Compound ΔG∘f (kJ/mol) A +387.7 B +550.7 C +402.0 Use the data given here to calculate the values of ΔG∘rxn at 25 ∘ C for the reaction described by the equation A+B↽−−⇀C If ΔH∘rxn and ΔS∘rxn are both negative values, what drives the spontaneous reaction and in what direction at standard conditions? ΔG∘rxn= kJ The spontaneous reaction is enthalpy-driven to the left. enthalpy-driven to the right. entropy-driven to the right. entropy-driven to the left.
(A) ΔH is positive and ΔS is negative → (2) Spontaneous in reverse at all temperatures
(B) ΔH is positive and ΔS is positive → (3) Spontaneous as written above a certain temperature
(C) ΔH is negative and ΔS is positive → (1) Spontaneous as written at all temperatures
(D) ΔH negative and ΔS is negative → (4) Spontaneous as written below a certain temperature
According to Gibb's equation:
= Gibbs free energy
= enthalpy change
= entropy change
T = temperature in Kelvin
As we know that:
= +ve, reaction is non spontaneous
= -ve, reaction is spontaneous
= 0, reaction is in equilibrium
(A) ΔH is positive and ΔS is negative.
The reaction is non-spontaneous at all temperatures or spontaneous in reverse at all temperatures.
(B) ΔH is positive and ΔS is positive.
(at low temperature) (non-spontaneous)
(at high temperature) (spontaneous)
The reaction is spontaneous as written above a certain temperature.
(C) ΔH is negative and ΔS is positive.
The reaction is spontaneous as written at all temperatures
(D) ΔH is negative and ΔS is negative.
(at high temperature) (non-spontaneous)
(at low temperature) (spontaneous)
The reaction is spontaneous as written below a certain temperature.
The basis to solve this question is the Gibbs Free Energy relation between the change in enthalpy , ΔH, and the change in entropy ΔS, through the relation
ΔG = ΔH - TΔS ( T is temperature )
For a given reaction ΔG must be negative for the reaction to be positive.
So to solve our question we have to determine the sign for the change in free energy studying whether the enthalpy is positive or positive, and doing the same with the term TΔS.
A. ΔH positive, ΔS negative ⇒ ΔG : always positive non spontaneous.
ΔH positive, ΔS positive ⇒ If TΔS > ΔH, ΔG is negative spontaneous
If TΔS < ΔH, ΔG is positive nonspontaneous
C. ΔH negative, ΔS positive ⇒ ΔG: always is negative spontaneous
ΔH negative, ΔS negative ⇒ If TΔS < ΔH, ΔG is negative spontaneous
If TΔS > ΔH, ΔG is positive nonspontaneous
1. Spontaneous as written at all temperatures: C
2. Spontaneous in reverse at all temperatures: A
3. Spontaneous as written above a certain temperature : B
4. Spontaneous as written below a certain temperature : D
The release of free energy drives the spontaneous reaction.
Spontaneity can be determined using the change in Gibbs free energy (the thermodynamic potencial):
delta G=delta H – T*delta S
where delta H is the enthalpy and delta S is the entropy.
The direction (the sign) of delta G depends of the changes of enthalpy and entropy. If delta G is negative then the process is spontaneous.
In our case, both delta H and delta S are negative values, the process as said is spontaneous which means that it may proceed in the forward direction.
ΔG=ΔH-TΔS, where ΔH and ΔS is the enthalpy and entropy of the reaction, and T is the temperature at which the reaction happens.
So if ΔH and ΔS are both positive, in order to make ΔG negative, T must be very high. In this way, TΔS would be larger than ΔH.
Therefore, a reaction with a positive ΔH and ΔS will be spontaneous at high temperatures.
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