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

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, #DeltaH#, the change in entropy, #DeltaS#, and the temperature at which the reaction takes place, #T#

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

At standard state conditions, i.e. at a pressure of #"1 atm"#, you can use the notation #DeltaG^@#, #DeltaH^@#, and #DeltaS^@#.

Now, what does the sign of #DeltaG# tell you?

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

If you break this down by looking at the equation for #DeltaG#, you can say that

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 #T#, which must be expressed in Kelvin!

More specifically, if #T# is high enough, the term #T * DeltaS# will become bigger in magnitude than #DeltaH#, which will cause #DeltaG<0#.

If not, then #DeltaG>0# at that given #T#.

So, plug in your values and solve for #DeltaG# - 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 #"298 K"#.