Lanthanoids And Actinides

(1): Thermal Decomposition of Potassium Permanganate (KMnO 4) - On heating, potassium permanganate (KMnO 4) decomposes to form potassium manganate (K 2 MnO 4), manganese dioxide (MnO 2), and oxygen gas (O 2). - This reaction is used in laboratories as a source of oxygen. The thermal decomposition reaction is: 2KMnO 4 heat \to K 2 MnO 4 + MnO 2 + O 2 (2): Redox Reaction of Dichromate and Iodide in Acidic Medium - Dichromate (Cr 2 O 7 2-) acts as an oxidizing agent and oxidizes iodide (I -) to iodine (I 2) in acidic medium. - The chromium in dichromate is reduced from Cr 6+ to Cr 3+ . - The reaction follows the principles of redox balancing, maintaining charge and mass balance. The balanced redox reaction is: Cr 2 O 7 2- + 6I - + 14H + \to 2Cr 3+ + 3I 2 + 7H 2 O

To solve the problem, we need to calculate the limiting molar conductivity ( $%5CLambda_m%5E0$ ) of acetic acid and its degree of dissociation ( $%5Calpha$ ), given its molar conductivity ( $%5CLambda_m$ ) as 48.1 $%5COmega%5E%7B-1%7D%20%5C%2C%20%5Ctext%7Bcm%7D%5E2%20%5C%2C%20%5Ctext%7Bmol%7D%5E%7B-1%7D$ , and the limiting molar conductivities: $%5CLambda_m%5E0%20(%5Ctext%7BHCl%7D)%20%3D%20426$ , $%5CLambda_m%5E0%20(%5Ctext%7BNaCl%7D)%20%3D%20126$ , and $%5CLambda_m%5E0%20(%5Ctext%7BCH%7D_3%5Ctext%7BCOONa%7D)%20%3D%2091%20%5C%2C%20%5COmega%5E%7B-1%7D%20%5C%2C%20%5Ctext%7Bcm%7D%5E2%20%5C%2C%20%5Ctext%7Bmol%7D%5E%7B-1%7D$ .

1. Calculate $%5CLambda_m%5E0$ for Acetic Acid:
Acetic acid ( $%5Ctext%7BCH%7D_3%5Ctext%7BCOOH%7D$ ) dissociates into $%5Ctext%7BH%7D%5E%2B$ and $%5Ctext%7BCH%7D_3%5Ctext%7BCOO%7D%5E-$ . We use Kohlrausch’s law to find $%5CLambda_m%5E0%20(%5Ctext%7BCH%7D_3%5Ctext%7BCOOH%7D)$ : $%5CLambda_m%5E0%20(%5Ctext%7BCH%7D_3%5Ctext%7BCOOH%7D)%20%3D%20%5Clambda_m%5E0%20(%5Ctext%7BH%7D%5E%2B)%20%2B%20%5Clambda_m%5E0%20(%5Ctext%7BCH%7D_3%5Ctext%7BCOO%7D%5E-)$ .
We can find these ionic conductivities using the given data:
- $%5CLambda_m%5E0%20(%5Ctext%7BHCl%7D)%20%3D%20%5Clambda_m%5E0%20(%5Ctext%7BH%7D%5E%2B)%20%2B%20%5Clambda_m%5E0%20(%5Ctext%7BCl%7D%5E-)%20%3D%20426$ ,
- $%5CLambda_m%5E0%20(%5Ctext%7BNaCl%7D)%20%3D%20%5Clambda_m%5E0%20(%5Ctext%7BNa%7D%5E%2B)%20%2B%20%5Clambda_m%5E0%20(%5Ctext%7BCl%7D%5E-)%20%3D%20126$ ,
- $%5CLambda_m%5E0%20(%5Ctext%7BCH%7D_3%5Ctext%7BCOONa%7D)%20%3D%20%5Clambda_m%5E0%20(%5Ctext%7BNa%7D%5E%2B)%20%2B%20%5Clambda_m%5E0%20(%5Ctext%7BCH%7D_3%5Ctext%7BCOO%7D%5E-)%20%3D%2091$ .
Subtract the second equation from the first to eliminate $%5Clambda_m%5E0%20(%5Ctext%7BCl%7D%5E-)$ :
$%5Clambda_m%5E0%20(%5Ctext%7BH%7D%5E%2B)%20%2B%20%5Clambda_m%5E0%20(%5Ctext%7BCl%7D%5E-)%20-%20(%5Clambda_m%5E0%20(%5Ctext%7BNa%7D%5E%2B)%20%2B%20%5Clambda_m%5E0%20(%5Ctext%7BCl%7D%5E-))%20%3D%20426%20-%20126$ ,
$%5Clambda_m%5E0%20(%5Ctext%7BH%7D%5E%2B)%20-%20%5Clambda_m%5E0%20(%5Ctext%7BNa%7D%5E%2B)%20%3D%20300$ .
Now use the third equation with $%5Clambda_m%5E0%20(%5Ctext%7BNa%7D%5E%2B)$ :
From the third equation, $%5Clambda_m%5E0%20(%5Ctext%7BCH%7D_3%5Ctext%7BCOO%7D%5E-)%20%3D%2091%20-%20%5Clambda_m%5E0%20(%5Ctext%7BNa%7D%5E%2B)$ .
We need $%5Clambda_m%5E0%20(%5Ctext%7BH%7D%5E%2B)%20%2B%20%5Clambda_m%5E0%20(%5Ctext%7BCH%7D_3%5Ctext%7BCOO%7D%5E-)$ :
Add $%5Clambda_m%5E0%20(%5Ctext%7BH%7D%5E%2B)%20-%20%5Clambda_m%5E0%20(%5Ctext%7BNa%7D%5E%2B)%20%3D%20300$ to $%5Clambda_m%5E0%20(%5Ctext%7BNa%7D%5E%2B)%20%2B%20%5Clambda_m%5E0%20(%5Ctext%7BCH%7D_3%5Ctext%7BCOO%7D%5E-)%20%3D%2091$ :
$%5Clambda_m%5E0%20(%5Ctext%7BH%7D%5E%2B)%20%2B%20%5Clambda_m%5E0%20(%5Ctext%7BCH%7D_3%5Ctext%7BCOO%7D%5E-)%20%3D%20300%20%2B%2091%20%3D%20391$ .
Thus, $%5CLambda_m%5E0%20(%5Ctext%7BCH%7D_3%5Ctext%7BCOOH%7D)%20%3D%20391%20%5C%2C%20%5COmega%5E%7B-1%7D%20%5C%2C%20%5Ctext%7Bcm%7D%5E2%20%5C%2C%20%5Ctext%7Bmol%7D%5E%7B-1%7D$ .

2. Calculate the Degree of Dissociation ( $%5Calpha$ ):
The degree of dissociation is given by $%5Calpha%20%3D%20%5Cfrac%7B%5CLambda_m%7D%7B%5CLambda_m%5E0%7D$ :

$%5Calpha%20%3D%20%5Cfrac%7B48.1%7D%7B391%7D%20%5Capprox%200.123$ .

Final Answer:
The limiting molar conductivity $%5CLambda_m%5E0$ for acetic acid is $391%20%5C%2C%20%5COmega%5E%7B-1%7D%20%5C%2C%20%5Ctext%7Bcm%7D%5E2%20%5C%2C%20%5Ctext%7Bmol%7D%5E%7B-1%7D$ , and the degree of dissociation $%5Calpha$ is approximately $0.123$ .

High-Yield Trend

Chapter Questions
2 MCQs

Official Solution

Official Solution

Lanthanoids And Actinides

High-Yield Trend

Chapter Questions 2 MCQs

Official Solution

Official Solution

Chapter Questions
2 MCQs