Thermodynamics

Step 1: Recall the Bohr–Bury rules for electronic configuration:

Maximum electrons in a shell = .
The outermost shell can have a maximum of 8 electrons.
The penultimate shell can have a maximum of 18 electrons, but if the outermost shell has electrons, the penultimate shell cannot have more than 8 electrons.

Step 2: Examine option (A): Here, the outermost shell has 1 electron, but the penultimate shell has 13 electrons, which violates the Bohr–Bury rule.
Step 3: Check the remaining options:

(B) — valid as per Bohr–Bury rules.
(C) — valid.
(D) — valid.

Step 4: Therefore, the incorrect electronic arrangement is 2, 8, 13, 1.

Step 1: For a reversible process at constant temperature, the change in entropy is given by
Step 2: Given:
Step 3: Rearranging the formula to find temperature:
Step 4: Substituting values, Hence, the temperature of the sample is

Step 1: Intensive properties are those properties which do not depend on the amount of substance present in the system.
Step 2: Pressure remains the same irrespective of the size or quantity of the system, hence it is an intensive property.
Step 3: Internal energy, heat capacity, and enthalpy depend on the amount of substance present, therefore they are extensive properties.

Step 1: When an inert gas is added at constant pressure, the volume of the system increases.
Step 2: Increase in volume leads to a decrease in partial pressures of all reacting gases.
Step 3: According to Le Chatelier’s principle, the equilibrium shifts towards the side with greater number of moles of gas.
Step 4: Since the reactant side has more gaseous moles, equilibrium shifts towards the left.
Step 5: As a result, less is formed.
Step 6: The equilibrium constant depends only on temperature and remains unchanged.

Step 1: The average kinetic energy of gas molecules is given by:
Step 2: The above expression shows that average kinetic energy depends only on temperature and not on:
Nature of the gas
Molecular mass of the gas
Amount of the gas
Step 3: Since both and are at the same temperature, their average kinetic energies are equal.

Step 1: Convert enthalpy of formation of liquid hydrazine to gaseous state.
Step 2: Write formation reaction for gaseous hydrazine:
Step 3: Enthalpy of formation using bond energies: Bonds broken: Total =
Step 4: Bonds formed in :
Step 5: Apply enthalpy relation:
Step 6: Considering standard approximations in bond energy data, the closest correct option is:

Step 1: The van’t Hoff equation relating equilibrium constant and temperature is:
Step 2: Hence, the slope of the graph between and is:
Step 3: The straight line is given to be at , so:
Step 4: Substituting:
Step 5: Using :

Step 1: At boiling point, the phase equilibrium condition is:
Step 2: Using the relation: At equilibrium,
Step 3: Convert enthalpy to joules:
Step 4: Substitute the given values:
Step 5: Hence, the boiling point of the liquid is:

Step 1: Let the bond dissociation energies be expressed using the given ratio. Given ratio: Let:
Step 2: Write the expression for enthalpy of formation of . Formation reaction:
Step 3: Substitute the bond energies.
Step 4: Use the given value of .
Step 5: Identify the required quantity.
Hence, the bond dissociation energy of is

Step 1: Use the relation between Gibbs free energy and emf. For an electrochemical reaction, where = number of electrons transferred, = Faraday constant, = emf of the cell.
Step 2: Determine the number of electrons transferred. Reaction: Oxidation state change: From the reaction, mol Al are oxidised:
Step 3: Substitute given values. Given:
Step 4: Calculate emf. But this value corresponds to one mole of formation being mole of . Since the given is per mole of , total electrons involved are 8. Thus,
Hence, the minimum emf required is

Step 1: Write Raoult’s law for an ideal binary solution. Total vapour pressure of solution: Step 2: Substitute the given values. Given: Step 3: Apply Raoult’s law. Step 4: Simplify the equation. Hence, the vapour pressure of pure ‘B’ is .

Step 1: Count the contribution of atom A. Atom A occupies the corners of the cube. There are corner atoms and each corner atom is shared by unit cells. Step 2: Count the contribution of atom B. Atom B occupies the face centres of the cube. There are face-centred atoms and each face-centred atom is shared by unit cells. Step 3: Write the atomic ratio. Step 4: Write the empirical formula. The simplest whole number ratio gives the empirical formula:

Acidity is inversely related to the value.
A lower corresponds to a higher concentration of hydrogen ions and thus greater acidity.
The options are compared: .
The smallest value listed is .
Therefore, the solution with is the most acidic.