Some Basic Concepts in Chemistry for NEET UG Preparation

1. Stoichiometry

1.1 Mole Concept

• Formula: $n=\frac{m}{M}$
• • Explanation: The number of moles ($n$) is calculated by dividing the mass of the substance ($m$) by its molar mass ($M$). This formula is fundamental in stoichiometric calculations where amounts of substances involved in chemical reactions are determined.
  • Units: $m$ in grams (g), $M$ in grams per mole (g/mol), $n$ in moles (mol).
  • Example: Calculate the number of moles in 36 g of water (molar mass = 18 g/mol).
  • • Solution: $n=\frac{36}{18}=2$ mol
  • Common Mistake: Confusing the molar mass with the molecular mass and using incorrect units.

1.2 Avogadro’s Law

• Formula: $V\propto n$ or $V=k\cdot n$
• • Explanation: At constant temperature and pressure, the volume ($V$) of a gas is directly proportional to the number of moles ($n$). This law helps in determining the volumes of gases involved in reactions.
  • Units: Volume in liters (L), moles in mol.
  • Example: If 1 mole of a gas occupies 22.4 L at STP, then 2 moles will occupy $2\times 22.4=44.8$ L.
  • Common Mistake: Not considering the conditions of temperature and pressure when applying this law.

1.3 Percentage Composition

• Formula: $\%\text{ Element}=\frac{\text{Mass of Element in 1 mole of compound}}{\text{Molar Mass of compound}}\times 100$
• • Explanation: This formula calculates the mass percentage of each element in a compound, which is essential for empirical and molecular formula calculations.
  • Units: Mass in grams (g), percentage as %.
  • Example: Calculate the percentage of carbon in methane ($C{H}_4$, molar mass = 16 g/mol).
  • • Solution: $\%\text{C}=\frac{12}{16}\times 100=75\%$
  • Common Mistake: Incorrectly identifying the molar masses of elements.

2. Chemical Reactions

2.1 Balancing Chemical Equations

• Principle: Conservation of mass
• • Explanation: The total mass of reactants equals the total mass of products in a chemical reaction. This principle is applied when balancing chemical equations.
  • Example: Balance the equation: ${\text{C}}_2{\text{H}}_6+{\text{O}}_2\to {\text{CO}}_2+{\text{H}}_2\text{O}$.
  • • Balanced Equation: $2{\text{C}}_2{\text{H}}_6+7{\text{O}}_2\to 4{\text{CO}}_2+6{\text{H}}_2\text{O}$
  • Common Mistake: Failing to balance all elements, particularly hydrogen and oxygen, which often appear in multiple compounds.

2.2 Limiting Reagent

• Formula: Identify the reagent with the smallest mole ratio from the balanced equation.
• • Explanation: The limiting reagent determines the amount of product formed in a chemical reaction. It is the reactant that is completely consumed first.
  • Example: In the reaction $2{\text{H}}_2+{\text{O}}_2\to 2{\text{H}}_2\text{O}$, if you have 5 moles of $H_2$ and 2 moles of $O_2$, $H_2$ is the limiting reagent.
  • Common Mistake: Misidentifying the limiting reagent by not correctly applying mole ratios.

3. Thermodynamics

3.1 First Law of Thermodynamics

• Formula: $\Delta U=q+W$
• • Explanation: The change in internal energy ($\Delta U$) of a system is equal to the heat ($q$) added to the system plus the work ($W$) done on the system.
  • Units: Energy in joules (J), heat in joules (J), work in joules (J).
  • Example: If 100 J of heat is added to a system and 50 J of work is done on it, the change in internal energy is $\Delta U=100+50=150\text{ J}$.
  • Common Mistake: Incorrectly assigning signs to work and heat, leading to errors in calculating internal energy.

3.2 Enthalpy Change ($\Delta H$)

• Formula: $\Delta H=\Delta U+P\Delta V$
• • Explanation: The enthalpy change ($\Delta H$) is the heat change at constant pressure, where $\Delta U$ is the internal energy change and $P\Delta V$ is the pressure-volume work.
  • Units: Enthalpy in joules (J), pressure in pascals (Pa), volume in cubic meters ($m^3$).
  • Example: Calculate $\Delta H$ for a reaction where $\Delta U=200$ J, $P=1\times 10^5$ Pa, and $\Delta V=0.002m^3$.
  • • Solution: $\Delta H=200+(1\times 10^5\times 0.002)=400\text{ J}$.
  • Common Mistake: Forgetting to convert units properly when calculating pressure-volume work.

4. Physical Chemistry

4.1 Ideal Gas Law

• Formula: $PV=nRT$
• • Explanation: The relationship between pressure ($P$), volume ($V$), number of moles ($n$), and temperature ($T$) for an ideal gas is described by this equation.
  • Units: Pressure in pascals (Pa), volume in cubic meters ($m^3$), temperature in kelvins (K), gas constant $R=8.314$ J/mol·K.
  • Example: Calculate the pressure of 2 moles of an ideal gas at 300 K occupying 0.05 $m^3$.
  • • Solution: $P=\frac{nRT}{V}=\frac{2\times 8.314\times 300}{0.05}=9976.8\text{ Pa}$.
  • Common Mistake: Not using the correct units for $R$ or failing to convert temperature to kelvins.

5. Practice Questions

1. Calculate the number of moles of carbon dioxide produced when 5 g of carbon is burned in excess oxygen.
2. • Solution: Use $n=\frac{m}{M}$ where $M_{{\text{CO}}_2}=44\text{ g/mol}$.
2. Determine the limiting reagent when 10 g of hydrogen reacts with 80 g of oxygen.
3. • Solution: Compare moles of