Free energy, G=H-TS, is a state function that includes whether a reaction is spontaneous or non-spontaneous. If you think of TS as the part of the system's energy that is disordered already, then (H-TS) is the part of the system's energy that is still ordered and therefore free to cause spontaneous change by becoming disordered.
Also, `DeltaG=DeltaH-TDeltaS`
To see what this equation for free energy change has to do with spontaneity let us return to relationship.
`DeltaS_("total")=DeltaS_("sys")+DeltaS_("surr") = DeltaS + DeltaS_("surr")`
(It is generally understood that symbols without subscript refer to the system not the surroundings.)
`DeltaS_("surr")=-(DeltaH)/T`, where `DeltaH` is the heat gained by then system at constant pressure.
`DeltaS_("total") = DeltaS -(DeltaH)/T`
`rArr TDeltaH_("total")=DeltaH-TDeltaS`
`rArr -TDeltaS_("total") =DeltaH-TDeltaS`
i.e. `DeltaG=-TDeltaS_("total")`
From second law of thermodynamics, a reaction is spontaneous if `DeltaS_("total")` is positive, non-spontanous if `DeltaS_("total")` is negative and at equilibrium if `DeltaS_("total")` is zero.
Since, `-TDeltaS=DeltaG` and since `DeltaG` and `DeltaS` have opposite signs, we can restate the thermodynamic criterion for the spontaneity of a reaction carried out at constant temperature and pressure.
If `DeltaG lt 0`, the reaction is spontaneous.
If `DeltaG gt 0`, the reaction is non-spontanous.
If `DeltaG=0`, the reaction is at equilibrium.
In the equation, `DeltaG=DeltaH-TDeltaS`, temperature is a weighting factor that determine the relative importance of enthalpy contribution to `DeltaG`.
Read the above paragraph carefully and answer the following questions based on above comprehension:
For the spontaneity of a reaction, which statement is true?

Options

A. `DeltaG =+ve, DeltaH=+ve`
B. `DeltaH =+ve, DeltaS=-ve`
C. `DeltaG=-ve, DeltaS=-ve`
D. `DeltaH=-ve, DeltaS=+ve`

Explanation

To determine the spontaneity of a reaction based on the given information about Gibbs free energy (ΔG), we can analyze the conditions under which ΔG is negative, zero, or positive. The relationship we are focusing on is:

\[ \Delta G = \Delta H - T \Delta S \]

Where:
- ΔG = change in Gibbs free energy
- ΔH = change in enthalpy
- T = temperature in Kelvin
- ΔS = change in entropy

### Step-by-Step Solution:

1. **Understanding the Conditions for Spontaneity**:
- A reaction is spontaneous if ΔG < 0.
- A reaction is non-spontaneous if ΔG > 0.
- A reaction is at equilibrium if ΔG = 0.

2. **Analyzing the Given Statements**:
- We need to evaluate the four options provided in the question regarding ΔG and ΔS.

3. **Option A: ΔG Positive and ΔH Positive**:
- If ΔG > 0, the reaction is non-spontaneous. Therefore, this option is incorrect.

4. **Option B: ΔH Positive and ΔS Negative**:
- If ΔH > 0 and ΔS < 0, then ΔG = ΔH - TΔS would likely be positive (since TΔS would also be negative, making ΔG more positive). Thus, this option is also incorrect.

5. **Option C: ΔG Negative and ΔS Negative**:
- If ΔG < 0 but ΔS < 0, we can analyze ΔG = ΔH - TΔS. Here, ΔH could be negative enough to make ΔG negative despite ΔS being negative. However, this situation is less favorable for spontaneity. Therefore, this option is also incorrect.

6. **Option D: ΔH Negative and ΔS Positive**:
- If ΔH < 0 and ΔS > 0, then ΔG = ΔH - TΔS.
- Since ΔH is negative and ΔS is positive, TΔS is also positive. This means ΔG can be negative (spontaneous) because a negative ΔH will outweigh the positive TΔS. Thus, this option is correct.

### Conclusion:
The correct statement for the spontaneity of a reaction is **Option D: ΔH Negative and ΔS Positive**.