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

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

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 .

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

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 .

Define: Reversible process

A thermodynamic process is called a reversible process if the process can revert the system and surroundings to their original state. If a system undergoes a change from state 1 to state 2 , then it would be called a reversible process if it can reverse itself completely from state 2 to state 1 , with no changes anywhere else in the universe. Reversible processes are very slow in nature. Reversible processes are idealized notion, they do not exist in reality. A quasi-static isothermal expansion of an ideal gas in a cylinder fitted with a frictionless piston is an example of a reversible process.