Which of the following conditions will most enhance the spontaneity of a reaction?
\(\Delta\)H is negative and greater than T\(\Delta\)S
\(\Delta\)H is negative and \(\Delta\)S = 0
\(\Delta\)H is positive and less than T\(\Delta\)S
\(\Delta\)H is positive and equal to T\(\Delta\)S
Explanation
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Discussions (15)
option A,B,C are correct in they own right but since the question said in what case will the highest value of ∆G be obtained, then option A is the most correct.
let us assume that (∆H = 10 and T∆S = 5)
Option A
∆G = ∆H - T∆S
∆G = -10 -(5)
∆G = -10 - 5 = -15
option B
∆G = ∆H - T∆S
∆G = -10 - 0 = - 10
option C
∆G = ∆H - T∆S
∆G = 5 - (10) = -5
option D is totally Off
so from my explanation, you will see that ∆G has the highest negative value when ∆H is negative and greater than T∆S
the correct answer is C
For the reaction to be spontaneous, gibb's free energy (∆G) MUST be negative
The only condition to make ∆G negative in the above question is if ∆H is positive and greater than T∆S
i.e ∆G= ∆H-T∆S
ΔH is negative and greater than TΔS
In this case, both ΔH and TΔS are negative, but ΔH is greater in magnitude than TΔS. When ΔH is negative and greater than TΔS, the negative contribution from ΔH dominates, leading to a negative ΔG (since ΔG=ΔH−TΔS). A negative ΔG indicates that the reaction is spontaneous. Thus, option A is the most favorable condition for enhancing spontaneity.
