Class 11th

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: option (i) 

If the equation is multiplied by 2, the equilibrium constant for the new equation is the square of K it means we must do the square of 5 that is 25. Then and on reversing the reaction the value of the equilibrium constant is inversed means the value of K comes out to be 1/25= 0.04. 

This is a short answer type question as classified in NCERT Exemplar

D₂O has greater intermolecular force of attraction. This is because D₂O exhibits higher values for melting point, enthalpy of vaporization, and viscosity compared to H₂O, indicating stronger intermolecular interactions needed to overcome these properties. 

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: option (ii)

 Greater the boiling point, lower will be the vapour pressure.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option(i)

The given reaction is :-

N₂ (g) + 3H₂ (g) ?2NH₃ (g)

A/C to vart Hoff's eqn,

$\frac{\mathrm{d}\mathrm{l}\mathrm{n}\mathrm{K}}{\mathrm{d}\mathrm{T}\mathrm{}}\mathrm{}$ = $\frac{?\mathrm{M}}{\mathrm{R}\mathrm{T}2}$

K is dependent on temperature. As temperature remains constant at equilibrium, so K will remain constant in the given conditions.

This is a short answer type question as classified in NCERT Exemplar

In H₂O₂ oxygen is in -1 oxidation state which has tendency to become -2 that is why it is a better oxidising agent than water.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option(i)

When the reaction is half completed, [A] = [B].
Thus, [B][A]=1
?G⁰=−RTInK=−RTIn(1)=0

This is a short answer type question as classified in NCERT Exemplar

2H₂O₂ → 2H₂O+O₂

5 volume H₂O₂ means IL of H₂O₂ will give 5L of O₂ at STP.

On the basis of the equation it is clear that 22.7 L of O₂ is produced by 68 g of H₂O₂.

5L of O₂ is produced by = $\frac{68g*51}{22.71}$  = $\frac{3400}{227}$ = g of H₂O₂=14.9g

i.e. 15 g H₂O₂ dissolved in 1 L solution will give 5 L oxygen or 1.5 g H₂O₂/100 mL solution will give 500 mL oxygen. Thus 15 g/L or 1.5% solution is known as the 5V solution of H₂O₂.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option(iii)

The equation for weak acid and weak base is given as-

pH=7+ $\frac{1}{2}$  ?(pKa?−pKb?)

K_a=1.8*10^-5

pK_a=−log (1.8*10-5) =5−log1.8 =4.744

pK_b?=−log (1.8*10-5) =4.744

pH=7+ $\frac{1}{2}$  ?(pKa?−pK_b?) =7

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: option (i)

pH for weak acid is given by the following equation:

pH= $\frac{1}{2}$ [pK_a−logC]

= $\frac{1}{2}$ [−log (1.74*10⁻⁵)−log0.01]

=3.4

This is a short answer type question as classified in NCERT Exemplar

Peroxodisulphate, obtained by electrolytic oxidation of acidified sulphate solutions at high current density, on hydrolysis yields hydrogen peroxide.

2HSO₄^- (aq) → HO₃SOOSO₃ 2HSO₄^- (aq)+2H⁺ (aq)+H₂O₂ (aq)

This method is now used for the laboratory preparation Of D₂O₂.

K₂S₂O₈ (s)+2D₂O (l) → 2KDSO₄ (aq)+D₂O₂ (l)