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:
A particular reaction has a negative value for the free energy change. Then at ordinary temperature.

Options

A. It has a large `-ve` value for the entropy change
B. It hs large `+ve` value for entropy change
C. It has a small `+ve` for enthalpy change
D. It has a `+ve` value for the entropy change and a `-ve` value for enthalpy change

Explanation

To solve the problem regarding the Gibbs free energy change (ΔG) for a particular reaction that has a negative value, we will analyze the implications of this negative value on the enthalpy (ΔH) and entropy (ΔS) changes of the system.

### Step-by-Step Solution:

1. **Understanding the Gibbs Free Energy Equation**:
The Gibbs free energy change is given by the equation:
\[ \Delta G = \Delta H - T \Delta S \]
where:
- ΔG = change in Gibbs free energy
- ΔH = change in enthalpy
- T = temperature in Kelvin
- ΔS = change in entropy

2. **Given Condition**:
We know that ΔG is negative for the reaction at ordinary temperature (approximately 298 K).

3. **Analyzing the Equation**:
For ΔG to be negative:
\[ \Delta H - T \Delta S < 0 \]
This can be rearranged to:
\[ \Delta H < T \Delta S \]

4. **Considering the Signs of ΔH and ΔS**:
- If ΔH is negative (exothermic reaction), it contributes negatively to ΔG.
- If ΔS is positive (increase in disorder), it contributes positively to the term \(T \Delta S\).

5. **Combining the Effects**:
- A negative ΔH (exothermic) and a positive ΔS (increase in entropy) will ensure that ΔG remains negative.
- Conversely, if ΔH is positive (endothermic) and ΔS is negative (decrease in entropy), ΔG would be positive, making the reaction non-spontaneous.

6. **Conclusion**:
Therefore, for the reaction to have a negative Gibbs free energy change (ΔG < 0), it must have:
- **ΔH < 0** (negative value for enthalpy change)
- **ΔS > 0** (positive value for entropy change)

### Final Answer:
The correct statement is that the reaction has a negative value for the enthalpy change (ΔH < 0) and a positive value for the entropy change (ΔS > 0).