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Heat treated as a reactant or product, why temperature is the only change that moves the equilibrium constant, and the compromise it forces on industry.
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By the end of this lesson you will be able to predict which way an equilibrium shifts when it is heated or cooled from the sign of its enthalpy change, say whether the equilibrium constant rises or falls, and work backwards from two measured constants to the sign of the enthalpy change. You will also be able to explain why an industrial process such as the Haber process runs at a compromise temperature and why its catalyst does not change the yield.
Two things from earlier in the course meet here. From Chemistry 1: an exothermic reaction has a negative enthalpy change and gives heat to its surroundings, an endothermic one a positive change and takes heat in. From the last lesson: after a change of concentration the constant stays put and the mixture moves until the quotient matches it. Temperature breaks the second rule, and the first rule tells you which way.
The forward enthalpy change, $\Delta H$, is the heat change for the equation as written, left to right; the reverse reaction has the same size with the opposite sign. The endothermic direction of an equilibrium is whichever direction takes heat in, and the exothermic direction is the other one. A compromise condition is an operating temperature or pressure chosen because it balances two things that pull in opposite directions, such as yield and speed.
Write the Haber process with its heat as a product, because the forward reaction is exothermic:
$$\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3} + 92\ \text{kJ}$$
Raising the temperature supplies heat, and a system at equilibrium responds by using up what was added — so the reverse reaction, the one that absorbs heat, is favoured. Less ammonia at equilibrium. Cooling removes heat, and the forward reaction, which releases it, is favoured. More ammonia.
For an endothermic forward reaction heat goes on the left, and every conclusion flips. Dinitrogen tetroxide splitting into nitrogen dioxide takes in 58 kJ/mol, so warming it makes more of the brown gas.
The summary is short: heating favours the endothermic direction; cooling favours the exothermic direction. Which of the two is forward depends on the reaction, which is why no single rule about temperature and yield can exist.
There is one real difference from a concentration change. When ammonia is added, $K_c$ stays fixed and the mixture moves back to it. When the temperature changes, $K_c$ itself takes a new value — smaller for an exothermic forward reaction heated, larger for an endothermic one — and the unchanged mixture is suddenly not at equilibrium for the new constant. Adding a substance, removing one or squeezing the gas moves the position of an equilibrium and leaves the equilibrium constant exactly where it was. Only a change of temperature changes the constant itself.
Another way: picture
Think of heat as a hidden ingredient printed on one side of the recipe. Adding more of an ingredient drives the recipe towards using it up; taking it away drives the recipe towards making more of it. The only thing to look up is which side of the recipe the heat is printed on.
Another way: steps
To predict a temperature shift:
Heating speeds up both the forward and the reverse reaction, and it does not speed them up equally. The endothermic direction has the higher activation energy — its hump is climbed from the lower-energy side — and a reaction with a higher activation energy is more sensitive to temperature, because raising the temperature increases the share of collisions over a tall barrier by a bigger factor than over a short one.
So at a higher temperature the endothermic direction gains on the exothermic one, and the two rates only balance again with more of the endothermic direction's products present. A new balance point is a new constant.
Measured constants show it. For the Haber process $K_c$ is roughly $4 \times 10^{8}$ at room temperature and well below 1 at 500 °C; for the splitting of dinitrogen tetroxide it rises roughly fiftyfold between 25 °C and 100 °C. The direction of change is set by the sign of $\Delta H$ and nothing else.
If the Haber process only cared about yield, it would run cold: at room temperature the equilibrium lies almost entirely on the ammonia side. It does not, because at room temperature the reaction is so slow that equilibrium would never arrive. Nitrogen's triple bond gives the reaction a huge activation energy.
So the plant runs at about 400–450 °C. The equilibrium yield per pass is lower there — typically around 15% conversion — but it is reached in seconds rather than centuries, and the unreacted gases are recycled until nearly all of them become ammonia. The temperature is a compromise between yield and rate.
The iron catalyst is what makes that compromise workable, and it is worth being exact about why. A catalyst lowers the activation energy of the forward and the reverse reaction by the same amount, so both speed up by the same factor. It changes how quickly equilibrium is reached and does not change where equilibrium is: the constant and the equilibrium yield are exactly what they would be without it.
"Heating always increases the yield." It increases the yield of an endothermic reaction and decreases the yield of an exothermic one. The sign of $\Delta H$ decides, not a general rule.
"Heating only makes the reactions faster, so the position cannot change." Both directions do speed up, by different factors, and that difference is exactly why the constant moves.
"$K_c$ stays the same after a temperature change and the mixture moves back to it." That is the concentration rule. Temperature is the only change that gives the constant a new value.
"A catalyst shifts the equilibrium towards products." It speeds both directions equally and moves nothing; it only gets the mixture to its equilibrium sooner.
"The enthalpy change is for one mole of product." It is for the equation as written. The Haber value of $-92$ kJ/mol is for two moles of ammonia, so each mole of ammonia releases 46 kJ.
The Contact process, $\mathrm{2SO_2 + O_2 \rightleftharpoons 2SO_3}$, has $\Delta H = -196$ kJ/mol. The converter is made hotter.
Negative, so heat is written as a product.
Heating adds heat, so the reverse, heat-absorbing reaction is favoured and the position moves left.
The system uses up what was added.
Less sulfur trioxide at equilibrium, and $K_c$ is smaller at the higher temperature.
A shift to the left means a smaller constant.
For $\mathrm{N_2 + O_2 \rightleftharpoons 2NO}$, $K_c$ is tiny at room temperature and about $10^{-3}$ at 2000 °C inside an engine.
Compare the constant at the two temperatures.
The constant is larger when hotter, so heating favoured the forward reaction.
Heating always favours the endothermic direction.
So the forward reaction is endothermic — which is why hot engines make nitrogen monoxide and cool air does not.
The measurement tells you the sign of $\Delta H$.
Positive, so heat is written on the left as a reactant.
The sign places the heat.
Heating adds a reactant, so the position moves $\ldots$
…to the right, so the reformer runs hot, and its constant is larger at high temperature.
For nitrogen and oxygen forming nitrogen monoxide in a hot engine, $\mathrm{N_2(g) + O_2(g) \rightleftharpoons 2NO(g)}$, the forward enthalpy change is $181$ kJ/mol. The equilibrium mixture is heated at constant pressure. Which way does the position move?
The equilibrium constants of three reactions were measured at $300$ K and at $500$ K. For each reaction, say whether heating raises or lowers the constant, and whether the forward reaction is exothermic or endothermic.
| constant at 300 K | constant at 500 K | on heating the constant | the forward reaction is | |
|---|---|---|---|---|
| Reaction X | 8 | 24 | ||
| Reaction Y | 16 | 8 | ||
| Reaction Z | 18 | 3 |
An isomerisation $\mathrm{A \rightleftharpoons B}$ has an endothermic forward reaction. At $300$ K its constant is $1$, and the flask holds $150$ mmol of each. The flask is heated to $400$ K, where the constant is $4$. Use the net-reaction steps to bring the mixture to its new equilibrium.
This task has no paper form; do it on a device.
For the Haber process, $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$ with a forward enthalpy change of $-92$ kJ/mol, match each change with what it does.
| the constant falls and the position moves left | the constant rises and the position moves right | neither changes; equilibrium is simply reached sooner | the constant is unchanged and the position moves right | |
|---|---|---|---|---|
| The temperature is raised | ||||
| The temperature is lowered | ||||
| An iron catalyst is added | ||||
| More nitrogen is pumped in |
In the Haber process, $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$, $\Delta H = -92$ kJ/mol for the equation as written. Writing heat in as a product, how much heat is released when $6$ mol of ammonia forms?
Answer: unit: J / kJ
The equilibrium for making methanol from carbon monoxide and hydrogen, $\mathrm{CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g)}$, has a forward enthalpy change of $-91$ kJ/mol. It is cooled. Put the explanation in order.
Number the steps in order (write the number in the box):
A sealed glass tube holds brown nitrogen dioxide in equilibrium with colourless dinitrogen tetroxide: $\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$, $\Delta H = +58$ kJ/mol. The tube is placed in a mixture of ice and salt. What happens to its colour?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
An isomerisation $\mathrm{A \rightleftharpoons B}$ is at equilibrium with $40$ mmol of each, where its constant is $1$. The forward reaction is endothermic, and after heating the constant is $3$. How much B is there at the new equilibrium?
Answer: unit: mol / mmol
You can place heat in an equation and predict a temperature shift, and you know temperature is the one change that alters the constant. Say out loud why the Haber process is not run at room temperature even though cold favours ammonia. Next: pressure, and why it matters only when the moles of gas differ.
10. Your turn: steam reforming, $\mathrm{CH_4 + H_2O \rightleftharpoons CO + 3H_2}$, has $\Delta H = +206$ kJ/mol. Should the reformer run hot or cool for a good yield of hydrogen?, step 3