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.

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

Hess's law of constant heat summation states that the change in enthalpy for a reaction is the same whether the reaction takes place in one or a series of steps. It can also be stated as the enthalpy change for a chemical reaction is the same regardless of the path by which the reaction occurs. For example, consider following two paths for the preparation of methylene chloride Path I : CH 4 ( g ) + 2 Cl 2 ( g ) → CH 2 Cl 2 ( g ) + 2 HCl ( g ) ΔH 0 1 = − 202 . 3 kJ Path II : CH 4 ( g ) + Cl 2 ( g ) → CH 3 Cl ( g ) + HCl ( g ) ΔH 2 0 = − 98 . 3 kJ CH 3 Cl ( g ) + Cl 2 ( g ) → CH 2 Cl 2 ( g ) + HCl ( g ) ΔH 3 0 = − 104 . 0 kJ Adding two steps CH 4 ( g ) + 2 Cl 2 ( g ) → CH 2 Cl 2 ( g ) + 2 HCl ( g ) ΔH 0 = − 202 . 3 kJ Thus, whether we follow path I or path II, the enthalpy change of the reaction is same. ΔH 1 0 = ΔH 2 0 + ΔH 3 0 = − 202 . 3 kJ

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 .

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-2020>Q 12

EASY

12th Maharashtra Board

IMPORTANT

Earn 100

Define:
Reversible process

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 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: