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 .

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

Write the conditions for maximum work done by the system.

The maximum work done is the reversible isothermal expansion of an ideal gas. It can be calculated by the following formula, W = - 2 . 303 nRT log P 1 P 2 There are the following conditions for maximum work done by the system, 1 All the changes that are taking place in the system are thermodynamically reversible. 2 All the changes in the system take place in an infinite number of steps. 3 The driving force of the system during the change is infinitesimally greater than the opposing force. 4 The system is present in mechanical equilibrium with its surroundings.

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 ?

The combustion of formaldehyde takes place by the following reaction, HCHO + O 2 → CO 2 44 g + H 2 O Since heat evolves in the formation of 44 g of CO 2 = - 517 kJ Therefore, heat will be evolved in the formation of 22 g of CO 2 = - 517 44 × 22 kJ = - 258 . 5 kJ

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 + 1 2 O 2 g → SO 3 g

The given reaction, SO 2 g + 1 2 O 2 g → SO 3 g We know, work done can be calculated by using the formula, W = - ∆ n g RT ⋯ ⋯ ⋯ ⋯ ⋯ ( i ) Where, ∆ n g = Number of moles of gaseous products - Number of moles of gaseous reactants ∆ n g = 1 - 1 + 1 2 = - 1 2 R = Universal gas constant R = 8 . 314 J mol - 1 K - 1 T = Temperature T = 50 ° C = ( 50 + 273 ) K = 323 K Put all the values in equation i , W = - - 1 2 mol × 8 . 314 J mol - 1 K - 1 × 323 K W = + 1342 . 71 J W = + 1 . 342 KJ From the above, the calculated work done is positive then work is done on the system.

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 .

E M B I B E

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

EASY

12th Maharashtra Board

IMPORTANT

Earn 100

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?

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-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: