EquilibriumNEET Chemistry · Class 11 · NCERT Chapter 6

Solution

${\text{Na}}_2{\text{CO}}_3$ is a salt of a strong base ($\text{NaOH}$) and a weak acid (${\text{H}}_2{\text{CO}}_3$), which hydrolyzes to produce ${\text{OH}}^{-}$ ions, increasing the pH. Therefore, ${\text{Na}}_2{\text{CO}}_3$ will give the highest pH in water, so option (c) is correct.

Solution

The Gibbs free energy change is related to the solubility product by $\Delta G^{\circ }=-2.303RT\log K_{sp}$. Substituting the given values, $63.3\times 10^3=-2.303\times 8.314\times 298\times \log K_{sp}\Rightarrow \log K_{sp}=\frac{63.3\times 10^3}{-2.303\times 8.314\times 298}\Rightarrow \log K_{sp}=-11.09\Rightarrow K_{sp}=8.0\times 10^{-12}$. Thus, option (b) is correct.

Solution

According to Le Chatelier's principle, increasing pressure and decreasing temperature will shift the equilibrium in the forward direction for the exothermic reaction ${\text{N}}_2(g)+3{\text{H}}_2(g)\leftrightarrow 2{\text{NH}}_3(g)+\text{heat}$. NCERT XI chapter Equilibrium explains that increasing pressure favors the side with fewer moles of gas, and decreasing temperature favors the exothermic direction, so option (d) is correct.

Solution

64Gd= 54[Xe]6s2 4f7 5d1

Solution

N2(g)+O2(g)⇌2NO(g) ;K 1 2N2(g)+1 2 O2(g)⇌NO(g) ;K′ K= [NO]2 [N2] [O2] K′= NO [N2]1 2⁄ [O2]1 2⁄ ∴K′=√K

Solution

Strong acid with its salt can not form buffer solution. HClO4 and NaClO4 cannot act as an acidic buffer.

Solution

N1V1−N2V2=N.V. 0.1×1−0. 01×1=N×2 www.vedantu.com 52 [OH−]=NR=0. 09 2 =0.045 N pOH=−log (0.045)=1.35 ∴pH=14−pOH=14−1.35=12.65 1 49. D ecreasing order of stability of O2,O2 −,O2 + and O2 2− is: (1) O2 −> O2 2−>O2 +>O2 (2) O2 +>O2> O2 −> O2 2− (3) O2 2−> O2 −>O2> O2 + (4) O2> O2 +> O2 2−> O2 − Solution: (2) Given species: O2,O2 −,O2 +,O2 2− Total number of electrons O2⟶16e− O2 −⟶17e− O2 +⟶15e− O2 2−⟶18e− O2 + O2 Bond order 2. 5 2 O2 − O2 2− 1.5 1 Stability×Bond order Stability order [O2 +>O2>O2 −>O2 2−]

Solution

MY insoluble salt Ksp = s2 MY(s)⇌ M(aq) + + Y(aq) − NY3 insoluble salt Ksp = 4s3 NY3(s)⇌ N(aq) + + 3Y(aq) − ∴ S(MY)= √6.2 ×10−13 = 7.8 ×10−7 Solubility values ∴ S(NY3)= ( 6.2×10−13 4 ) 1 3⁄ = 5.2 ×10−5 ∴ S(MY)< S(NY3)

Solution

2 224 2 4 2s s Ag C O (s) 2 Ag (aq) C O (aq)+- +⇌ KSP = [Ag +]2 [C2O4 2–] [Ag+] = 2.2 × 10 –4 M ∴ 424 24 2.2 10[C O ] M 1.1 10 M2 --- ×== × ∴ KSP = (2.2 × 10 –4)2 (1.1 × 10 –4) = 5.324 × 10–12

Solution

(I) 2 3 22 3 1 3 22 [NH ]N3 H 2 N H ; K [N ] [H ] += ⇌ (II) 2 22 2 22 [NO]NO 2 N O ; K [N ] [O ]+= ⇌ (III) 2 22 2 3 1/ 2 22 [H O]1HO H O ; K2 [H ] [O ] +⎯ ⎯ →= (II + 3 × III – II) will give K 32 2 52NH O 2NO 3H O;2++ ⇌ ∴ 3 23 1KK K / K=×

Solution

The solubility of ${\text{BaSO}}_4$ is $2.42\times 10^{-3}{\text{g L}}^{-1}$. Converting to molarity, $S=\frac{2.42\times 10^{-3}{\text{g L}}^{-1}}{233{\text{g mol}}^{-1}}=1.04\times 10^{-5}{\text{mol L}}^{-1}$. The solubility product $K_{sp}=S^2=(1.04\times 10^{-5}{)}^2=1.08\times 10^{-10}{\text{mol}}^2{\text{L}}^{-2}$, so option (b) is correct.

Solution

The reaction is exothermic ($\Delta rH=-X\text{kJ}$), so lower temperature favors product formation according to Le Chatelier's principle. High pressure also favors the side with fewer moles of gas, which is the product side. Therefore, option (c) is correct.

Solution

Bin ∝ d (d ≤ R) And 𝐵0∝ 1 𝑑(𝑑>𝑅)

Solution

The correct structure of intermediate A can be determined by analyzing the reaction mechanism. In the given reaction, intermediate A forms through the attack of a nucleophile on the electrophilic carbon, leading to the formation of a new bond and the creation of a negative charge on the oxygen atom, as shown in option (c). This aligns with the concepts of nucleophilic addition in NCERT XII chapter Equilibrium.

Solution

The major product of the reaction is determined by the equilibrium position and the stability of the products. In this case, option (c) represents the most stable product, which is favored at equilibrium according to Le Chatelier's principle.

Solution

The relationship between the equilibrium constant $K_c$ and the standard Gibbs free energy change ${\Delta }_r{G}^{\circ }$ is given by ${\Delta }_r{G}^{\circ }=-RT\ln K_c$. Substituting the values, ${\Delta }_r{G}^{\circ }=-8.314{\text{J mol}}^{-1}{\text{K}}^{-1}\times 300\text{K}\times \ln (2\times 10^{13})$, so option (a) is correct.

Solution

For a salt of a weak base and a weak acid, the pH is given by $\text{pH}=7+\frac{1}{2}({\text{pK}}_a-{\text{pK}}_b)$. Substituting the values, $\text{pH}=7+\frac{1}{2}(4.77-3.27)=7+\frac{1}{2}(1.50)=7+0.75=7.75$, so option (c) is correct.

Solution

CH3COONa + CH3COOH, CH3COOH(pKa) = 4.57 0.1M 0.01M 50 mL 50 mL It is a mixture of weak acid and salt of its conjugate base. Hence it is acidic buffer. pH = a [Salt]pK log [Acid]+ = 4.57 + log 0.1 0.01 ⎛⎞ ⎜⎟⎝⎠ = 4.57 + 1 = 5.57

Solution

233O (g) 2O (g) 2 3 C 3 2 [O ]K [O ] = 2 3 59 3 3 C 2[O ] K [O ] 3 10 (0.04) −= = × × 2 63 64 3[O ] 1.9 10 19 10−−= × = × 32 3[O ] 4.38 10 −=× Concentration of O3 at equilibrium = 4.38 × 10–32 M

Solution

2A B + C, KC = 4 × 10–3 At a given time t, QC is to be calculated and been compared with KC. 33 C 2 3 2 [B][C] (2 10 )(2 10 )Q [A] (2 10 ) −− − ××== × QC = 1 As QC > KC, so reaction has a tendency to move backward.

Solution

2 W 1000M M V (in mL) ×= × 2M M V (in mL) 0.75 36.5 25W 1000 1000 ×× ××== = 0.684 g (Mass of HCl) 36.5g 40 g HCl NaOH HCl NaOH+ ⎯⎯ → + 36.5 g HCl reacts with NaOH = 40 g 0.684 g HCl reacts with 40NaOH 0.684 0.750 g36.5=× Amount of NaOH left = 1 g – 0.750 g = 0.250 g = 250 mg SECTION-B

Solution

During the preparation of Mohr’s salt, dilute sulphuric acid is added to prevent the hydrolysis of Fe 2+ ion. - 41 - NEET (UG)-2024 (Code-Q1)

Solution

2NO(g) ? N2(g) + O2(g) 22 c 2 [N ][O ]K [NO] = = 33 33 3 10 4.2 10 2.8 10 2.8 10 −− −− × × × × × × = 1.607 2NO(g) ? N2(g) + O2(g) t = 0 0.1 0 0 0.1 – 0.1α 0.05α 0.05α c 2 0.05 0.05K (0.1 0.1 ) α × α= −α c 2 0.05 0.05K 0.01(1 ) α × α= −α 1.607 = 22 2 (0.05) 0.01 (1 ) α −α 22 22 1.607 (0.1) (1 ) (0.05) α× =−α 1.27 0.1 1 0.05 α× =−α 2.541 α =−α α = 2.54 – 2.54α 3.54α = 2.54 2.54 0.7173.54α = = - 42 - NEET (UG)-2024 (Code-Q1)

Solution

H₃PO₄ is a stronger acid than H₂PO₄⁻ and HPO₄²⁻.

H₃PO₄ (aq) ⇌ H⁺ (aq) + H₂PO₄⁻ (aq) &nbsp;&nbsp; $K_{a_1}=7.5\times 10^{-3}$

H₂PO₄⁻ (aq) ⇌ H⁺ (aq) + HPO₄²⁻ (aq) &nbsp;&nbsp; $K_{a_2}=6.2\times 10^{-8}$

HPO₄²⁻ (aq) ⇌ H⁺ (aq) + PO₄³⁻ (aq) &nbsp;&nbsp; $K_{a_3}=1.7\times 10^{-12}$

$K_{a_1}>K_{a_2}>K_{a_3}$

$\log K=\log K_{a_1}+\log K_{a_2}+\log K_{a_3}$

Ans. (A), (B) and (C) only.

Solution

Degree of dissociation ($\alpha$) is given as
$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}$

${\Lambda }_m^{\circ }={\Lambda }_{+}^{\circ }+{\Lambda }_{-}^{\circ }$
= 349.6 + 50.4
= 400 S cm${}^2$ mol${}^{-1}$

$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}=\frac{90}{400}=0.225$

Solution

The reaction N₂(g) + O₂(g) ⇌ 2NO(g) has ΔH = +180.7 kJ/mol — endothermic — and the moles of gas are equal on both sides (Δn = 0).

Le Chatelier's principle:
• Higher temperature favours the endothermic forward reaction → MORE NO. (helpful)
• Pressure has no effect because Δn_gas = 0.
• Adding a catalyst speeds up both directions equally — does not shift equilibrium.

Hence higher yield of NO is obtained at high temperature. Per the official key, the matching option is the one stating 'high temperature' alone (or 'high temperature, any pressure').