# Question #23633

##### 1 Answer

#### Answer:

You need to know the standard change in entropy as well.

#### Explanation:

The standard change in Gibbs free energy can be calculated using the equation

#color(blue)(ul(color(black)(DeltaG^@ = DeltaH^@ - T * DeltaS^@)))#

Here

#DeltaG^@# is the standard change in Gibbs free energy#DeltaH^@# is the standard change in enthalpy#T# is theabsolute temperatureat which the reaction takes place#DeltaS^@# is the standard change in entropy

As you can see, all you have to do here is to plug in the values you have for

#DeltaG^@ = "128.4 kJ mol"^(-1) - "298 K" * DeltaS^@#

Now, it's important to keep track of the **units** you have for your values. The standard change in entropy is expressed in *joules per mole Kelvin*, or

So let's say that you have a standard change in entropy equal to

#DeltaG^@ = "128.4 kJ mol"^(-1) - 298 color(red)(cancel(color(black)("K"))) * xcolor(white)(.)"J mol"^(-1)color(red)(cancel(color(black)("K"^(-1))))#

This is equivalent to

#DeltaG^@ = "128.4 kJ mol"^(-1) - (298 * x)color(white)(.) "J mol"^(-1)#

In order to be able to *subtract* those two values, the units **must** match. So you will have to convert *joules per mole* to *kilojoules per mole* by using the conversion factor

#(1 color(red)(cancel(color(black)("J"))))/"1 mole" * "1 kJ"/(10^3color(red)(cancel(color(black)("J")))) = (10^(-3)color(white)(.)"kJ")/"1 mole" = 10^(-3)"kJ mol"^(-1)#

This means that you will have

#DeltaG^@ = "128.4 kJ mol"^(-1) - (298 * x) * 10^(-3)color(white)(.) "kJ mol"^(-1)#

The answer will thus take the form

#DeltaG^@ = (128.4 - (298 * x)/1000) color(white)(.)"kJ mol"^(-1)#

Where * joules per mole Kelvin*!