Question 5 of 6: Equilibrium Composition of Three Pentane Isomers

Question B2: Equilibrium Composition of Three Pentane Isomers (Part B — equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Three simultaneous formation equilibria from graphite and $H_2$ at $T=400$ K, $P=1$ atm, with standard Gibbs energies of formation $\Delta G^\circ_{f,i}$ (kJ/mol): n 40.195, iso 34.415, neo 37.640.

Find. The equilibrium mole fractions of the three isomers in the gas phase.

Approach. The three isomers share the identical formation reaction ($5C+6H_2\to$ isomer), so the graphite and hydrogen activities are common to all three; the isomer partial pressures therefore follow a Boltzmann-type weighting $\propto\exp(-\Delta G^\circ_{f,i}/RT)$, and normalizing gives the composition.

1. Equilibrium constant for each isomer. For $5C(s)+6H_2\leftrightarrow C_5H_{12}$, with graphite at unit activity, $$K_i=\frac{p_i}{p_{H_2}^{6}}=\exp\!\left(-\frac{\Delta G^\circ_{f,i}}{RT}\right).$$
2. Ratio of isomers eliminates the common terms. Dividing any two equilibria, the identical $p_{H_2}^{6}$ (and graphite) cancel, so $$\frac{p_i}{p_j}=\exp\!\left(-\frac{\Delta G^\circ_{f,i}-\Delta G^\circ_{f,j}}{RT}\right)\;\Rightarrow\; y_i\propto\exp\!\left(-\frac{\Delta G^\circ_{f,i}}{RT}\right).$$
3. Boltzmann weights at 400 K. With $RT=8.314(400)=3325.6$ J/mol, the unnormalized weights $w_i=\exp(-\Delta G^\circ_{f,i}/RT)$ are in the ratio $$w_{\text{n}}:w_{\text{iso}}:w_{\text{neo}}=\exp(-12.087):\exp(-10.349):\exp(-11.319),$$ i.e. $1.000:5.686:2.156$ after dividing through by $w_{\text{n}}$.
4. Normalize to mole fractions. Dividing by the sum $1.000+5.686+2.156=8.842$: $$\boxed{y_{\text{n}}=0.113,\quad y_{\text{iso}}=0.643,\quad y_{\text{neo}}=0.244.}$$

Isopentane, having the lowest Gibbs energy of formation, is the most stable and dominates the mixture; neopentane is intermediate and n-pentane the least favoured. The result is independent of the total pressure because every isomer is formed by the same mole-change reaction, so the pressure factors cancel in the ratios.