# 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: $Q = \setminus \frac{{\left({P}_{N {H}_{3}}\right)}^{2}}{{P}_{{N}_{2}} \cdot {\left({P}_{{H}_{2}}\right)}^{3}}$ from the reaction below ${N}_{2} + 3 {H}_{2} \setminus r i g h t \le f t h a r p \infty n s 2 N {H}_{3}$ My $Q$ is $\setminus \approx 230000$ (actual value: 230,356.6929) However, I do not know how to calculate for $\setminus \Delta {G}^{o}$. Should I be using the formula $\setminus \Delta {G}^{o} = - n F {E}_{c e l l}^{o}$ or $\setminus \Delta {G}^{o} = - R T \ln k$? I have Q, but this is not a battery so... (the top question in this picture) ##### 1 Answer
Jul 28, 2018

Based on the determined equilibrium constant, $\Delta {G}^{\circ} = - \text{82.13 kJ/mol}$.

You don't need to calculate $Q$, because you're at equilibrium already and can calculate ${K}_{P}$ right away (in implied units of $\text{atm}$).

${K}_{P} = \frac{{P}_{N {H}_{3}}^{2}}{{P}_{{N}_{2}} {P}_{{H}_{2}}^{3}}$

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

$= 2.304 \times {10}^{5}$

(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. ${K}_{P} = 5.38 \times {10}^{- 6}$ at $\text{823.15 K}$. Thus, ${K}_{P} \approx {10}^{- 5}$ at $\text{800 K}$.)

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

$\Delta G = \Delta {G}^{\circ} + R T \ln Q$

where $Q$ is the reaction quotient, i.e. the NOT-YET-equilibrium constant. $\Delta G$ is the change in Gibbs' free energy, and $\circ$ indicates standard conditions ($\text{1 M}$ concentrations, $\text{1 atm}$ surroundings, at a certain temperature).

At equilibrium, recall that $\Delta G = 0$ AND $Q = K$. Thus,

$\Delta {G}^{\circ} = - R T \ln K$

For gas-phase reactions, we define $K \equiv {K}_{P}$ at standard conditions, so

$\Delta {G}^{\circ} \left(\text{gas-phase}\right) = - R T \ln {K}_{P}$

Therefore,

$\textcolor{red}{\Delta {G}^{\circ}} = - \left(\text{0.008314 kJ/mol"cdot"K")("800 K}\right) \ln \left(2.304 \times {10}^{5}\right)$

$= \textcolor{red}{- \text{82.13 kJ/mol}}$

In reality, for this reaction at $\text{800 K}$, $\Delta {G}^{\circ} \approx \text{76.57 kJ/mol}$.