With the following informations, determine standard Gibb's free energy of formation of N 2 O 4 ( g ) . 1 2 N 2 ( g ) + 1 2 O 2 ( g ) ⟶ NO ( g ) ΔG ° = 86 . 6 kJ … ( i ) NO ( g ) + 1 2 O 2 ( g ) ⟶ NO 2 ( g ) ΔG ° = - 34 . 82 kJ … ( ii ) 2 NO 2 ( g ) ⟶ N 2 O 4 ( g ) ΔG ° = - 5 . 77 kJ … ( iii )

Given: 1 2 N 2 ( g ) + 1 2 O 2 ( g ) ⟶ NO ( g ) ΔG ° = 86 . 6 kJ … ( i ) NO ( g ) + 1 2 O 2 ( g ) ⟶ NO 2 ( g ) ΔG ° = - 34 . 82 kJ … ( ii ) 2 NO 2 ( g ) ⟶ N 2 O 4 ( g ) ΔG ° = - 5 . 77 kJ … ( iii ) Solution: Formation of any compound is made by elements those are present in natural state. So for doing this: Reaction: 2 × i + 2 × ii + iii = N 2 g + 2 O 2 g → N 2 O 4 g = 2 × 86 . 6 + 2 × - 34 . 82 + - 5 . 77 = 97 . 79 kJ

For the equilibrium: PCl 5 ( g ) ⇌ PCl 3 ( g ) + Cl 2 ( g ) at 298 K , equilibrium constant, K = 1 . 8 × 10 - 7 . What is ΔG 0 for the reaction?

PCl 5 ( g ) ⇌ PCl 3 ( g ) + Cl 2 ( g ) Relation between equilibrium constant and standard gibbs free energy is given by: ∆ G 0 = - 2 . 303 RT log K eq ∆ G 0 = - 2 . 303 × 8 . 314 × 298 log 1 . 8 × 10 - 7 ∆ G 0 = 38 . 484 kJ

State the second law of thermodynamics.

The second law of thermodynamics can be expressed in terms of entropy. It can be stated as “the entropy of an isolated system increases during a spontaneous process”. The total entropy change ( ∆ S total ) for the system and the surroundings of a spontaneous process is given by, ∆ S total = ∆ S system + ∆ S surrounding > 0 When a system is in equilibrium, the entropy is maximum and the change in entropy, ∆ S = 0 . Hence, entropy of a spontaneous process increases till it reaches a maximum and at equilibrium the change in entropy is zero.

The work done in ergs for the reversible expansion of one mole of an ideal gas from a volume of 10 litres to 20 litres at 25 ° C is

298 × 10 7 × 8 . 31 × 2 . 303 log 2 .

2 . 303 × 298 × 0 . 082 log 2 .

2 . 303 × 298 × 0 . 082 log 0 . 5 .

8 . 31 × 10 7 × 298 × 2 . 303 log 0 . 5 .

For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by:

with increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases.

with increase in temperature, the value of K for exothermic reaction decreases because the entropy change of the system is positive.

with increase in temperature, the value of K for endothermic reaction increases because the entropy change of the system is negative.

E M B I B E

Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 96

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IMPORTANT

Earn 100

One mole of liquid iron at its boiling point was vaporized in an oven at $3500 K .$ If iron boils at $3133 K$ and enthalpy of vaporization is $349 K J {mol}^{- 1},$ determine ${ΔS}_{system,} {ΔS}_{surrounding}$ and ${ΔS}_{universe} \cdot$ Oven is considered as surroundings).

Important Questions on Thermodynamics

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 97

Calculate the entropy change in surroundings when

1.00

mol of

H_{2} O (l)

is formed under standard conditions at

298 K .

Given

Δ_{r} H ° = - 286 kJ {mol}^{- 1} .

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 98

Order of increasing of entropy among given condition of substance is:

$(I)$ $1$ mole of $H_{2} O (l)$ at $298 K$ and $0.101 MPa$

$(II)$ $1$ mole of ice at $273 K$ and $0.101 MPa$

$(III)$ $1$ mole of $H_{2} (g)$ at $298 K$ and $1 atm$

$(IV)$ $1$ mole of $C_{2} H_{6} (g)$ at $298 K$ and $1 atm$

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 99

Oxygen and ozone are gases at standard temperature. Their molar entropies are in the sequence

O_{2} < O_{3} .

Using molecular properties, explain why ozone is more disordered than oxygen.

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 100

What will be the states of a chemical reaction occurring at constant pressure and temperature conditions when:

$(i) ΔG = 0$

$(ii) ΔG > 0$

$(iii) ΔG < 0$

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 101

A gaseous reactant $A$ forms two different product, in parallel reaction, $B$ and $C$ as follows:

$A ⟶ B; ΔH ° = - 3 kJ, ΔS ° = 20 {JK}^{- 1} A ⟶ C; ΔH ° = - 3.6 kJ, ΔS ° = 10 J K^{- 1}$

Discuss the relative stability of $B$ and $C$ on the basis of Gibb's free energy change at $27 ° C .$

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 102

With the following informations, determine standard Gibb's free energy of formation of $N_{2} O_{4} (g) .$

$\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ NO (g) ΔG ° = 86.6 kJ \dots (i)$

$NO (g) + \frac{1}{2} O_{2} (g) ⟶ {NO}_{2} (g) ΔG ° = - 34.82 kJ \dots (ii)$

$2 {NO}_{2} (g) ⟶ N_{2} O_{4} (g) ΔG ° = - 5.77 kJ \dots (iii)$

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 103

For the equilibrium:

{PCl}_{5} (g) ⇌ {PCl}_{3} (g) + {Cl}_{2} (g)

at

298 K,

equilibrium constant,

K = 1.8 \times 10^{- 7} .

What is

{ΔG}^{0}

for the reaction?

EASY

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 104

In which state, matter have highest entropy:

Learn and Prepare for any exam you want !

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One mole of liquid iron at its boiling point was vaporized in an oven at 3500 K. If iron boils at 3133 K and enthalpy of vaporization is 349 K J mol-1 , determine ΔSsystem, ΔSsurrounding and ΔSuniverse · Oven is considered as surroundings).

Important Questions on Thermodynamics

Learn and Prepare for any exam you want !

One mole of liquid iron at its boiling point was vaporized in an oven at $3500 K .$ If iron boils at $3133 K$ and enthalpy of vaporization is $349 K J {mol}^{- 1},$ determine ${ΔS}_{system,} {ΔS}_{surrounding}$ and ${ΔS}_{universe} \cdot$ Oven is considered as surroundings).