The following reaction is performed at 298 K . 2 N O g + O 2 g ⇌ 2 N O 2 g The standard free energy of the formation of N O g is 86 . 6 kJ mol - 1 at 298 K . What is the standard free energy of the formation of N O 2 g at 298 K ? ( K P = 1.6 × 10 12 )

0.5 [ 2 × 86,600 - R 298 ln ⁡ ( 1.6 × 10 12 ) ]

R 298 ln ⁡ 1.6 × 10 12 -86,600

86,600 + R 298 ln ⁡ ( 1.6 × 10 12 )

86,600 - ln ⁡ ( 1.6 × 10 12 ) R ( 298 )

MEDIUM

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The correct relationship between free energy and equilibrium constant K of a reaction is:

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Important Questions on Thermodynamics

HARD

Chemistry>Physical Chemistry>Thermodynamics>Gibbs Energy Change and Equilibrium

The following reaction is performed at

298 K

2 N O (g) + O_{2} (g) ⇌ 2 N O_{2} (g)

The standard free energy of the formation of

N O (g)

86.6 kJ {mol}^{- 1}

298 K

. What is the standard free energy of the formation of

N O_{2} (g)

298 K

(K_{P} = 1.6 \times 10^{12})

EASY

Chemistry>Physical Chemistry>Thermodynamics>Gibbs Energy Change and Equilibrium

The correct thermodynamic conditions for the spontaneous reaction at all temperatures is:

MEDIUM

Chemistry>Physical Chemistry>Thermodynamics>Gibbs Energy Change and Equilibrium

For the reaction;

A (l) \to 2 B (g)

Δ U = 2.1 k c a l, Δ S = 20 c a l K^{- 1}

300 K .

Hence

Δ G

k c a l

is _____________.

MEDIUM

Chemistry>Physical Chemistry>Thermodynamics>Gibbs Energy Change and Equilibrium

According to the following diagram, A reduces

B O_{2}

when the temperature is:
Question Image

EASY

Chemistry>Physical Chemistry>Thermodynamics>Gibbs Energy Change and Equilibrium

For a given reaction,

Δ H = 35.5 k J m o l^{- 1}

and

Δ S = 83.6 J K^{- 1} m o l^{- 1} .

The reaction is spontaneous at : (Assume that

Δ H

and

Δ S

do not vary with temperature)

MEDIUM

Chemistry>Physical Chemistry>Thermodynamics>Gibbs Energy Change and Equilibrium

Using the Gibbs change,

∆ G^{o} = + 63.3 k J,

for the following reaction,

A g_{2} C O_{3} (g) ⇌ 2 A g^{+} (a q) + C O_{3}^{2 -} (a q)

the

K_{s p}

A g_{2} C O_{3} (s)

in water at

25^{o} C

(R = 8.314 J K^{- 1} m o l^{- 1})

HARD

Chemistry>Physical Chemistry>Thermodynamics>Gibbs Energy Change and Equilibrium

The standard state Gibb's free energies of formation of

C

(graphite) and

C

(diamond) at

T = 298 K

are

Δ_{f} G^{o} [C (graphite] = 0 kJ {mol}^{- 1}

Δ_{f} G^{o} [C (diamond)] = 2 .9 kJ {mol}^{- 1}

The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [

C

(graphite)] to diamond [

C

(diamond)] reduces its volume by

2 \times 10^{- 6} m^{3} {mol}^{- 1}

. If

C

(graphite) is converted to

C

(diamond) isothermally at

T = 298 K

, the pressure at which

C

(graphite) is in equilibrium with

C

(diamond), is

[Useful information:

1 J = 1 kg m^{2} s^{- 2}, 1 Pa = 1 kg m^{- 1} s^{- 2}; 1 bar = 10^{5} Pa

]

MEDIUM

Chemistry>Physical Chemistry>Thermodynamics>Gibbs Energy Change and Equilibrium

When the heat of a reaction at constant pressure is

- 25 \times 10^{- 3} cal

and entropy change for the reaction is

7.4 cal {deg}^{- 1}

, it is predicted that the reaction at

25^{0} C

MEDIUM

Chemistry>Physical Chemistry>Thermodynamics>Gibbs Energy Change and Equilibrium

A process has

Δ H = 200 J m o l^{- 1}

and

Δ S = 40 J K^{- 1} m o l^{- 1} .

Out of the values given below choose the minimum temperature above which the process will be spontaneous:

MEDIUM

Chemistry>Physical Chemistry>Thermodynamics>Gibbs Energy Change and Equilibrium

For the reaction,

A (g) + B (g) \to C (g) + D (g), ∆ H^{o}

and

∆ S^{o}

are, respectively,

- 29.8 k J m o l^{- 1}

and

- 0.100 k J K^{- 1} m o l^{- 1}

at 298 K. The equilibrium constant for the reaction at 298 k is:

MEDIUM

Chemistry>Physical Chemistry>Thermodynamics>Gibbs Energy Change and Equilibrium

For a dimerization reaction,

$2 A (g) \to A_{2} (g)$

at $298 K, {ΔU}^{-} = - 20 {kJmol}^{- 1}, {ΔS}^{-} = - 30 {JK}^{- 1} {mol}^{- 1}$ ,then the $Δ G^{-}$ will be ….. $J$ .

Give an answer to the nearest integer value.

HARD

Chemistry>Physical Chemistry>Thermodynamics>Gibbs Energy Change and Equilibrium

The surface of copper gets tarnished by the formation of copper oxide. $N_{2}$ gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, the $N_{2}$ gas contains 1 mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below:

$2 Cu (s) + H_{2} O (g) \to {Cu}_{2} O (s) + H_{2} (g)$

$P_{H_{2}}$ Is the minimum partial pressure of $H_{2}$ (in bar) needed to prevent the oxidation at $1250 K$ . The magnitude of value of $\ln (P_{H_{2}})$ is ____.

(Given: total pressure = 1 bar, R (universal gas constant) $= 8 J K^{- 1} {mol}^{- 1}, \ln (10) = 2 .3 . Cu (s) and {Cu}_{2} O (s)$ are mutually immiscible.

At $1250 K : 2 Cu (s) + \frac{1}{2} O_{2} (g) \to {Cu}_{2} O (s); ∆ G^{⊝} = - 78,000 J {mol}^{- 1}$

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g); ∆ G^{⊝} = - 1,78,000 J {mol}^{- 1};$ )

Round off the answer up to the nearest integer.

MEDIUM

Chemistry>Physical Chemistry>Thermodynamics>Gibbs Energy Change and Equilibrium

If for a certain reaction

Δ_{r} H

30 {kJmol}^{- 1}

450 K,

the value of

Δ_{r} S

(in

{JK}^{- 1} {mol}^{- 1}

) for which the same reaction will be spontaneous at the same temperature is

EASY

Chemistry>Physical Chemistry>Thermodynamics>Gibbs Energy Change and Equilibrium

A process will be spontaneous at all temperatures if:

EASY

Chemistry>Physical Chemistry>Thermodynamics>Gibbs Energy Change and Equilibrium

For the reaction,

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l), ΔH = - 285.8 kJ {mol}^{- 1}$

$ΔS = - 0.163 kJ {mol}^{- 1} K^{- 1}$ .

What is the value of free energy change, at $27 ° C$ for the reaction?

EASY

Chemistry>Physical Chemistry>Thermodynamics>Gibbs Energy Change and Equilibrium

For a reaction to be spontaneous at all the temperatures, what are the required thermodynamic quantities?

MEDIUM

Chemistry>Physical Chemistry>Thermodynamics>Gibbs Energy Change and Equilibrium

For the reaction.

2 H_{2} (g) + O_{2} {(g)}_{} ⟶ 2 H_{2} O {(g)}_{}, at 300 K, ΔG

and

ΔH

of water are

- 228.4 kJ . {mol}^{- 1}

and

- 241.60 kJ . {mol}^{- 1}

, respectively. Then calculate the value of change in entropy for the given reaction.

EASY

Chemistry>Physical Chemistry>Thermodynamics>Gibbs Energy Change and Equilibrium

A reaction at 1 bar is non-spontaneous at low temperature but becomes spontaneous at high temperature. Identify the correct statement about the reaction among the following:

EASY

Chemistry>Physical Chemistry>Thermodynamics>Gibbs Energy Change and Equilibrium

Which of the following can help predict the rate of a reaction if the standard Gibbs free energy of reaction

(Δ G^{0})

is known?

HARD

Chemistry>Physical Chemistry>Thermodynamics>Gibbs Energy Change and Equilibrium

The increase of pressure on

i c e ⇌ w a t e r

system at constant temperature will lead to:

The correct relationship between free energy and equilibrium constant K of a reaction is:

Important Questions on Thermodynamics

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