Question

Is a positive change in Gibbs free energy $\left(∆G\right)$ spontaneous?

Answer 1

Step 1: Where $\mathbf{∆}\mathit{G}$ is take negative , $\mathbf{∆}\mathit{G}\mathbf{<}\mathbf{0}$

Gibbs free energy:

1. Gibbs Free energy is a system's thermodynamic quantity that represents the workable energy. It is employed to ascertain if a reaction is spontaneous or not. In other words, spontaneous processes are those that happen "naturally," whereas nonspontaneous processes do not.
2. By "naturally," I mean that a reaction will take place in a system even in the absence of a net inflow of free energy from the environment. For instance, ice will melt spontaneously at ${10}^{0}C$and $1atm$, but not at $-{10}^{0}C$ and $1atm$.
3. In this case $∆G$ releases energy , which means that they can proceed without an energy input so the process is spontaneous.

Step 2: Where $\mathbf{∆}\mathit{G}$ is take positive , $\mathbf{∆}\mathit{G}\mathbf{>}\mathbf{0}$

1. In this case $∆G$need an input of energy in order to take place the process is non-spontaneous.

Step 3: Equation for Gibbs energy

1. The following equation states that the change in enthalpy$\left(∆H\right)$ and the change in entropy $\left(∆S\right)$determine the change in Gibbs free energy $\left(∆G\right)$ for a system:

$$\begin{gathered} \Rightarrow \Delta G=\Delta H-T\Delta S \\ \Rightarrow \Delta G^0=\Delta H^0-T\Delta S^0 \end{gathered}$$

1. Where, $∆G=$ change in Gibbs free energy, $∆H=$change in enthalpy, $∆S=$change in entropy
2. Whether the situation is normal or unusual, the relationship still holds true. Regarding whether a reaction will be spontaneous (i.e., when $∆G<0$), there are a few generalizations that we can draw.
3. A negative value for$∆G$ and a spontaneous reaction are both produced by a negative value for $∆H$ and a positive value for$∆S$ And at least one of them (negative $∆H$or positive $∆S$) must be true for a reaction to even stand a chance of being spontaneous.
4. Enthalpy change, the first term in the computation of $∆G$, is frequently the major term in the equation for reactions and circumstances. This is why we frequently assume that the majority of exothermic reactions (negative $∆H$) will occur spontaneously and the majority of endothermic reactions (positive $∆H$) will not, but we cannot be assured of this.
5. $-T∆S$ serves as the second term in the computation of$∆G$.$∆S$is frequently much smaller than $∆H$, which explains why$∆H$ is frequently the dominant factor in the equation. However, this term also includes temperature, and as the temperature rises, this term, and $∆S$ in particular, become more significant.

Step 4: The following can be summed up in terms of when a reaction is spontaneous.

If $∆H$ is Negative and$∆S$ is Positive

1. A reaction will never be spontaneous at any temperature if $∆H$ is positive and $∆S$ is negative. $∆G$ will always be negative, alternatively, you could state that the reverse reaction is spontaneous at all temperatures.

If $∆H$ and $∆S$ are Both Negative

1. When both$∆H$ and$∆S$are negative, the reaction is only spontaneous at "low temperatures" since $∆G$ will only be negative below a specific temperature threshold.

If $∆H$ and $∆S$ are Both Positive

1. When both$∆H$ and $∆S$ are positive, the reaction is only spontaneous at high temperatures because $∆G$ will only become negative above a specific temperature.

Hence , a positive change in Gibbs's free energy $\left(∆G\right)$ is a non-spontaneous process .