Test: JEE Main 35 Year PYQs- Equilibrium - JEE MCQ

Let the heat capacity of insulated beaker be C.
Mass of aqueous content in expt. 1 = (100 + 100) × 1 = 200 g

⇒ ± Total heat capacity = (C + 200 × 4.2) J/K Moles of acid, base neutralised in expt. 1 = 0.1 × 1 = 0.1

⇒ Heat released in expt. 1 = 0.1 × 57 = 5.7 KJ = 5.7 × 1000 J

⇒ 5.7 × 1000 = (C + 200 × 4.2) × Δ´T.
5.7 × 1000 = (C + 200 + 4.2) × 5.7

⇒ (C + 200 × 4.2) = 1000
In second experiment,

Total mass of aqueous content = 200 g

⇒ Total heat capacity = (C + 200 × 4.2) = 1000

⇒ Heat released = 1000 × 5.6 = 5600 J.
Overall, only 0.1 mol of CH₃COOH undergo neutralization.

⇒ Δ H_{neutralization} of CH₃COOH = -5600/0.1

= – 56000 J/mol
= – 56 KJ/mol.

⇒ Δ H_{neutralization} of CH₃COOH = 57 – 56 = 1 KJ/mol

Final solution contain 0.1 mole of CH₃COOH² and CH₃COONa each.
Hence it is a buffer solution.

(A) Correct statement.
As on decrease in pressure reaction move indirection where no. of gaseous molecules increase.
(B) Correct statement At the start of reaction Q_P < K_P so dissociation of X₂ take place spontaneousely.
(C) Incorrect statement as

The statement-1 is clearly wrong in context to LeChateliers principle, which states that “increase in temperature shifts the equilibrium in the forward direction of those reactions which proceed with absorption of heat (endothermic reactions), and in the backward direction of those reactions which proceed with the evolution of heat (exothermic reactions).” Statement -2 is clearly true again according to Lechatelier principle.

TIPS/Formulae :

Among oxyacids, the acidic character increases with increase in oxidation state of the central atom.
O.S. of  N in HNO₃ = + 5
O.S.of  N in HNO₂  = + 3

thus HNO₃ stronger  than HNO₂. Hence assertion is correct.

The assertion is true but the reason is wrong as can be clearly seen from the above structures.

We know that for every chemical reaction at equilibrium, Gibb's free energy (ΔG = 0) is zero.
However standard Gibb's free energy (ΔG°) may or may not be zero. Thus statement 1 is False.
For a spontaneous reaction, at constant temperature and pressure, the reaction proceeds in the direction in which ΔG is < 0 i.e. in the direction of decreasing Gibb's energy (G) so the statement 2 is True.
Thus the only such option is (d) which is correct answer.

NOTE :  A buffer is a solution of weak acid and its salt with strong base and vice versa.
HCl is strong acid and NaCl is its salt with strong base. pH is less than 7 due to HCl.

(HSO₄)^– can accept and donate a proton
(HSO₄)^– + H⁺ → H₂SO₄  (acting as base)
(HSO₄)^– – H⁺ → SO₄^2–.   (acting as acid)

In this reaction the ratio of number of moles of reactants to products is same i.e. 2 : 2, hence change in volume will not alter the number of moles.

K_p = K_c (RT)^Δn;

pH of an acidic solution should be less than 7. The reason is that from H₂O. [H⁺] = 10^–7M  which cannot be neglected in comparison to 10^–8M. The pH can be calculated as.
from acid, [H⁺]  = 10^–8M.

from H₂O, [H+] = 10^–7M
∴ Total [H⁺] = 10–8 + 10^–7 = 10^–8 (1 + 10) = 11× 10^–8
∴ pH = – log [H⁺] = –log 11×10–8 = –[log11 + 8 log 10]            
= –[1.0414 – 8] = 6.9586

Due to exothermicity of reaction low or optimum temperature will be required. Since 3 moles are changing to 2 moles.
∴ High pressure will be required.

The rain water after thunderstorm contains dissolved acid and therefore the pH is less than rain water without thunderstorm.

NOTE : Conjugate acid-base differ by H⁺

 The solids have con centration unity

Where  s is the solubility of MX₂ then K_sp = 4s³; s × (2s)² = 4×10^–12 = 4s³; s = 1 × 10^–4
∴ [M⁺⁺] = s = 1[M⁺⁺] = 10 × 10^{– 4}

The reaction given is an exothermic reaction thus accordingly to Lechatalier’s principle lowering of temperature, addition of F₂ and  or Cl₂ favour the for ward direction and hence the production of ClF₃.

For the reaction:- 

Given K_c = 1.8 × 10^–6   at 184ºC
R = 0.0831 kj/mol. k
K_p = 1.8 × 10^–6  × 0.0831 × 457 = 6.836 × 10^–6
[∵ 184°C = (273 + 184) = 457 k, Δn = (2 + 1, –1) = 1]
Hence it is clear that  K_p >  K_c

On solving,  [H⁺] = 3.98 × 10^–6

Conjugate acid-base pair differ by only one proton.

Conjugate base of OH^– is O2^–

Then 0.5 + x + x = 2x + 0.5 = 0.84  (given) ⇒ x = 0.17 atm.

Total moles after dissociation 1 – x + x + x = 1 + x

= mole fraction of PCl₃ × Total pressure