Which oxidizing agent, O 2 , O 3 or H 2 O 2 , will generate the greatest amount of energy for 1 mole of H 2 ( g ) ? (i) 2 H 2 ( g ) + O 2 ( g ) → 2 H 2 O ( g ) ; ΔH = - 483 . 6 kJ (ii) 3 H 2 ( g ) + O 3 ( g ) → 3 H 2 O ( g ) ; ΔH = - 868 . 2 kJ (iii) H 2 ( g ) + H 2 O 2 ( g ) → 2 H 2 O ( g ) ; ΔH = - 347 . 3 kJ Which of the above given reactions will generate the greatest amount of energy on a total mass basis of reactants that may be used in rocket propulsion?

Given equations: (i) 2 H 2 ( g ) + O 2 ( g ) → 2 H 2 O ( g ) ; ΔH = - 483 . 6 kJ (ii) 3 H 2 ( g ) + O 3 ( g ) → 3 H 2 O ( g ) ; ΔH = - 868 . 2 kJ (iii) H 2 ( g ) + H 2 O 2 ( g ) → 2 H 2 O ( g ) ; ΔH = - 347 . 3 kJ We have to find which oxidizing agent will generate the greatest amount of energy for 1 mol of H 2 . For 2 mol of H 2 in reaction (i), ∆ H = - 483 . 6 kJ . So, for 1 mol of H 2 , ∆ H = - 483 . 6 2 kJ mol = - 241 . 8 kJ mol For 3 mol of H 2 in reaction (ii), ∆ H = - 868 . 2 kJ . So, for 1 mol of H 2 , ∆ H = - 868 . 2 3 kJ mol = - 289 . 4 kJ mol For 1 mol of H 2 in reaction (iii), ∆ H = - 347 . 3 kJ mol . The change in enthalpy ( ∆ H ) of H 2 O 2 is less. So, it will generate the greatest amount of energy for 1 mol of H 2 ( g ) .

The Edison storage cell is represented as Fe ( s ) | FeO ( s ) | KOH ( aq ) Ni 2 O 3 ( s ) NiO ( s ) . The half cell reactions are: Ni 2 O 3 s + H 2 O l + 2 e - → 2 NiO s + 2 OH aq - E 0 = + 0 . 40 V FeO s + H 2 O l + 2 e - → Fe s + 2 OH aq - E 0 = - 0 . 87 V What is the cell emf? How does it depend on the concentration of KOH ?

Actual half reactions are: Reaction at anode: Fe + 2 OH - → FeO + H 2 O + 2 e - (oxidation) Reaction at cathode: Ni 2 O 3 + H 2 O + 2 e - → 2 NiO + 2 OH - (reduction) So, the cell net reaction is: Fe + Ni 2 O 3 → FeO + 2 NiO E cell = E cell ∘ - 2 . 303 RT nF logQ Temperature, T = 298 K Faraday's constant F = 96500 coulomb mol - 1 Universal gas constant R = 8 . 314 Joule K - 1 mol - 1 Reaction quotient, Q = concentration of products concentration of reactants Substituting in equation we get, E cell = E cell ∘ - 0 . 0591 2 log [ NiO ] 2 [ FeO ] [ Fe ] [ Ni 2 O 3 ] [ NiO ] 2 [ FeO ] [ Fe ] [ Ni 2 O 3 ] = 1 as all are solid E cell = E cell ∘ - 0 . 0591 2 log 1 log 1 = 0 So , E cell = E cell ∘ = 0 . 40 - ( - 0 . 87 ) = 1 . 27 volt Thus, the emf of cell is independent of KOH concentration.

The Edison storage cell is represented as Fe s | FeO s | KOH aq Ni 2 O 3 s Ni s . The half cell reactions are Ni 2 O 3 s + H 2 O l + 2 e - → 2 NiO s + 2 OH aq - E 0 = + 0 . 40 V FeO s + H 2 O l + 2 e - → Fe s + 2 OH aq - E 0 = - 0 . 87 V What is the maximum amount of electrical energy that can be obtained from one mole of Ni 2 O 3 ?

Actual half reactions are: Reaction at anode: Fe + 2 OH - → FeO + H 2 O + 2 e - (oxidation) Reaction at cathode: Ni 2 O 3 + H 2 O + 2 e - → 2 NiO + 2 OH - (reduction) So, the cell net reaction is: Fe + Ni 2 O 3 → FeO + 2 NiO E cell = E cell ∘ = 0 . 40 - ( - 0 . 87 ) = 1 . 27 volt Maximum amount of electrical energy= nFE ∘ = 2 × 96500 × 1 . 27 = 245 . 11 kJ

HARD

JEE Main

IMPORTANT

Earn 100

The two liquids $A$ and $B$ have the same molecular weight and form an ideal solution. The solution has the vapour pressure $700 mm Hg$ at $80 ° C$ . The above solution is distilled without reflux till $3 / 4$ of the solution is collected as condensate. The composition of the condensate is $x'_{A} = 0.75$ and that of the residue is ${x^{''}}_{A} = 0.3 .$ If the vapour pressure of the residue at $80 ° C$ is $600 mm Hg$ , calculate $x_{A}, P_{A}^{0}$ and $P_{B}^{0}$ .

Important Questions on Miscellaneous Problems for Revision

MEDIUM

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 21 - Miscellaneous Problems for Revision>Exercise 1>Q 47

A solution containing compound $X$ in water and a solution containing urea in water were put in a closed system. By doing this, some water vapour was removed from one solution and got condensed in the other. It is found that when both the solutions were at equilibrium vapour pressure, one solution contains $2 %$ of $X$ and the other $5 %$ by weight. Find the molecular weight of $X$ in grams.

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 21 - Miscellaneous Problems for Revision>Exercise 1>Q 48

The heat of formation for

{Fe}_{2} O_{3}

from simple substances is

- 821.32 kJ / mol

298 K

and standard pressure and that of

{Al}_{2} O_{3}

- 1675.60 kJ / mol

under the same conditions. Calculate the heat of reaction for reduction of

1 mol

{Fe}_{2} O_{3}

with metallic aluminium in

kJ / mol

MEDIUM

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 21 - Miscellaneous Problems for Revision>Exercise 1>Q 49

The heat of combustion of graphite at $298 K$ is $- 393.795 kJ / mol$ , while that of diamond's combustion at the same temperature is $- 395.692 kJ / mol$ . The specific heats for these substances are $720.83$ and $505.58 J {kg}^{- 1} K^{- 1}$ , respectively. Calculate the heat of graphite's transformation into diamond in $kJ / mol$ at $298 K$ .

EASY

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 21 - Miscellaneous Problems for Revision>Exercise 1>Q 50

Which oxidizing agent, $O_{2}, O_{3}$ or $H_{2} O_{2}$ , will generate the greatest amount of energy for $1$ mole of $H_{2} (g)$ ?

(i) $2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (g); ΔH = - 483.6 kJ$

(ii) $3 H_{2} (g) + O_{3} (g) \to 3 H_{2} O (g); ΔH = - 868.2 kJ$

(iii) $H_{2} (g) + H_{2} O_{2} (g) \to 2 H_{2} O (g); ΔH = - 347.3 kJ$

Which of the above given reactions will generate the greatest amount of energy on a total mass basis of reactants that may be used in rocket propulsion?

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 21 - Miscellaneous Problems for Revision>Exercise 1>Q 51

For the reaction ${[Ag (CN)_{2}]}^{-} ⇌ {Ag}^{+} + 2 {CN}^{-},$ the equilibrium constant at $25 ° C$ is $4 \times 10^{- 19}$ . Calculate the silver ion concentration in a solution which was originally $0.1$ molar in $KCN$ and $0.03$ molar in ${AgNO}_{3}$ . If your answer is $y \times 10^{- 18}, then y is$ :

MEDIUM

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 21 - Miscellaneous Problems for Revision>Exercise 1>Q 52

The Edison storage cell is represented as $Fe (s) | FeO (s) | KOH (aq) |{Ni}_{2} O_{3} (s)| Ni (s)$ .

The half cell reactions are:

${Ni}_{2} O_{3} (s) + H_{2} O (l) + 2 e = 2 NiO (s) + 2 {OH}^{-}$

$E^{0} = + 0.40 V$

$FeO (s) + H_{2} O (l) + 2 e = Fe (s) + 2 {OH}^{-}$

$E^{0} = - 0.87 V$

What is the cell reaction?

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 21 - Miscellaneous Problems for Revision>Exercise 1>Q 52

The Edison storage cell is represented as $Fe (s) | FeO (s) | KOH (aq) |{Ni}_{2} O_{3} (s)| NiO (s)$ .

The half cell reactions are:

${Ni}_{2} {O_{3}}_{(s)} + H_{2} O_{(l)} + 2 e^{-} \to 2 {NiO}_{(s)} + 2 {OH}_{(aq)}^{-}$

$E^{0} = + 0.40 V$

${FeO}_{(s)} + H_{2} O_{(l)} + 2 e^{-} \to {Fe}_{(s)} + 2 {OH}_{(aq)}^{-}$

$E^{0} = - 0.87 V$

What is the cell emf? How does it depend on the concentration of $KOH$ ?

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 21 - Miscellaneous Problems for Revision>Exercise 1>Q 52

The Edison storage cell is represented as ${Fe}_{(s)} | {FeO}_{(s)} | {KOH}_{(aq)} |{Ni}_{2} O_{3 (s)}| {Ni}_{(s)}$ .

The half cell reactions are

${Ni}_{2} {O_{3}}_{(s)} + H_{2} O_{(l)} + 2 e^{-} \to 2 {NiO}_{(s)} + 2 {OH}_{(aq)}^{-}$

$E^{0} = + 0.40 V$

${FeO}_{(s)} + H_{2} O_{(l)} + 2 e^{-} \to {Fe}_{(s)} + 2 {OH}_{(aq)}^{-}$

$E^{0} = - 0.87 V$

What is the maximum amount of electrical energy that can be obtained from one mole of ${Ni}_{2} O_{3} ?$

Important Questions on Miscellaneous Problems for Revision

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