Using the data (all values are in kilocalories per mole at 25 ° C ) given below, calculate the bond energy of C - C and C - H bonds. ΔH ∘ combustion (ethane) = - 372 . 0 ΔH ∘ combustion (propane) = - 530 . 0 ΔH ∘ for C (graphite) → C ( g ) = 172 . 0 Bond energy of H - H = 104 . 0 Δ f H ∘ for H 2 O ( l ) = - 68 . 0 Δ f H ∘ for CO 2 ( g ) = - 94 . 0

The combustion of ethane is given by: C 2 H 6 g + 7 2 O 2 → 2 CO 2 g + 3 H 2 O l ; ∆ H r = - 372 . 0 kcal ∆ H r = ∑ ∆ H products - ∑ ∆ H reactants ∆ H r = [ 2 Δ f H ∘ CO 2 + 3 Δ f H ∘ H 2 O ] - 7 2 Δ f H ∘ O 2 + Δ f H ∘ C 2 H 6 Δ f H ∘ C 2 H 6 = 2 Δ f H ∘ CO 2 + 3 Δ f H ∘ H 2 O - ∆ H r Δ f H ∘ C 2 H 6 = 2 × - 94 . 0 + 3 × - 68 . 0 - - 372 . 0 = - 20 kcal . Similarly, for the combustion of propane C 3 H 8 g + 5 O 2 → 3 CO 2 g + 4 H 2 O l ; ∆ H r = - 530 . 0 kcal ∆ H r = 3 Δ f H ∘ CO 2 + 4 Δ f H ∘ H 2 O - Δ f H ∘ O 2 + Δ f H ∘ C 3 H 8 Δ f H ∘ C 3 H 8 = 3 Δ f H ∘ CO 2 + 4 Δ f H ∘ H 2 O - ∆ H r Δ f H ∘ C 3 H 8 = 3 × - 94 . 0 + 4 × - 68 . 0 - 530 . 0 = - 24 kcal . Given, C Graphite → C g , ∆ H = 172 . 0 kcal … equation 1 2 H g → H 2 g , ∆ H = 104 . 0 kcal … equation 2 The equation for formation of ethane is given by: 2 C g + 3 H 2 g → C 2 H 6 ; ∆ H r = - 20 kcal ∆ H r = ∑ ∆ H products - ∑ ∆ H reactants ∆ H r = Δ f H ∘ C 2 H 6 - [ 2 Δ f H ∘ C g + 3 Δ f H ∘ H 2 g ] - 20 = Δ f H ∘ C 2 H 6 - 2 × - 172 + 3 × - 104 Δ f H ∘ C 2 H 6 = - 676 kcal Similarly, for the formation of propane 3 C g + 4 H 2 g → C 3 H 8 ; ∆ H r = - 24 kcal ∆ H r = ∑ ∆ H products - ∑ ∆ H reactants ∆ H r = Δ f H ∘ C 3 H 8 - 3 Δ f H ∘ C g + 4 Δ f H ∘ H 2 g - 24 = Δ f H ∘ C 3 H 8 - 3 × - 172 + 4 × - 104 Δ f H ∘ C 3 H 8 = - 956 . 0 kcal Let the enthalpy of formation of C - C bond be x and C - H bond be y . Then, for ethane, we have x + 6 y = 676 … i Hence, for propane, we will have 2 x + 8 y = 956 … ii Solving i & ii , we get x = 82 and y = 99 Hence, bond energy of C - C bond is 82 kcal and C - H bond is 99 kcal .

Calculate the resonance energy of N 2 O from the following data: ∆ H f o of N 2 O = 82 kJ mol - 1 , N ≡ N 946 kJ mol - 1 ; N = N , 418 kJ mol - 1 ; O = O , 498 kJ mol - 1 ; N - O , 607 kJ mol - 1

Formation of N 2 O , N 2 g + 1 2 O 2 g → N 2 O g i.e., N ≡ N g + 1 2 O = O g → N = N - O ∆ H f ° calculated = ∑ B . E . ( Reactants ) - ∑ B . E . ( products ) = B . E . N ≡ N + 1 2 B . E . ( O = O ) - B . E . N = N + B . E . ( N - O ) = 946 + 1 2 498 - 418 + 607 kJ mol - 1 = 170 kJ mol - 1 Resonance energy = ∆ f H ° observed - ∆ f H ° calculated = 82 - 170 kJ mol - 1 = - 88 kJ mol - 1

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Important Questions on Thermodynamics

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 5 - Thermodynamics>DPP 5.4>Q 1

Using the data (all values are in kilocalories per mole at $25 ° C$ ) given below, calculate the bond energy of $C - C$ and $C - H$ bonds.

${ΔH}^{\circ}$ combustion (ethane) $= - 372.0$

${ΔH}^{\circ}$ combustion (propane) $= - 530.0$

${ΔH}^{\circ}$ for $C$ (graphite) $\to C (g) = 172.0$

Bond energy of $H - H = 104.0$

$Δ_{f} H^{\circ}$ for $H_{2} O (l) = - 68.0$

$Δ_{f} H^{\circ}$ for ${CO}_{2} (g) = - 94.0$

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 5 - Thermodynamics>DPP 5.4>Q 2

What is the

S - S

bond energy if

{ΔH}_{f} ° (Et - S - Et) = - 147 kJ / mol; {ΔH}_{f} ° (Et - S - S - Et) = - 202 kJ / mol

and

{ΔH}_{f}^{o} S (g) = + 233 kJ / mol

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JEE Main

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 5 - Thermodynamics>DPP 5.4>Q 3

Bond energies of $F_{2}$ and ${Cl}_{2}$ are $153.0$ and $2424 kJ {mol}^{- 1}$ respectively. The heat liberated in the reaction is $111.0 kJ$ Find out the bond energy of $F - Cl$ .

$1 / 2 F_{2} + 1 / 2 {Cl}_{2} \to FCl$

MEDIUM

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 5 - Thermodynamics>DPP 5.4>Q 4

Calculate the resonance energy of

N_{2} O

from the following data:

$∆ H_{f}^{o}$ of $N_{2} O = 82 kJ {mol}^{- 1},$

$N \equiv N 946 kJ {mol}^{- 1};$

$N = N, 418 kJ {mol}^{- 1};$

$O = O, 498 kJ {mol}^{- 1};$

$N - O, 607 kJ {mol}^{- 1}$

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JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 5 - Thermodynamics>DPP 5.4>Q 5

The enthalpy of combustion of

H_{2} (g)

298 K

to give

H_{2} O (g)

- 249 kJ {mol}^{- 1}

and bond enthalpies of

H - H

and

O = O

are

433 kJ {mol}^{-}

and

492 kJ {mol}^{- 1}

, respectively. What will be the bond enthalpy of

O - H

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JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 5 - Thermodynamics>DPP 5.4>Q 6

Calculate

{ΔH}_{r}^{o}

for the reaction

{CH}_{2} {Cl}_{2} (g) \to C (g) + 2 H (g) + 2 Cl (g)

. The average bond enthalpies of

C - H

and

C - Cl

bonds are

414 kJ {mol}^{- 1}

are

330 kJ {mol}^{- 1}

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IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 5 - Thermodynamics>DPP 5.4>Q 7

Calculate the average $O - H$ bond energy in $H_{2} O$ with the help of the following data.

$(1) H_{2} O (l) \to H_{2} O (g); ∆ H_{1} = + 40.6 kJ {mol}^{- 1} (2) 2 H (g) \to H_{2} (g); ∆ H_{2} = - 435.0 kJ {mol}^{- 1} (3) O_{2} (g) \to 2 O (g); ∆ H_{3} = + 489.6 kJ {mol}^{- 1} (4) 2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l); ∆ H_{4} = - 571.6 kJ {mol}^{- 1}$

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JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 5 - Thermodynamics>DPP 5.4>Q 8

Heat of hydrogenation of ethene is

x_{1}

and that of benzene is

x_{2}

. Hence, what will be the resonance energy?

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Important Questions on Thermodynamics

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