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 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.


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Which of the following statements is correct for a reversible process in a state of equilibrium?

  • ΔG = -2.303RT log K
  • ΔG = 2.303RT log K

  • ΔG0 = -2.303RT log K

  • ΔG0 = -2.303RT log K


C.

ΔG0 = -2.303RT log K

 ΔG =ΔG0+2.303RT log K log Q
At equilibrium when Δ G = 0 and Q = K
 then ΔG = ΔG0 +2.303 RT log K = 0

ΔG0 = -2.303 RT log K

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The difference between ΔH and ΔU ( ΔH- ΔU), when the combustion of one mole heptane(I) is carried out a temperature T, is equal to:

  • -3RT

  • -4RT

  • 3RT

  • 4R


B.

-4RT

C7H16(l) + 11O2 → 7CO2 + 8H2O

Δngas = 7-11

ΔH = ΔU + ΔnRT

ΔH- ΔV = -4RT


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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 be negative


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. 
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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

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