Question

Question #23633

Answer 1

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 the absolute temperature at 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 `DeltaH^@`, `T`, and `DeltaS^@`. Since you're missing the value for the standard change in entropy, I'll just express the answer in terms of `DeltaS^@`

`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 `"J mol"^(-1)"K"^(-1)`.

So let's say that you have a standard change in entropy equal to `x` `"J mol"^(-1)"K"^(-1)`. Plug this into the equation to find

`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 `x` is the value of the standard change in entropy expressed in joules per mole Kelvin!