﻿ Which of the following statement is correct for the spontaneous absorption of a gas? from Chemistry Thermodynamics Class 12 Nagaland Board

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For the reaction X2O4 (l) -->  2XO2 (g) ΔU = 2.1 kcal, ΔS = 20 cal K-1 at 300 K hence ΔG is

• 2.7 kcal

• -2.7kcal

• 9.3 kcal

• -9.3 kcal

B.

-2.7kcal

The change in Gibbs free energy is given by

ΔG = ΔH-TΔS

Where, ΔH = enthalpy of the reaction

ΔS = entropy of reaction
Thus, in order to determine ΔG, the values of ΔH must be known. The value of ΔH can be calculated by the equation
ΔH = ΔU + ΔngRT
Where (ΔU) = change in internal energy

Δng = (number of moles of gaseous products)-(number of moles of gaseous reactant) = 2-0 =2
R = gas constant = 2 cal
But, ΔH= Δu +  ΔngRT
Δu  =2.1 kcal = 2.1 x 103 cal

[1kcal = 103 cal]
therefore,
ΔH = (2.1 x 103) +(2x2x300) =3300 cal
Hence, ΔG = ΔH-TΔS
ΔG = (3300)-(300 x20)
ΔG =-2700 cal
ΔG =-2.7 Kcal

The change in Gibbs free energy is given by

ΔG = ΔH-TΔS

Where, ΔH = enthalpy of the reaction

ΔS = entropy of reaction
Thus, in order to determine ΔG, the values of ΔH must be known. The value of ΔH can be calculated by the equation
ΔH = ΔU + ΔngRT
Where (ΔU) = change in internal energy

Δng = (number of moles of gaseous products)-(number of moles of gaseous reactant) = 2-0 =2
R = gas constant = 2 cal
But, ΔH= Δu +  ΔngRT
Δu  =2.1 kcal = 2.1 x 103 cal

[1kcal = 103 cal]
therefore,
ΔH = (2.1 x 103) +(2x2x300) =3300 cal
Hence, ΔG = ΔH-TΔS
ΔG = (3300)-(300 x20)
ΔG =-2700 cal
ΔG =-2.7 Kcal

4144 Views

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

• ΔH> 0 and Δ S< 0
• ΔH< 0 and ΔS > 0
• ΔH ΔS >0
• ΔH < 0 and ΔS=0

B.

ΔH< 0 and ΔS > 0

According to the Gibbs-Helmholtz reaction for spontaneity as,

ΔG =ΔH-TΔS

For reaction to being spontaneous, ΔG must be negative.
For this ΔH should be negative ΔS should be positive.
therefore,

ΔHΔS >0.

According to the Gibbs-Helmholtz reaction for spontaneity as,

ΔG =ΔH-TΔS

For reaction to being spontaneous, ΔG must be negative.
For this ΔH should be negative ΔS should be positive.
therefore,

ΔHΔS >0.

1844 Views

The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 dm3 to a volume of 100 dm3 at 27°C is

• 38.3 J mol-1 K-1

• 35.8 J  mol-1 K-1

• 32.3 J  mol-1 K-1

• 42.3 J mol-1 K-1

A.

38.3 J mol-1 K-1  789 Views

# Which of the following statement is correct for the spontaneous absorption of a gas? ΔS is negative and therefore, ΔH should be highly positive ΔS is negative and therefore, ΔH should be highly negative ΔS is positive and therefore, ΔH should be negative ΔS is positive and therefore, ΔH should also be highly positive

B.

ΔS is negative and therefore, ΔH should be highly negative
ΔS [change in entropy] and ΔH [change in enthalpy] are related by the equation

ΔG = ΔH-TΔS
[here, ΔG = change in Gibbs free energy]
For adsorption of a gas, ΔS is negative because randomness decreases. Thus, in order to make ΔG negative [for spontaneous reaction], ΔH must be highly negative. Hence, for the adsorption of gas, if ΔS is negative, therefore,ΔH should be highly negative.
ΔS [change in entropy] and ΔH [change in enthalpy] are related by the equation

ΔG = ΔH-TΔS
[here, ΔG = change in Gibbs free energy]
For adsorption of a gas, ΔS is negative because randomness decreases. Thus, in order to make ΔG negative [for spontaneous reaction], ΔH must be highly negative. Hence, for the adsorption of gas, if ΔS is negative, therefore,ΔH should be highly negative.
1139 Views