Chemistry

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

Mass of H₂O₂ = 68 g.

Mol. Mass of H₂O₂=34gmol^-1

∴1L of M solution of H₂O₂ will contain H₂O₂=34*5 g

∴2L of 5 M solution will contain H₂O₂=34*5*2=340g

or 200 mL of 5 solution will contain H₂O₂ = $\frac{340}{2000}$ $*$  200 = 34 g

Now 2H₂O₂ → 2H₂O+O₂

           64 g  32g

Now 68 g of H₂O₂ on decomposition will give O₂=32g

∴34g of H₂O₂ on decomposition will give O₂ = $\frac{34}{68}$ $*$    34 =16 g

(i) Industrial preparation: H₂O₂ is prepared by the auto-oxidation of 2- alkylanthraquinols

2-ethylanthraquinol ↔ H₂O₂⁺ oxidised product

2Fe₂+ (aq)+2H+ (aq)+H₂O₂ → Fe3 (aq)+2H₂O (l)

PbS (s)+4H₂O₂ (l) → PbSO4 (s)+4H₂O (l)

(ii) Reducing action of hydrogen peroxide 

2MnO₄^-+6H⁺+5H₂O₂→ 2Mn²⁺+8H₂O+5H₂

HOCl+H₂O₂ → H₃O⁺+Cl^-+O₂

Oxidising action of hydrogen peroxide 

2Fe²⁺+H₂O₂→ 2Fe³⁺+2OH^-

Mn²⁺+H₂O₂ → Mn⁴⁺+2OH^-

Acidic properties of H₂O₂

I₂+H₂O₂+2OH^-→ 2I^-+2H₂O+H₂

2MnO₄^-+3H₂O₂→ 2MnO₂+3O₂+2H₂O+2OH^-

This is a Long Answer Type Questions as classified in NCERT Exemplar

Ans: According to Le Chatelier's principle, when we raise the temperature, it shifts the equilibrium to left and decreases the equilibrium concentration of ammonia since it is an exothermic reaction. In other words, low temperature and high pressure is favourable for high yield of ammonia. There will be no change in equilibria on addition of argon (Ar).

This is a Long Answer Type Questions as classified in NCERT Exemplar

Ans: The values of K_c and Q_c are itself sufficient to explain the direction of reaction and less than or greater than one another decides the direction in which reaction will proceed as follows-

(i) As Qc < Kc, the reaction proceeds in the forward direction. 

(ii) If Qc > Kc, the reaction will proceed in the direction of reactants (reverse reaction). 

(iii) If Qc = Kc, no net reaction occurs.

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

D₂O can be prepared by prolonged electrolysis of water. Due to high molecular mass D₂O differs from water.

NaOH+D₂O → NaOD+HOD

HCl+D₂O → DCl+ HOD

NH₄Cl+D₂O → NH₃DCl+HOD

Physical Properties of H₂O and D₂O

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

Atomic hydrogen is highly reactive, whereas molecular hydrogen is rather inert. Bond dissociation enthalpy determines the chemical behaviour of dihydrogen (and, for that matter, any molecule) to a great extent. For a single bond between two atoms of any element, the H-H bond dissociation enthalpy is the highest. As a result, molecular hydrogen reacts only with a few elements.

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

Atomic hydrogen is highly reactive, whereas molecular hydrogen is rather inert. Bond dissociation enthalpy determines the chemical behaviour of dihydrogen (and, for that matter, any molecule) to a great extent. For a single bond between two atoms of any element, the H-H bond dissociation enthalpy is the highest. As a result, molecular hydrogen reacts only with a few elements.

This is a Assertion Type Questions as classified in NCERT Exemplar

Ans: Option (iv)

Both the statements given in the assertion and reason are false.

The energy required to break the first O-H bond is not same than that of the second O-H bond in water molecule, that means the bond enthalpies of the two O-Hare different. This is because the electronic environment around the oxygen atom in water molecules is not the same even after the O-H bond breaks.