Titration of NaOH ( Titrate ) with CH3COOH (Titrant)

On adding CH3COOH solution to NaOH solution OH–will be replaced by CH3COO– ions, so conductance of solution decreases. After complete neutralisation further added CH3COOH will remain 

undissociated because it is a weak acid and there is also common ion effect on acetate ions. So number of ions in solution will remain almost constant therefore conductance of solution will remain constant.

Derivation of Van't Hoff equation.

This equation gives the quantitative temperature dependance of equilibrium constant K. 

The relation between standard free energy change ∆G° and equilibrium constant is 

∆G° = – RT ln K ……(1) 

We know that, 

∆G° = ∆H° – T∆S° -(2)  

Substituting (1) in equation (2)  

– RT ln K = ∆H° – T∆S° 

Rearranging, 

ln K = − Δ H °/R T + T ΔS °/R T

ln K = − Δ 𝐻 °/ 𝑅 𝑇 + Δ 𝑆 °/ 𝑅   ….(3) 

Differentiating equation (3) with respect to temperature  

d(in K) = d(in K) / dT = ∆H° / RT2

Equation (4) is known as differential form of van't Hoff equation. 

On integrating the equation (4), between T1 and T2 with the respective equilibrium constants and K2 

Equation is known as integrated form of van't Hoff equation.

Why Bond length of O-O is greater in H2O2 than O2F2?

Electronegativity of F is much more than hydrogen and also hybridization of oxygen atoms in of H₂O₂ and O₂F₂ both have (sp3) same and hence we can apply bent’s rule.

According bent’s rule more eletronegative atoms reduce % s-character (or increases % p-character vice versa) of those hybride orbital in which they attach. So in case of O₂F₂ % s-character of those hybride orbital decrease which have  flourine while % p-character increases in same way % s-character of  remaing hybride orbital inceases and p-character decreases hence its bond length also (bond length is directly proportional to % p-character) decreases.  hence O-O bond length in O₂F₂ is shorter thane H₂O₂.

“तलाश” मेरी क़लम से …

गुजरी है करीब से कई बहारे , किसी की महक से ये वजूद महका ही नहीँ ,

न जाने कहाँ है कौन है ओ । जिसकी तलाश मे दर बदर मै भटकता हूँ ,

कही तो ये तलाश ख़त्म होगीं , इस सफ़र की ,जो मंजिलें कहीँ खो गयी है ,

कभी तो मिलेंगी इसी इंतज़ार मे , चिलमन गिराये बैठा हूँ ॥ मै तेरी तलाश मे , हाँ तेरी तलाश मे , हाँ तेरी तलाश मे  ॥

(Written by: सुनील कुमार "सकल")

The activation energy of a reaction is 225 k Cal mol-1 and the value of rate constant at 40°C is 1.8 ×10^−5 s^−1 . Calculate the frequency factor, A.

SOLUTION:

Here, we are given that

Ea = 22.5 kcal mol-1 = 22500 cal mol-1

T = 40°C = 40 + 273 = 313 K

k = 1.8 × 10-5 sec-1

Substituting the values in the equation

log A = log (1.8) −5 + (15.7089)

log A = 10.9642

A = antilog (10.9642)

A = 9.208 × 101^0 collisions s^−1