibrium)}}{\text{reactants (at equilibrium)}} = \frac{4}{1}
$$
Where K has no unit.
example
if increase concentration of reactant A -> forward rate increase -> then after some time backward rate increase.
exm
forward rate jumps -> More products form -> product concentration increases -> backward rate increases -> New equilibrium reached.
This is Le Chatelier's Principle:
- system responds to disturbance by opposing it.
Now, at equilibrium
Forward rate = Backward rate.
$$
\text{forward rate} = k_f [\text{A}]^a [\text{B}]^b
$$
$$
\text{backward rate} = k_r [\text{C}]^c [\text{D}]^d
$$
at equilibrium:
$$
k_f [\text{A}]^a [\text{B}]^b = k_r [\text{C}]^c [\text{D}]^d
$$
re arrange
$$
\frac{[\text{C}]^c [\text{D}]^d}{[\text{A}]^a [\text{B}]^b} = \frac{k_f}{k_r}
$$
The ratio is K.
If $k_f = k_b$, what would K be ?
- from eqⁿ its 1.
But not like products and reactants are in equal quantity.
because $K = 1$ means:
$$
\frac{[\text{C}]^c [\text{D}]^d}{[\text{A}]^a [\text{B}]^b} = 1
$$
So,
$$
[\text{C}]^c [\text{D}]^d = [\text{A}]^a [\text{B}]^b
$$
exm
$$
\text{A} \rightleftharpoons \text{B}
$$
$$
K = \frac{[\text{B}]}{[\text{A}]} = 1
$$
But:
$$
\text{A} \rightleftharpoons 2\text{B}
$$
Then $K = \frac{[\text{B}]^2}{[\text{A}]}$, so $[\text{B}]^2 = [\text{A}]$
Now, what actually determines $k_f$ & $k_r$ ?
They are determined by:
- Activation energy
- Temperature
- Molecular orientation & collision probability
What ? Here is eqⁿ
$$
k = A e^{-\frac{E_a}{RT}} \quad \text{(Arrhenius equation)}
$$
where:
- $A$ = frequency factor (collision + orientation)
- $E_a$ = activation energy
- $R$ = gas constant
- $T$ = temperature
Formula:
$k = A e^{-E_a/RT}$
Now, what the hell is this ?
here A is how often molecules collide correctly)
what is activation energy
- It is minimum energy molecules need to react.
what is $e$ ? -> fundamental mathematical constant approximately equals to 2.71828
why there is negative sign ?
-> look at exponent
$$
-E_a / RT
$$
- If $E_a$ increase fraction gets bigger.
- But negative sign controls it
as:
if exponent = $-1 \implies e^{(-1)} = 0.37$
= $-10 \implies$ basically microscopic
So,
High $E_a \implies$ very negative exponent $\implies$ exponent becomes tiny $\implies$ value of $k$ drop $\implies$ reaction slow $\implies$ because minimum energy molecules need to react is very high.
Now Temperature
- If it increases $E_a/RT$ becomes smaller
- Exponent becomes less negative.
- Exponential term increase sharply &
reaction speeds up.
So, backward & forward reactions usually have different activation energies.
That's why
$$
k_f \neq k_b
$$
Now let us deep dive.
$E$ = energy of a chemical species.
specifically:
- represents Gibbs free energy ($G$) in Thermodynamics or potential energy.
So,
$E_A$ = energy of reactant A
$E_B$ = energy of product B
$E_{TS}$ = Energy of transition state.
unit = $\text{kJ/mol}$
Now what is Transition state ?
- highest-energy arrangement of atoms during reaction.
- The highest energy at femtoseconds where bonds are partially broken & formed.
- energy always higher than both reactants & products.
Now Activation energy
Forward reaction
$$
E_a^{\text{forward}} = E_{TS} - E_A
$$
means how much extra energy does A need to reach Transition state.
Same, backward
$$
E_a^{\text{backward}} = E_{TS} - E_B
$$
example
Suppose
- $E_A = 100$
- $E_B = 50$
- $E_{TS} = 150$
Now calculate
Forward:
$$
E_a^{\text{forward}} = E_{TS} - E_A
$$
$$
= 150 - 100
$$
$$
= 50
$$
(Here $E_a$ is activation energy not energy of species)
Same
$$
E_a^{\text{backward}} = E_{TS} - E_B
$$
$$
= 150 - 50
$$
$$
= 50
$$
Now physics ?
$$
\Delta G = G_{\text{products}} - G_{\text{reactants}}
$$
simply : How much lower or higher is the product energy compared to reactants ?
what ?
$$
G_{\text{product}} < G_{\text{reactants}}
$$
remember nature wants low energy state.
So here -> products will more &
vice versa.
Now how does this connect to equilibrium constant $K$ ?
eqⁿ
$$
\Delta G^\circ = -RT \ln K
$$
where
- $R = 8.314\text{ J/mol}\cdot\text{K}$
- $T$ = Temperature (Kelvin)
- $K$ = equilibrium constant
- $\Delta G^\circ$ = standard free energy change
how ?
take $R$ & $T$ constant
Now if $K = 1$ then $\log 1 = 0$ so $\Delta G^\circ = 0$.
- equilibrium mixture is balance.
if $K = 10$, $\log 10 = 2.3$ so $\Delta G^\circ = -RT(2.3)$
So $\Delta G^\circ < 0$, so less system energy.
As $K$ increases,
$\log K$ increases,
So $-RT \log K$ become more negative.
So,
$$
\text{Large } K \implies \text{large negative } \Delta G^\circ
$$
products are lower in energy. System prefers products.
Vice versa : $K < 1$
$\ln(0.1) = -2.3$
So $\Delta G^\circ = -RT(-2.3) = +RT(2.3)$ So $\Delta G^\circ > 0$
small $K \implies$ large positive $\Delta G^\circ$
so products are higher in energy. System prefers reactants.
rewritten eqⁿ
$$
K = e^{-\Delta G^\circ / RT}
$$
Now ask yourself:
can a reaction have very small $k_f$ & large $K$ ?
Answer is yes!
$K$ means:
$$
\Delta G^\circ = -RT \ln K
$$
If $\Delta G < 0 \implies$ large $K \implies$ product favored
If $\Delta G > 0 \implies$ small $K \implies$ reactants favored
Now what $k$ ? (small $k$)
$$
k = A e^{-E_a/RT}
$$
$k$ edge depends on activation energy & temperature.
Now
example: Rusting of iron
Iron + oxygen -> rust.
- Thermodynamically favorable (large $K$)
- But very slow.
Now if $\Delta G$ is very negative but $E_a$ is very high, what happens ?
Answer this yourself before:
Answer is:
- if $\Delta G < 0$ then reaction will be more spontaneous, so it will be more stable.
- High $E_a$ means reaction will need energy to become product
- reaction become slower.
- exm : Rusting
What ?
take Graphite
It is more stable than Diamond at room temperature.
So why it won't convert into Graphite ?
The hell $\Delta G$ is lower but $E_a$ is too much high that practically impossible.
Impossible says I am possible.
But Not here, this possibility takes more time then a sun will die & form again.
Now stable.
- There are two types:
Thermodynamic stability
- Lower energy state.
- system at minimum free energy.
- exm : Graphite has lower free energy than diamond.
- So graphite is thermodynamically more stable.
- or $\Delta G$ has very less value.
kinetic stability
- Hard to change because barrier is high.
- exm : why diamond don't become graphite even graphite is more stable.
- Because it's activation energy ($E_a$) is too high.
- So diamond is kinetically stable.
- or $E_a$ is massive.