For the reaction: N 2 O 4 ( g ) → 2 NO 2 ( g ) ∆ H ° = + 57 . 24 kJ , ∆ S ° = 175 . 8 JK - 1 At what temperature the reaction will be spontaneous?

Given reaction, N 2 O 4 g → 2 NO 2 g Also, given, ∆ H ∘ = + 57 . 24 kJ ∆ S ∘ = 175 . 8 Jk – 1 As we know, the relation between standard Gibbs free energy and standard entropy is ∆ G ∘ = ∆ H ∘ - T ∆ S ∘ For spontaneous reactions, ∆ G ∘ is always negative. ∆ H ∘ - T ∆ S ∘ < 0 ∆ H ∘ < T ∆ S ∘ T ∆ S ° > ∆ H ∘ Therefore, T > ∆ H ° ∆ S ° T > 57240 175 . 8 T > 325 . 6 K The reaction will be spontaneous above 325 . 6 K .

Standard enthalpy of combustion Calculate the enthalpy change for the reaction: N 2 g + 3 H 2 g → 2 NH 3 g The bond enthalpies are: Bond N ≡ N H - H N - H ∆ H ° kJ mol - 1 946 435 389

Standard enthalpy of combustion is defined as the enthalpy change which occurs when one mole of a substance is completely burned in oxygen under standard conditions. For example, the burning of hydrogen gas. H 2 g + 1 2 O 2 g → H 2 O l The given reaction, N 2 g + 3 H 2 g → 2 NH 3 g Given, ∆ H ° N ≡ N = 946 kJ mol - 1 ∆ H ° H - H = 435 kJ mol - 1 ∆ H ° N - H = 946 kJ mol - 1 We know that the standard enthalpy of the reaction = ∑ ∆ H reactant - ∑ ∆ H product = ∆ H N ≡ N + 3 ∆ H H - H - 6 ∆ H N - H = 946 + 3 × 435 - 6 × 389 = 946 + 1305 - 2334 = - 83 kJmol - 1 Therefore, the enthalpy change for the given reaction is - 83 kJ mol - 1 .

5 moles of helium expand isothermally and reversibly from a pressure 40 × 10 - 5 Nm - 2 to 4 × 10 - 5 Nm - 2 at 300 K. Calculate the work done, change in internal energy and heat absorbed during the expansion. ( R = 8 . 314 JK - 1 mol - 1 )

We know, W max = - 2 . 303 nRTlog 10 P 1 P 2 On putting the given values, W max = - 2 . 303 × 5 × 8 . 314 × 300 log 10 ( 40 × 10 - 5 ) 4 × 10 - 5 = - 28 . 72 kJ According to first law of thermodynamics, ∆ U = q + W For an isothermal process, ∆ T = 0 Thus, change in internal energy ∆ U = 0 Heat absorbed = q ∆ U = q + W q = - W ⇒ q = 28 . 72

Standard enthalpy of combustion Calculate the enthalpy change for the reaction: N 2 g + 3 H 2 g → 2 NH 3 g The bond enthalpies are: Bond N ≡ N H - H N - H ∆ H ° kJ mol - 1 946 435 389

Standard enthalpy of combustion is defined as the enthalpy change which occurs when one mole of a substance is completely burned in oxygen under standard conditions. For example, the burning of hydrogen gas. H 2 g + 1 2 O 2 g → H 2 O l The given reaction, N 2 g + 3 H 2 g → 2 NH 3 g Given, ∆ H ° N ≡ N = 946 kJ mol - 1 ∆ H ° H - H = 435 kJ mol - 1 ∆ H ° N - H = 946 kJ mol - 1 We know that the standard enthalpy of the reaction = ∑ ∆ H reactant - ∑ ∆ H product = ∆ H N ≡ N + 3 ∆ H H - H - 6 ∆ H N - H = 946 + 3 × 435 - 6 × 389 = 946 + 1305 - 2334 = - 83 kJmol - 1 Therefore, the enthalpy change for the given reaction is - 83 kJ mol - 1 .

Write features of reversible process.

Features of a reversible process: 1. The driving and opposing forces differ from each other, only infinitesimally. 2. During the course of the process, it can be reversed at any point, by making infinitesimal change in conditions. 3. The process takes place in an infinite number of steps. 4. A reversible process results in obtaining maximum work. 5. The process takes place so slowly that, the system is always in temperature-pressure equilibrium with its surroundings.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 3 - Chemical Thermodynamics and Energetics>Maharashtra Board-2019>Q 6

HARD

12th Maharashtra Board

IMPORTANT

Earn 100

Write the conditions for maximum work done by the system.

Important Questions on Chemical Thermodynamics and Energetics

HARD

12th Maharashtra Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 3 - Chemical Thermodynamics and Energetics>Maharashtra Board-2019>Q 7

Calculate the work done in the following reaction at $50 ° C$ . State whether work is done on the system or by the system.

${SO}_{2 (g)} + \frac{1}{2} O_{2 (g)} \to {SO}_{3 (g)}$

HARD

12th Maharashtra Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 3 - Chemical Thermodynamics and Energetics>Maharashtra Board-2019>Q 8

The standard enthalpy of combustion of formaldehyde is

∆_{c} H ° = - 517 kJ

. How much heat will be evolved in the formation of

22 g

of

{CO}_{2}

?

EASY

12th Maharashtra Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 3 - Chemical Thermodynamics and Energetics>Maharashtra Board-2020>Q 9

Calculate the work done during the reactions represented by the following thermochemical equation at

300 K

:

\underset{(g)}{{CH}_{4}} + \underset{(g)}{2 O_{2}} \to \underset{(g)}{{CO}_{2}} + \underset{(l)}{2 H_{2} O}

(R = 8.314 {JK}^{- 1} {mol}^{- 1})

EASY

12th Maharashtra Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 3 - Chemical Thermodynamics and Energetics>Maharashtra Board-2020>Q 10

For the reaction:
$\underset{(g)}{N_{2} O_{4}} \to 2 \underset{(g)}{{NO}_{2}}$

$(∆ H^{°} = + 57.24 kJ, ∆ S^{°} = 175.8 {JK}^{- 1})$
At what temperature the reaction will be spontaneous?

HARD

12th Maharashtra Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 3 - Chemical Thermodynamics and Energetics>Maharashtra Board-2020>Q 11

State and explain Hess’s law of constant heat summation.

EASY

12th Maharashtra Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 3 - Chemical Thermodynamics and Energetics>Maharashtra Board-2020>Q 12

Define:
Reversible process

HARD

12th Maharashtra Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 3 - Chemical Thermodynamics and Energetics>Maharashtra Board-2020>Q 13

Standard enthalpy of combustion
Calculate the enthalpy change for the reaction:
${N_{2}}_{(g)} + 3 H_{2 (g)} \to 2 {NH}_{3 (g)}$
The bond enthalpies are: