CBSE
Gujarat Board
Haryana Board
Class 10
Class 12
The correct thermodynamic conditions for Â the spontaneous reaction at all temperatures is
B.
Î”H< 0 andÂ Î”S > 0According to the Gibbs-HelmholtzÂ reaction for spontaneity as,
Î”G =Î”H-TÎ”S
For reaction to being spontaneous,Â Î”G must be negative.
For thisÂ Î”H should be negativeÂ Î”S should be positive.
therefore,
Î”HÎ”S >0.
According to the Gibbs-HelmholtzÂ reaction for spontaneity as,
Î”G =Î”H-TÎ”S
For reaction to being spontaneous,Â Î”G must be negative.
For thisÂ Î”H should be negativeÂ Î”S should be positive.
therefore,
Î”HÎ”S >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 27Â°C is
38.3 J mol^{-1} K^{-1}
35.8 J Â mol^{-1}Â K^{-1}
32.3 J Â mol^{-1}Â K^{-1}
42.3 JÂ mol^{-1}Â K^{-1}
A.
38.3 J mol^{-1} K^{-1}
Which of the following statement is correct for the spontaneous absorption of a gas?
B.
Î”S is negative and therefore,Â Î”H should be highly negativeFor the reaction X_{2}O_{4} (l) --> Â 2XO_{2} (g)Â Î”U = 2.1 kcal,Â Î”S = 20 cal K^{-1} at 300 K henceÂ Î”G is
2.7 kcal
-2.7kcal
9.3 kcalÂ
-9.3 kcal
B.
-2.7kcal
The change in Gibbs free energy is given by
Î”G =Â Î”H-TÎ”S
Where,Â Î”H = enthalpy of the reaction
Î”S = entropy of reaction
Thus, in order to determineÂ Î”G, the values ofÂ Î”H must be known. The value ofÂ Î”H can be calculated by the equation
Î”H =Â Î”U +Â Î”n_{g}RT
Where (Î”U) = change in internal energy
Î”n_{g} = (number of moles of gaseous products)-(number of moles of gaseous reactant) = 2-0 =2
R = gas constant = 2 cal
But,Â Î”H=Â Î”u +Â Â Î”n_{g}RTÂ
Î”u Â =2.1 kcal = 2.1 x 10^{3} cal
[1kcal = 10^{3} cal]
therefore,
Î”H = (2.1 x 10^{3}) +(2x2x300) =3300 cal
Hence,Â Î”G =Â Î”H-TÎ”S
Â Î”G = (3300)-(300 x20)
Â Î”G =-2700 cal
Â Î”G =-2.7 Kcal
The change in Gibbs free energy is given by
Î”G =Â Î”H-TÎ”S
Where,Â Î”H = enthalpy of the reaction
Î”S = entropy of reaction
Thus, in order to determineÂ Î”G, the values ofÂ Î”H must be known. The value ofÂ Î”H can be calculated by the equation
Î”H =Â Î”U +Â Î”n_{g}RT
Where (Î”U) = change in internal energy
Î”n_{g} = (number of moles of gaseous products)-(number of moles of gaseous reactant) = 2-0 =2
R = gas constant = 2 cal
But,Â Î”H=Â Î”u +Â Â Î”n_{g}RTÂ
Î”u Â =2.1 kcal = 2.1 x 10^{3} cal
[1kcal = 10^{3} cal]
therefore,
Î”H = (2.1 x 10^{3}) +(2x2x300) =3300 cal
Hence,Â Î”G =Â Î”H-TÎ”S
Â Î”G = (3300)-(300 x20)
Â Î”G =-2700 cal
Â Î”G =-2.7 Kcal