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Thermodynamics

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Chemistry Part I

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Chemistry

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Class 10 Class 12

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

increment straight S space equals space nR space In space straight V subscript 2 over straight V subscript 1
equals space 2.303 space nR space log space straight V subscript 2 over straight V subscript 1
space equals space 2.303 space straight x space 2 space straight x space 8.314 space straight x space log space 100 over 10
equals space 38.3 space straight J space mol to the power of negative 1 end exponent straight k to the power of negative 1 end exponent
increment straight S space equals space nR space In space straight V subscript 2 over straight V subscript 1
equals space 2.303 space nR space log space straight V subscript 2 over straight V subscript 1
space equals space 2.303 space straight x space 2 space straight x space 8.314 space straight x space log space 100 over 10
equals space 38.3 space straight J space mol to the power of negative 1 end exponent straight k to the power of negative 1 end exponent
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

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