Chemistry NCERT Exemplar Solutions Class 12th Chapter Four

BCl₃ – electron deficient molecule

SF₆ – expanded octet molecule

NO – odd electron bond containing molecule

H₂SO₄ – expanded octet molecule

Sol. 

$P_T=X_A\left(P_A^0-P_B^0\right)+P_B^0$

ATQ

$550=\frac{1}{4}\left({\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0\right)+{\mathrm{P}}_{\mathrm{B}}^0$

$2200={\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0+4{\mathrm{P}}_{\mathrm{B}}^0$

$560=\frac{1}{5}\left({\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0\right)+{\mathrm{P}}_{\mathrm{B}}^0$

$2200={\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0+5{\mathrm{P}}_{\mathrm{B}}^0$

${\mathrm{P}}_{\mathrm{A}}^0+3{\mathrm{P}}_{\mathrm{B}}^0=2200$

$\frac{{\mathrm{P}}_{\mathrm{A}}^0\pm 4{\mathrm{P}}_{\mathrm{B}}^0=2800}{{\mathrm{P}}_{\mathrm{B}}^0=600}$

${\mathrm{P}}_{\mathrm{A}}^0=400\mathrm{m}\mathrm{m}\mathrm{H}\mathrm{g}$

Sol. $4\mathrm{L}\mathrm{i}+{\mathrm{O}}_2?2{\mathrm{L}\mathrm{i}}_2\mathrm{O}\mathrm{}$ oxide

$2\mathrm{N}\mathrm{a}+{\mathrm{O}}_2?{\mathrm{N}\mathrm{a}}_2{\mathrm{O}}_2$ peroxide

$\mathrm{K}+{\mathrm{O}}_2?{\mathrm{K}\mathrm{O}}_2\mathrm{}$ superoxide

Sol. $\mathrm{U}$ and $\mathrm{H}$ are temperature dependent

${\mathrm{C}}_{\mathrm{P},\mathrm{m}}-{\mathrm{C}}_{\mathrm{V},\mathrm{m}}=\mathrm{R}$  (for 1 mole of ideal gas)

$\mathrm{d}\mathrm{U}={\mathrm{C}}_{\mathrm{V}}\mathrm{d}\mathrm{t}$

Aromatic species are most stable, here

Initial temperature ; T₁ = 300 K

Final temperature ; T₂ = 309 K

Rate constant gets doubled i.e K₂ = 2K₁

$Usinglo{g}_{10}\frac{k_2}{k_1}=\frac{Ea}{2.3R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$

$log2=\frac{Ea}{2.3*8.3}\left[\frac{9}{300*309}\right]$

$E_a=\frac{2.3*8.3*300*309*log2}{9}J/mo\mathcal{l}$

$E_a=58.98kJ/mol$    

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Ans: Correct option B

If A→B then the concentration of both reactants and the products vary exponentially with time. But, in option B graph the reactant concentration decreases exponentially and the product concentration increases.

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Ans: Correct option C

Let's start with what a pseudo-first-order response is.

Although the pseudo-first-order reaction looks to be an order, it belongs to another order. It's a second-order reaction because it involves two reactants.

Let's have a look at a reaction.

CH₃Br + OH→CH₃OH + Br^- 

So, the rate law for the reaction is

Rate = k [OH] [CH₃Br]

Rate = k [OH^- ] [CH₃Br] = k (constant) [CH₃Br] = K' [CH₃Br]

Only the concentration of CH₃Br will change during the reaction, and the rate will be determined by the reaction's modifications.

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Correct option C    

In the presence of a catalyst, the value of ΔG cannot be changed for any reaction.

ΔG=−RtlnQ

Where Q is the Reaction Quotient, which is determined by the product and reactant concentrations. As a result, ΔG has no connection to the catalyst. Only when the reaction is spontaneous, which must be negative, is it checked. As a result, ΔG cannot be changed.