Comprehensive NEET Chemistry Formulae Guide: The S-Block Elements

1. Stoichiometry and Chemical Reactions

1.1 Mole Concept

• Mole: $1\text{ mole}=6.022\times 10^{23}\text{ particles}$ (Avogadro's number)
• Molar Mass: Mass of 1 mole of a substance (g/mol)
• Number of Moles (n): $n=\frac{\text{Given mass}}{\text{Molar mass}}$

1.2 Balancing Chemical Equations

• Ensure the same number of atoms for each element on both sides of the equation.
• Use stoichiometric coefficients to balance the equation.

1.3 Law of Conservation of Mass

• The total mass of reactants equals the total mass of products.

1.4 Empirical and Molecular Formulae

• Empirical Formula: Simplest whole-number ratio of atoms in a compound.
• Molecular Formula: Actual number of atoms of each element in a molecule.
• Relation: $\text{Molecular formula}=(\text{Empirical formula})\times n$ where $n=\frac{\text{Molar mass}}{\text{Empirical formula mass}}$

1.5 Limiting Reagent

• The reactant that is completely consumed in a reaction, limiting the amount of product formed.
• Calculation: Determine moles of each reactant and compare with the stoichiometric ratio.

Did You Know?

The concept of the mole was first introduced by Wilhelm Ostwald in 1893.

2. Thermodynamics

2.1 First Law of Thermodynamics

• Law: $\Delta U=q+W$
• • $\Delta U$ = change in internal energy
  • $q$ = heat added to the system
  • $W$ = work done on the system

2.2 Enthalpy (H)

• Definition: $H=U+PV$
• Change in Enthalpy: $\Delta H=\Delta U+P\Delta V$
• • At constant pressure, $\Delta H=q_p$ (heat at constant pressure)

2.3 Hess's Law

• The total enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps.

2.4 Gibbs Free Energy (G)

• Definition: $G=H-TS$
• Gibbs Free Energy Change: $\Delta G=\Delta H-T\Delta S$
• • $\Delta G<0$: Spontaneous process
  • $\Delta G>0$: Non-spontaneous process
  • $\Delta G=0$: System is in equilibrium

Real-life Application:

Gibbs free energy is crucial in predicting the feasibility of chemical reactions, especially in industrial processes like the Haber process for ammonia synthesis.

3. Chemical Kinetics

3.1 Rate of Reaction

• Average Rate: $\text{Rate}=\frac{\Delta [\text{Reactant}]}{\Delta t}$
• Instantaneous Rate: $\text{Rate}=-\frac{d[\text{Reactant}]}{dt}$

3.2 Rate Law

• Expression: $\text{Rate}=k[\text{A}{]}^m[\text{B}{]}^n$
• • $k$ = rate constant
  • $m,n$ = reaction orders with respect to each reactant

3.3 Arrhenius Equation

• Equation: $k=A{e}^{\frac{-E_a}{RT}}$
• • $k$ = rate constant
  • $E_a$ = activation energy
  • $R$ = gas constant
  • $T$ = temperature in Kelvin
  • $A$= pre-exponential factor

NEET Problem-Solving Strategy:

When dealing with kinetics problems, always check units to ensure correct application of the rate law. Units of the rate constant vary depending on the reaction order.

4. Chemical Equilibrium

4.1 Equilibrium Constant (K)

• For a general reaction: $aA+bB\leftrightarrow cC+dD{K}_c=\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}$
• • $K_p=K_c(RT{)}^{\Delta n}$ for gaseous reactions
  • $\Delta n=\text{(moles of gaseous products)}-\text{(moles of gaseous reactants)}$

4.2 Le Chatelier's Principle

• A system at equilibrium will adjust to counteract the effect of any change in conditions (concentration, pressure, temperature).

4.3 Reaction Quotient (Q)

• Expression: Same as $K_c$ but for non-equilibrium conditions.
• Comparison:
• • If $Q<K_c$: Reaction proceeds forward.
  • If $Q>K_c$: Reaction proceeds backward.

Common Misconception:

Students often confuse $K_c$ and $K_p$. Remember, $K_c$ involves concentrations, while $K_p$ is used for gaseous systems and involves partial pressures.

5. Example Problems

Problem 1: Stoichiometry

• Given: 4.4 g of CO reacts with 7.1 g of O₂ to form CO₂.
• Find: Limiting reagent and amount of CO₂ formed.
• Solution:
• 1. Calculate moles: $n_{CO}=\frac{4.4}{28}=0.157\text{ moles}$, $n_{O_2}=\frac{7.1}{32}=0.222\text{ moles}$.
  2. Reaction: $2CO+O_2\to 2C{O}_2$.
  3. Stoichiometric ratio $\frac{0.157}{2}:\frac{0.222}{1}$ indicates CO is limiting.
  4. Amount of $C{O}_2$ formed = 0.157 moles, i.e., 6.88 g.

Problem 2: Thermodynamics

• Given: $\Delta H=-100kJ$, $\Delta S=-200J/K$.
• Find: $\Delta G$ at 300 K.
• Solution: $\Delta G=\Delta H-T\Delta S=-100-(300\times -0.2)=-40kJ$.
• Interpretation: Reaction is spontaneous at this temperature.

This guide covers essential chemistry formulae and principles relevant for NEET preparation, focusing on stoichiometry, chemical reactions, thermodynamics, and physical chemistry. It includes explanations, example applications, and tips to avoid common mistakes.