Table of Contents

$N_2 (g) + 3H_2 (g)$ $\; \xrightarrow[\; \ 200-300^\circ C, \; 200 \, atm \;]{\; \; Fe \; \; \text{(catalyst)} \;}$ $2NH_3 (g)}$

$2H_2 (g) + O_2 (g)$ $\; \xrightarrow[\;\ NaOH, \; 200^\circ C \;]{\; \;Cu \;/ \; {MO} \;}$ $2H_2O$

Class 12 Chemistry Formula Library (LaTeX Ready)

1. Mole Concept

$n = \frac{W}{M}$
जहाँ, n = Moles, W = Mass of substance, M = Molar Mass
$n = \frac{N}{N_A}$

$n=\frac{N}{N_A}$
जहाँ, N = Number of particles, N_A = Avogadro’s Number

2. Chemical Kinetics

👉 First-order reaction rate constant $k = \frac{2.303}{t} \log \frac{a}{a-x}$

👉 Half-life formula $t_{1/2} = \frac{0.693}{k}$

3. Electrochemistry

👉 Nernst Equation $E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q$

👉 Relation between Gibbs free energy & EMF $\Delta G = -nFE_{cell}$

4. Solid State

👉 Density of unit cell $d = \frac{zM}{a^3 N_A}$

5. Thermodynamics

👉 Gibbs Free Energy $\Delta G = \Delta H - T\Delta S$

👉 Heat absorbed $q = mc\Delta T$

6. Equilibrium

👉 Law of mass action $K_c = \frac{[Products]^{x}}{[Reactants]^{y}}$

👉 pH formula $pH = -\log [H^+]$

7. Organic Chemistry

👉 Dehydration of ethanol

$C_2H_5OH$ $\;\xrightarrow{\; \text{conc.}H_2SO_4, \; 443K \;}$ $C_2H_4 + H_2O}$

👉 Esterification

$CH_3COOH + C_2H_5OH$ $\;\xrightarrow{\;\text{Conc.}H_2SO_4\;}$ $CH_3COOC_2H_5 + H_2O$

$2H_2 (g) + O_2 (g)$ $\; \xrightarrow[\;\ NaOH, \; 200^\circ C \;]{\; \;Cu \;/ \; {MO} \;}$ $2H_2O$

8. Important Constants

👉 Universal Gas Constant $R = 8.314 \, J\, mol^{-1} K^{-1}$

👉 Avogadro’s Number $N_A = 6.022 \times 10^{23}$

$Mg + 2HCl$ $\; \xrightarrow{\;\text{iron}H_2SO_4 \;}$ $MgCl_2 + H_2$

Q.1. Calculate the percentage strength and strength in grams per litre of 10 volume hydrogen peroxide solution.

Solution. Hydrogen peroxide decomposes on heating according to the equation.
$2H_2O_2 \;\xrightarrow \;2H_2O + O_2$

$2KI +H_2SO_4 + H_2O_2 \;\xrightarrow{400K} \; K_2SO_4 + I_2 + 2H_2O$

What is called mole ?

The amount of any substance which contains 6.023 × 10²³ constituting particles is called 1 mole of this substance. For example 1 mole of sodium element contains 6.022 × 10²³ atoms, 1 mole of water contains 6.022 × 10²³ molecules of H₂O.

Important methods for determination of Moles:

If mass is given of any substance: $\text{Mole of the given substance}$ = $\dfrac{\text{Given Mass}}{\text{Molar mass}}$

Example: No. of moles in 2.3 gram sodium, No. of moles = $\frac{\text{given mass}}{\text{Molar mass}}$ or $\frac{2.3}{23}$ = 0.1 mole

If no. of particles given, then $\text{Moles} = \dfrac{\text{Given number of molecules}}{6.022 \times 10^{23}}$

Q.No. of moles in 6.022 × 10²² molecules of water.

As No. of molecules of water = 6.022 × 10²² is given, we shall apply the above formula in following ways: Moles in water = $\frac {6.022\times 10^{22}}{6.22\times 10^{23}}$ = 0.1mole

If Volume of any gas at STP given, then Mole of the given gas at the same condition = $\frac {\text{Volume of the gas in L}}{22.4 L}$

Example : Number of moles in 44.8 L of CO2, No. of moles = $\frac{44.8}{22.4}$ = 2L

My formula Know Everyone: Class 12th Chemistry