Which of the given conditions leads to a negative free energy value (spontaneous reaction)?
1. Highly positive enthalpy and highly negative entropy
2. Slightly negative enthalpy and highly negative entropy
3. Highly positive enthalpy and slightly positive entropy
4. Highly negative enthalpy and highly positive entropy
thakns dr strash
To determine which condition leads to a negative free energy value (spontaneous reaction), we need to consider the relationship between enthalpy (ΔH) and entropy (ΔS) in the Gibbs free energy equation:
ΔG = ΔH - TΔS
where ΔG is the change in free energy, ΔH is the change in enthalpy, ΔS is the change in entropy, and T is the temperature in Kelvin.
For a reaction to be spontaneous (ΔG < 0), we need the term -TΔS to dominate over ΔH, so that the overall value of ΔG is negative.
Now let's analyze each of the given conditions:
1. Highly positive enthalpy and highly negative entropy:
In this case, the ΔH term is positive and the -TΔS term is also negative. If ΔH dominates over -TΔS, ΔG would be positive, indicating a non-spontaneous reaction. Therefore, this condition does not lead to a negative ΔG value.
2. Slightly negative enthalpy and highly negative entropy:
Here, the ΔH term is slightly negative and the -TΔS term is still negative. If ΔH dominates over -TΔS, ΔG would be positive, indicating a non-spontaneous reaction. Thus, this condition also does not lead to a negative ΔG.
3. Highly positive enthalpy and slightly positive entropy:
In this case, the ΔH term is positive and the ΔS term is slightly positive. Although both terms are of the same sign, they are both positive. As a result, ΔH would still dominate over ΔS, causing ΔG to be positive. Hence, this condition does not lead to a negative ΔG.
4. Highly negative enthalpy and highly positive entropy:
For this condition, the ΔH term is highly negative, and the ΔS term is highly positive. As both terms have the same sign, but ΔS is dominating in magnitude, the -TΔS term would outweigh ΔH, resulting in a negative ΔG value. Therefore, this condition leads to a negative ΔG, indicating a spontaneous reaction.
So, option 4, "Highly negative enthalpy and highly positive entropy," corresponds to the condition that leads to a negative free energy value and a spontaneous reaction.
Use dGo = dHo - TdSo, then plug in each scenario listed as answers and see which is more likely to give a negative dGo. Negative dGo means a spontaneous reaction. Here is how I would do answer 1.
dG = dH - TdS
........+++ - (---) so that give + value for dH and a + value for TdS which gives a + value for dG which means non-spontaneity. So answer 1 can't be right. Do the same for answers 2,3,4 and decide which is more likely to give a - value for dG.