Which of the following statements is correct for a reversible process in a state of equilibrium?
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
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
Which of the following statement is correct for the spontaneous absorption of a gas?
B.
ΔS is negative and therefore, ΔH should be highly negative ΔS [change in entropy] and ΔH [change in enthalpy] are related by the equationThe 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.
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