Degree of dissociation and simultaneous equilibria

The degree of dissociation

When a gas splits, the degree of dissociation $\alpha$ is the fraction of it that has split at equilibrium. It runs from $0$ (none) to $1$ (all). Exams ask you to find it from $K_p$, or to find $K_p$ from it.

Always build a small table: moles at the start, the change, and moles at equilibrium. Then turn moles into partial pressures with the total pressure $P$.

One molecule into two: N2O4

Start with $1$ mol of $\text{N}_2\text{O}_4$. At equilibrium there are $1 - \alpha$ mol of it and $2\alpha$ mol of $\text{NO}_2$, so $1 + \alpha$ mol in all.

Solving for $\alpha$ gives a neat result:

Worked example. At some temperature $K_p = 0.45\ \text{bar}$. Find $\alpha$ at a total pressure of $1.20\ \text{bar}$.

1. Put in the numbers. $\alpha = \sqrt{\dfrac{0.45}{0.45 + 4.80}} = \sqrt{0.0857}$.
2. Answer. $\alpha = 0.293$, so about $29\ \%$ has split.

At $0.30\ \text{bar}$ the same gas gives $\alpha = 0.522$. Lower pressure, more splitting.

One molecule into two different ones: PCl5

The table is the same, but now each product has $\alpha$ mol. So

With $K_p = 2.10\ \text{bar}$ at $P = 3.00\ \text{bar}$, $\alpha = \sqrt{\dfrac{2.10}{5.10}} = 0.642$.

Pressure and added gases

Both formulas have $P$ in the bottom, so raising the pressure lowers $\alpha$. This is Le Chatelier's principle in numbers: the side with fewer gas molecules is favoured.

An inert gas such as argon needs care:

• At constant volume, the partial pressures of the reacting gases do not change. Nothing shifts.
• At constant total pressure, the reacting gases are diluted. Their partial pressures fall, so more splits, just as at a lower pressure.

Alpha from the vapour density

Splitting makes more molecules, so the gas gets lighter for its volume. The vapour density $d$ (molar mass over $2$) falls.

If each molecule splits into $n$ particles, the moles go from $1$ to $1 + (n - 1)\alpha$. The vapour density falls in the same ratio:

Here $D$ is the vapour density with no splitting.

Worked example. $\text{PCl}_5$ has a molar mass of $208.5\ \text{g mol}^{-1}$, so $D = 104.25$. Heated, its vapour density is $62.0$.

1. Count the particles. One $\text{PCl}_5$ gives two molecules, so $n = 2$.
2. Find alpha. $\alpha = \dfrac{104.25 - 62.0}{62.0} = 0.681$.

Two equilibria that share a gas

Sometimes two reactions in the same vessel make the same gas. Then that gas has one partial pressure, and it appears in both constants.

Worked example. Two solids, $\text{AX}$ and $\text{BX}$, sit in one evacuated flask. Each gives off the gas $\text{X}$:

1. Name the pressures. Let $p_\text{A} = a$ and $p_\text{B} = b$. Every A or B comes with one X, so $p_\text{X} = a + b$.
2. Write both constants. $a(a + b) = 0.24$ and $b(a + b) = 0.40$.
3. Add them. $(a + b)^2 = 0.64$, so $p_\text{X} = 0.80\ \text{bar}$.
4. Each gas. $a = \dfrac{0.24}{0.80} = 0.30\ \text{bar}$ and $b = \dfrac{0.40}{0.80} = 0.50\ \text{bar}$.
5. Total. $0.30 + 0.50 + 0.80 = 1.60\ \text{bar}$.

The shared gas holds each reaction back. Alone, the first solid would give $p_\text{X} = \sqrt{0.24} = 0.49\ \text{bar}$ of each gas.

Try it in Python

When a formula is hard to rearrange, a computer can still find $\alpha$. The program halves an interval again and again until $K_p$ from the guess matches the real one. This is bisection.

Common mistakes

• Using moles in place of partial pressures. Multiply each mole fraction by the total pressure first.
• Forgetting that the total moles change. It is $1 + \alpha$ for one-into-two, not $1$.
• Mixing up the two formulas. One-into-two of the same gas gives $4\alpha^2$; two different products give $\alpha^2$.
• Dividing by $D$ in the vapour density formula. The measured $d$ goes at the bottom.
• Thinking argon always shifts an equilibrium. At constant volume it changes nothing.

Remember these

• $\text{A} \rightleftharpoons 2\text{B}$: $K_p = \dfrac{4\alpha^2 P}{1 - \alpha^2}$. $\text{AB} \rightleftharpoons \text{A} + \text{B}$: $K_p = \dfrac{\alpha^2 P}{1 - \alpha^2}$.
• Higher total pressure, smaller $\alpha$.
• Vapour density: $\alpha = \dfrac{D - d}{(n - 1)\,d}$.
• A shared gas has one partial pressure; write both constants with it.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A gas $\text{Y}_2$ splits into $2\text{Y}$. At $2.0\ \text{bar}$ it is $20\ \%$ split. What is $K_p$?

2. Argon is pumped into a flask of $\text{PCl}_5$ at equilibrium, at constant volume. Does $\alpha$ change?

3. A gas splits one into three particles. Its vapour density falls from $40$ to $25$. What is $\alpha$?