CBSE
Gujarat Board
Haryana Board
Class 10
Class 12
Which of the following statement is correct for the spontaneous absorption of a gas?
B.
ΔS is negative and therefore, ΔH should be highly negativeThe correct thermodynamic conditions for the spontaneous reaction at all temperatures is
B.
ΔH< 0 and ΔS > 0According 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.
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 dm^{3} 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}
For the reaction X_{2}O_{4} (l) --> 2XO_{2} (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 + Δn_{g}RT
Where (ΔU) = change in internal energy
Δn_{g} = (number of moles of gaseous products)-(number of moles of gaseous reactant) = 2-0 =2
R = gas constant = 2 cal
But, ΔH= Δu + Δn_{g}RT
Δu =2.1 kcal = 2.1 x 10^{3} cal
[1kcal = 10^{3} cal]
therefore,
ΔH = (2.1 x 10^{3}) +(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 + Δn_{g}RT
Where (ΔU) = change in internal energy
Δn_{g} = (number of moles of gaseous products)-(number of moles of gaseous reactant) = 2-0 =2
R = gas constant = 2 cal
But, ΔH= Δu + Δn_{g}RT
Δu =2.1 kcal = 2.1 x 10^{3} cal
[1kcal = 10^{3} cal]
therefore,
ΔH = (2.1 x 10^{3}) +(2x2x300) =3300 cal
Hence, ΔG = ΔH-TΔS
ΔG = (3300)-(300 x20)
ΔG =-2700 cal
ΔG =-2.7 Kcal