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Why compressing a gas equilibrium favours the side with fewer moles of gas, why equal counts mean no shift, and why an inert gas at fixed volume changes nothing.
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By the end of this lesson you will be able to balance a gas equilibrium, count the moles of gas on each side, and predict which way the position moves when the pressure is raised or lowered, including the case where it does not move at all. You will also be able to explain the prediction with the reaction quotient, say why an inert gas added at fixed volume leaves the equilibrium alone, and describe how pressure is chosen in the Haber process.
From Chemistry 1: at a fixed temperature the pressure of a gas depends on how many molecules are in how much space, so equal volumes of gases hold equal numbers of moles, and halving the volume doubles the pressure. From the last two lessons: a disturbance to an equilibrium is partly undone by a net reaction. Put together, those predict what squeezing a gas equilibrium does.
The total pressure of a gas mixture is the sum of the partial pressures of its gases, and each partial pressure is proportional to that gas's moles in the container. Compression reduces the volume and raises every partial pressure; expansion does the opposite. The moles of gas on a side of an equation is the sum of the coefficients of the gaseous species on that side — solids and liquids do not count. An inert gas, such as argon, takes no part in the reaction.
Squeeze the Haber mixture into half the volume:
$$\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$$
There are four moles of gas on the left and two on the right. Every forward reaction turns four gas molecules into two, and so lowers the pressure; every reverse reaction raises it. After compression the pressure has jumped, and the system responds in the direction that lowers it — the forward direction. The position moves right and the equilibrium mixture holds more ammonia.
The rule is: raising the pressure favours the side with fewer moles of gas; lowering it favours the side with more. And the rule that goes with it: if both sides have the same number of moles of gas, pressure has no effect on the position. For $\mathrm{H_2 + I_2 \rightleftharpoons 2HI}$ there are two moles on each side, and squeezing it changes the concentrations of everything but not the proportions.
The count must come from the balanced equation, and only gases count. In $\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}$ the left side holds no gas and the right side one mole, so compression pushes the position left, towards the solid limestone.
Why does it work in terms of the constant? For the Haber process $K_c = [\mathrm{NH_3}]^2 \div ([\mathrm{N_2}][\mathrm{H_2}]^3)$. Halving the volume doubles every concentration: the top grows by $2^2 = 4$ and the bottom by $2^4 = 16$, so $Q_c$ falls to a quarter of $K_c$ and the forward reaction runs. With equal powers top and bottom the factors cancel, which is exactly the equal-moles case.
Another way: picture
Think of a crowded lift. When more people squeeze in, the crowd eases if some of them pair up and hold hands in a tight group, taking less room. A reaction that turns four gas molecules into two is that pairing up; one that turns two into two cannot help, however it rearranges.
Another way: steps
To predict a pressure shift:
Pump argon into the rigid steel vessel of a Haber equilibrium. The total pressure rises — there are more molecules in the same space — and a learner reading the rule too fast predicts a shift to the right.
Nothing moves. The argon takes no part in the reaction and does not appear in the expression for $K_c$. The vessel's volume has not changed, so the moles of nitrogen, hydrogen and ammonia per litre are exactly what they were, the quotient is unchanged, and the mixture is still at equilibrium.
The lesson in that is precise: pressure shifts an equilibrium only through the concentrations of the reacting gases. Compressing the vessel changes them; adding an unreactive gas at fixed volume does not. (Adding argon while letting the vessel expand to keep the total pressure constant does dilute the reacting gases, and then the mixture shifts towards the side with more moles — the same rule, applied to what really changed.)
Four moles of gas become two, so high pressure favours ammonia, and the higher the better for yield. Real plants run at roughly 150 to 250 atmospheres, not thousands, for reasons that are about money and safety rather than chemistry: every extra atmosphere needs thicker pipes and vessels, bigger compressors and more electricity to drive them, and a failure at high pressure is more dangerous.
High pressure has a second benefit that has nothing to do with the position: squeezing the gases raises their concentrations, so collisions are more frequent and the rate is higher too. Temperature pulls the other way — cooler gives more ammonia but too slowly — and the iron catalyst makes a moderate temperature fast enough. The operating conditions are a single decision about yield, rate and cost, and each of the three lessons on shifting an equilibrium supplies one part of it.
"Pressure favours the side with the bigger molecules." Molecules are not crushed. Pressure depends on how many gas particles there are, so the count from the coefficients decides, whatever the formulas look like.
"Counting atoms instead of molecules." Both sides of a balanced equation have the same atoms. What differs is how many molecules they are grouped into, and that is what the coefficients count.
"Any increase in total pressure shifts the equilibrium." An inert gas added at fixed volume raises the total and changes no concentration of a reacting gas, so nothing moves.
"Solids and liquids count." Only gases contribute to the pressure, so only gaseous species are counted.
"Compression changes $K_c$." It changes $Q_c$. The constant is still the temperature's to change, and the mixture moves until the quotient is back at it.
Steam reforming, $\mathrm{CH_4(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g)}$, is run at higher pressure.
Balance first, then count.
Left: $1 + 1 = 2$ moles of gas. Right: $1 + 3 = 4$ moles of gas.
Every species is a gas, so every coefficient counts.
Higher pressure favours the side with fewer moles, the left, so less hydrogen at equilibrium — which is why reformers run at only modest pressures.
Fewer gas molecules means lower pressure, which is the direction the system moves.
$\mathrm{N_2O_4 \rightleftharpoons 2NO_2}$ has $K_c = [\mathrm{NO_2}]^2 \div [\mathrm{N_2O_4}]$. The volume is halved.
Halving the volume doubles every concentration.
The top grows by $2^2 = 4$, the bottom by $2$, so $Q_c$ becomes twice $K_c$.
Different powers top and bottom, so the factors do not cancel.
$Q_c > K_c$, so the reverse reaction runs: the position moves left, towards one mole of gas from two, as the counting rule said.
The count and the quotient always agree.
Left: $2 + 1 = 3$ moles of gas. Right: $2$ moles of gas.
Coefficients of the gases only.
Higher pressure favours the side with fewer moles, which is $\ldots$
…the right, so compressing the mixture makes more nitrogen dioxide at equilibrium.
Every substance in hydrogen reacting with iodine vapour is a gas. Balance the equation, because the count of gas molecules on each side is what decides the effect of pressure.
This task has no paper form; do it on a device.
For making methanol from carbon monoxide and hydrogen, $\mathrm{CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g)}$, by how many moles does the amount of gas change when the forward reaction happens once as written? Give products minus reactants.
Answer: mol
The equilibrium mixture for the Haber process, $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$, is compressed to half its volume at constant temperature. Which way does the position move?
For each equilibrium, count the moles of gas on each side and say which way the position moves when the mixture is compressed.
| moles of gas on the left | moles of gas on the right | on compression the position moves | |
|---|---|---|---|
| $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$ | |||
| $\mathrm{CH_4(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g)}$ | |||
| $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ |
A gas equilibrium mixture is at $100$ kPa. At constant temperature its volume is divided by $3$. What is the total pressure straight after the squeeze, before the equilibrium has had time to respond?
Answer: unit: Pa / atm / kPa
Match each change to the way the equilibrium position moves.
| moves to the right | moves to the left | does not move | |
|---|---|---|---|
| $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$, volume halved | |||
| $\mathrm{N_2O_4 \rightleftharpoons 2NO_2}$, volume halved | |||
| $\mathrm{H_2 + I_2 \rightleftharpoons 2HI}$, volume halved | |||
| $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$, argon pumped into the rigid vessel | |||
| $\mathrm{2SO_2 + O_2 \rightleftharpoons 2SO_3}$, volume doubled |
A sealed bottle of sparkling water holds carbon dioxide at about $4$ atm above the liquid, in equilibrium with dissolved carbon dioxide: $\mathrm{CO_2(g) \rightleftharpoons CO_2(aq)}$. The cap is taken off and the gas above the liquid drops to about $1$ atm. What happens?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A rigid vessel at a fixed temperature starts with $2$ mol of nitrogen and $6$ mol of hydrogen at a total pressure of $1600$ kPa. By the time equilibrium is reached, $1.5$ mol of nitrogen has reacted: $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$. What is the total pressure at equilibrium?
Answer: unit: Pa / atm / kPa
You can predict a pressure shift from the moles of gas in a balanced equation. Say out loud why squeezing hydrogen and iodine vapour changes nothing while squeezing nitrogen and hydrogen makes more ammonia. That completes the equilibrium unit: two opposing rates, a constant that describes where they balance, and three kinds of disturbance, only one of which moves the constant. Next: acids and bases, where the same constant measures how far an acid gives up its protons.
10. Your turn: $\mathrm{2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)}$ is compressed. Which way does it move?, step 3