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

##### 1 Answer

#### 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*, *change in entropy*,

#color(blue)(DeltaG = DeltaH - T * DeltaS)#

At *standard state conditions*, i.e. at a pressure of

Now, what does the sign of

In order for a reaction to be **spontaneous** at a given temperature, you **must** have **non-spontaneous**.

If you break this down by looking at the equation for

This means that you can have

#DeltaH<0# ,#DeltaS>0 -># *spontaneous at any temperature*#DeltaH>0# ,#DeltaS<0 -># *non-spontaneous regardless of temperature*#DeltaH>0# ,#DeltaS>0 -># *spontaneous at a certain temperature range*#DeltaH<0# ,#DeltaS<0 -># *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 **must** be expressed in Kelvin!

More specifically, if **high enough**, the term

If not, then

So, plug in your values and solve for **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")))#

#DeltaG = color(green)(-"105.4 kJ")#

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

So, this particular reaction is *spontaneous* at