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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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

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

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

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

Thermodynamics