# Fora particular reaction, ΔH = 120.5 kJ and ΔS = 758.2 J/K. What is ΔG for this reaction at 298 K?

Dec 3, 2015

$\Delta G = - \text{105.4 kJ}$

#### Explanation:

As you know, the change in Gibbs free energy for a given chemical reaction tells you whether or not that reaction is spontaneous at a given temperature or not.

The change in Gibbs free energy is calculated using the change in enthalpy, $\Delta H$, the change in entropy, $\Delta S$, and the temperature at which the reaction takes place, $T$

$\textcolor{b l u e}{\Delta G = \Delta H - T \cdot \Delta S}$

At standard state conditions, i.e. at a pressure of $\text{1 atm}$, you can use the notation $\Delta {G}^{\circ}$, $\Delta {H}^{\circ}$, and $\Delta {S}^{\circ}$.

Now, what does the sign of $\Delta G$ tell you?

In order for a reaction to be spontaneous at a given temperature, you must have $\Delta G < 0$. When $\Delta G > 0$, the reaction is non-spontaneous.

If you break this down by looking at the equation for $\Delta G$, you can say that

This means that you can have

• $\Delta H < 0$, $\Delta S > 0 \to$ spontaneous at any temperature
• $\Delta H > 0$, $\Delta S < 0 \to$ non-spontaneous regardless of temperature
• $\Delta H > 0$, $\Delta S > 0 \to$ spontaneous at a certain temperature range
• $\Delta H < 0$, $\Delta S < 0 \to$ spontaneous at a certain temperature range

In your case, you have a positive change in enthalpy and a positive change in entropy. This means that the spontaneity of the reaction will depend on the temperature $T$, which must be expressed in Kelvin!

More specifically, if $T$ is high enough, the term $T \cdot \Delta S$ will become bigger in magnitude than $\Delta H$, which will cause $\Delta G < 0$.

If not, then $\Delta G > 0$ at that given $T$.

So, plug in your values and solve for $\Delta G$ - do not forget to convert the change in entropy from joules per Kelvin to kilojoules per Kelvin

DeltaG = "120.5 kJ" - 298 color(red)(cancel(color(black)("K"))) * 758.2 * 10^(-3)"kJ"/color(red)(cancel(color(black)("K")))

$\Delta G = \textcolor{g r e e n}{- \text{105.4 kJ}}$

I'll leave the answer rounded to four sig figs.

So, this particular reaction is spontaneous at $\text{298 K}$.