Question

What is the change in Gibbs' free energy for `"C"(s) + "O"_2(g) -> "CO"_2(g) + "393.51 kJ/mol"`?

The `∆S^@` for the reaction,

`"C"(s) + "O"_2(g) -> "CO"_2(g) + "393.51 kJ/mol"`,

at `"308.62 K"` is `"0.004 kJ/mol"cdot"K"`.

Find the value of `∆G^@` at `"308.62 K"`.

What would be the answer in `"kJ/mol"`? Thanks!

Answer 1

Follow the steps below

Explanation:

1. Look at the reaction and determine the `DeltaH^@`
   Since the reaction is exothermic: `DeltaH^@= -393.51 ("kJ")/("mol")`

2. Given `DeltaS^@ = 0.004 ("kJ")/("mol"*"K")`

3. Given `T= "308.62 K"`

4. Use the formula: `DeltaG^@=DeltaH^@-TDeltaS^@`

5. Plug in givens: `DeltaG^@=-393.51 ("kJ")/("mol")-(308.62 cancel("K"))*(0.004 ("kJ")/("mol"*cancel("K")))`

`= -394.74 ("kJ")/("mol")`

Answer 2

`-394.74` `" kJ"/"mol"`

Explanation:

To solve this question, use the standard-state free energy of reaction equation:

`Delta"G"_0=DeltaH_0 - TDeltaS_0`

Given that you were given all the values for enthalpy, temperature, and entropy:

`Delta"G"_0=(-393.51 " kJ"/"mol") - (308.62" K")(0.004 "kJ"/("mol" *"K"))`

`Delta"G"_0= -394.74` `" kJ"/"mol"`

Since the value Delta `"G"_0` is negative, this reaction is thermodynamically favorable.