Question

For the following reaction run at 800 K, a mixture at equlibrium contains `P_(N_2)=0.0040` atm, `P_(H_2)=0.063` atm, and `P_(NH_3)=0.48` atm. Calculate `\DeltaG^o` at 800 K?

What I have so far: `Q=\frac{(P_(NH_3))^2}{P_(N_2)*(P_(H_2))^3}` from the reaction below

`N_2+3H_2\rightleftharpoons2NH_3`
My `Q` is `\approx230000` (actual value: 230,356.6929)

---

However, I do not know how to calculate for `\DeltaG^o`. Should I be using the formula `\DeltaG^o=-nFE_(cell)^o` or `\DeltaG^o=-RTlnk`? I have Q, but this is not a battery so...

(the top question in this picture)

Answer 1

Based on the determined equilibrium constant, `DeltaG^@ = -"82.13 kJ/mol"`.

---

You don't need to calculate `Q`, because you're at equilibrium already and can calculate `K_P` right away (in implied units of `"atm"`).

`K_P = (P_(NH_3)^2)/(P_(N_2)P_(H_2)^3)`

`= ("0.48 atm"//"1 atm")^2/(("0.0040 atm"//"1 atm")("0.063 atm"//"1 atm")^3)`

`= 2.304 xx 10^5`

(FYI, this equilibrium constant makes no physical sense, because this reaction requires excess reactants in the industry to push it forward and it becomes LESS favorable at higher temperatures. `K_P = 5.38 xx 10^(-6)` at `"823.15 K"`. Thus, `K_P ~~ 10^(-5)` at `"800 K"`.)

For now we'll go with this one... In general,

`DeltaG = DeltaG^@ + RTlnQ`

where `Q` is the reaction quotient, i.e. the NOT-YET-equilibrium constant. `DeltaG` is the change in Gibbs' free energy, and `@` indicates standard conditions (`"1 M"` concentrations, `"1 atm"` surroundings, at a certain temperature).

At equilibrium, recall that `DeltaG = 0` AND `Q = K`. Thus,

`DeltaG^@ = -RTlnK`

For gas-phase reactions, we define `K -= K_P` at standard conditions, so

`DeltaG^@ ("gas-phase") = -RTlnK_P`

Therefore,

`color(red)(DeltaG^@) = -("0.008314 kJ/mol"cdot"K")("800 K")ln(2.304 xx 10^5)`

`= color(red)(-"82.13 kJ/mol")`

In reality, for this reaction at `"800 K"`, `DeltaG^@ ~~ "76.57 kJ/mol"`.