IB Chemistry SL — Rates of Reaction & Equilibrium

Complete Study Guide

Topics Covered

1. Collision Theory & Activation Energy
2. Factors Affecting Rate of Reaction
3. Catalysts
4. Measuring Rate of Reaction
5. Dynamic Equilibrium & Le Chatelier’s Principle
6. Equilibrium Constant $K_c$
7. Industrial Application — The Haber Process
8. Practice MCQs

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1. Collision Theory & Activation Energy

Reactivity 2.2 — How fast? Rates of Reaction

The rate of reaction is a measure of how quickly reactants are converted into products. More precisely:

$\text{Rate of reaction} = \frac{\Delta[\text{concentration}]}{\Delta t}$

The units of rate are typically $\text{mol dm}^{-3}\text{ s}^{-1}$ (moles per decimetre cubed per second). The concentration of reactants decreases over time (so the rate for reactants is technically negative, but we report it as a positive number), while the concentration of products increases.

Collision Theory

For a chemical reaction to occur, particles must collide. But not every collision leads to a reaction — only effective collisions produce products. An effective collision requires two conditions to be met simultaneously:

1. Sufficient energy — the colliding particles must together have energy equal to or greater than the activation energy ($E_a$)
2. Correct orientation — the particles must collide at the right angle so that the reactive parts of the molecules interact

Activation Energy and the Maxwell-Boltzmann Distribution

The activation energy ($E_a$) is the minimum energy that colliding particles must possess for a reaction to occur. Think of it as an energy barrier that reactants must climb before they can become products.

Not all molecules in a sample have the same energy at a given temperature. The Maxwell-Boltzmann distribution shows how the energies of molecules are spread across a population at a fixed temperature.

Key features of the Maxwell-Boltzmann distribution curve:

Feature	Description
x-axis	Kinetic energy of molecules
y-axis	Number of molecules (or fraction of molecules) with that energy
Shape	Starts at zero (no molecules have zero energy), rises to a peak, then tails off to the right — the tail never reaches zero
Peak	Represents the most probable energy
Mean energy	Slightly to the right of the peak
$E_a$ line	A vertical line on the x-axis; only molecules to the RIGHT of this line have enough energy to react
Shaded area	The area under the curve to the right of $E_a$ represents the proportion of molecules that CAN react

What happens when temperature increases:

When temperature rises, the distribution curve changes in two important ways:

• The peak shifts to the right (higher energy) and flattens slightly (fewer molecules at the most probable energy)
• The tail extends further to the right
• The area under the curve to the right of $E_a$ increases significantly — meaning a much larger proportion of molecules now have sufficient energy to react

This is the fundamental reason why increasing temperature increases the rate: more molecules exceed $E_a$, so more collisions are effective.

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2. Factors Affecting Rate of Reaction

Five key factors affect the rate of a chemical reaction. For each, the explanation must be in terms of collision theory — specifically, how the factor changes the frequency or energy of effective collisions.

Summary Table

Factor	Change	Effect on Rate	Collision Theory Explanation
Temperature	Increase	Increases	More molecules have energy ≥ $E_a$; collisions are more frequent AND more energetic
Concentration (solutions)	Increase	Increases	More particles per unit volume → more frequent collisions
Pressure (gases)	Increase	Increases	Equivalent to increasing concentration of gas particles
Surface area (solids)	Increase (smaller pieces)	Increases	More particles exposed at the surface → more collisions possible
Catalyst	Add catalyst	Increases	Provides alternative pathway with lower $E_a$ (see Section 3)

Temperature

Increasing temperature gives molecules more kinetic energy. This has two effects:

1. Molecules move faster, so collisions occur more frequently
2. More importantly, a larger proportion of molecules now exceed $E_a$ (seen as increased area under Maxwell-Boltzmann curve to the right of $E_a$)

Both effects increase the rate, but the increase in the fraction of molecules exceeding $E_a$ is the dominant explanation in IB exams.

Concentration

In a solution, increasing the concentration means more solute particles are present in the same volume of solvent. Particles are closer together and collide more frequently. Since the frequency of collisions increases, the frequency of effective collisions also increases, and the rate rises.

Pressure (for gases)

Increasing pressure on a gas squeezes the same number of gas molecules into a smaller volume — this is equivalent to increasing concentration. Particles are closer together, collide more frequently, and the rate increases.

Surface Area

A solid reactant can only react at its surface — interior particles are not exposed. Breaking a solid into smaller pieces (e.g. powder vs lump) dramatically increases the total surface area without changing the amount of substance. More surface particles are exposed to collisions with the other reactant, increasing collision frequency and therefore rate.

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3. Catalysts

Definition and Action

A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the overall process. The catalyst is regenerated at the end of the reaction — it participates in the mechanism but is not used up.

Catalysts increase rate by providing an alternative reaction pathway that has a lower activation energy ($E_a$). This is the key IB explanation. The catalyst does NOT:

• Change the energy of the reactants
• Change the energy of the products
• Change the enthalpy change of reaction ($\Delta H$)
• Shift the position of equilibrium (it speeds up forward and reverse equally)

On a Maxwell-Boltzmann diagram, adding a catalyst shifts the $E_a$ line to the left (lower energy). This means a larger proportion of molecules now have energy ≥ $E_a$ — the shaded area under the curve to the right of the new $E_a$ is larger — so more collisions are effective and the rate increases.

Homogeneous vs Heterogeneous Catalysts

Type	Definition	Example
Homogeneous	Catalyst and reactants are in the same phase (same physical state)	$\text{H}^+(\text{aq})$ catalysing ester hydrolysis in aqueous solution; $\text{NO}(\text{g})$ catalysing oxidation of $\text{SO}_2(\text{g})$ in the atmosphere
Heterogeneous	Catalyst and reactants are in different phases	Iron (solid) catalyst in the Haber Process; $\text{Pt}$ (solid) in catalytic converter; $\text{V}_2\text{O}_5$ (solid) in Contact Process

Heterogeneous catalysts work by adsorption — reactant molecules are adsorbed onto the catalyst surface, weakening bonds and reducing $E_a$. After reaction, products desorb from the surface, freeing active sites for more reactant molecules.

Enzyme Catalysts

Enzymes are biological catalysts — large protein molecules that catalyse specific biochemical reactions in living organisms. Each enzyme has an active site with a specific shape that fits only one (or a few) substrate molecules (the lock-and-key model). Enzymes are extremely efficient and specific, operating under mild conditions (body temperature ~37 °C, near-neutral pH). In the IB, you should know that enzymes are:

• Biological (protein-based)
• Homogeneous catalysts (both enzyme and substrate are typically in aqueous solution)
• Highly specific (one enzyme, one substrate)
• Sensitive to temperature and pH (denaturation above ~40 °C)

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4. Measuring Rate of Reaction

Rate Expression

Rate of reaction can be defined as:

$\text{Rate} = \frac{\Delta[\text{concentration}]}{\Delta t}$

In practice, “concentration of a species” can be substituted by any quantity proportional to concentration:

• Volume of gas produced (collected over water in a gas syringe or inverted measuring cylinder)
• Mass lost (if a gas is produced and escapes — measure on a balance)
• Colour intensity (if a coloured species is produced or consumed — use a colorimeter)
• Conductivity (if ions are produced or consumed)
• Optical rotation (for certain reactions involving chiral molecules)

Initial Rate vs Average Rate

Concept	Definition	How to find from graph
Average rate	Total change in concentration divided by total time elapsed	Gradient of the straight line connecting two points on a concentration-time curve
Initial rate	Rate at time $t = 0$ (when the reaction just starts)	Gradient of the tangent drawn at $t = 0$ on a concentration-time curve

The initial rate is the fastest rate during the reaction (for reactions where concentration of reactants decreases over time), because the reactant concentrations are at their highest at $t = 0$.

Rate-Concentration Graphs (Qualitative)

While IB SL does not require knowledge of rate laws or rate orders mathematically, you should be able to interpret graphs qualitatively:

Graph Shape	What it means
Rate stays constant as concentration changes (horizontal line)	Zero order — rate is independent of this reactant’s concentration
Rate increases linearly with concentration (straight line through origin)	First order — rate is directly proportional to concentration
Rate increases as a curve (exponential-like) with concentration	Second order — rate is proportional to concentration squared

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5. Dynamic Equilibrium & Le Chatelier’s Principle

Reactivity 2.3 — How Far? Equilibrium

Many reactions are reversible — products can react to re-form reactants. We write reversible reactions with a double arrow ($\rightleftharpoons$):

$\text{A} + \text{B} \rightleftharpoons \text{C} + \text{D}$

What Is Dynamic Equilibrium?

At the start of a reversible reaction, only reactants are present. As they react, products build up. The forward reaction slows (as reactant concentrations fall) and the reverse reaction speeds up (as product concentrations rise). Eventually, a state is reached where:

• The rate of the forward reaction equals the rate of the reverse reaction
• The concentrations of all species remain constant (but are NOT necessarily equal)
• The reaction is still occurring in both directions — it is dynamic (not static)

This state is called dynamic equilibrium.

Le Chatelier’s Principle

Le Chatelier’s Principle: When a system at equilibrium is subjected to a change (a “disturbance” or “stress”), the system will respond by shifting in the direction that opposes the change, in order to re-establish equilibrium.

The “shift” means the rate of one direction temporarily exceeds the other until a new equilibrium is established.

Applying Le Chatelier’s Principle

Consider the general equilibrium:

$\text{A}(\text{g}) + \text{B}(\text{g}) \rightleftharpoons \text{C}(\text{g}) + \text{D}(\text{g}) \quad \Delta H < 0 \text{ (exothermic)}$

Change 1: Concentration

Disturbance	System Response	Direction of Shift
Add more reactant $[\text{A}]$ or $[\text{B}]$	Equilibrium shifts to reduce added reactant; produces more $\text{C}$ and $\text{D}$	Shift right (forward)
Remove reactant $[\text{A}]$ or $[\text{B}]$	Equilibrium shifts to replace removed reactant	Shift left (reverse)
Add more product $[\text{C}]$ or $[\text{D}]$	Equilibrium shifts to reduce added product; consumes $\text{C}$ and $\text{D}$	Shift left (reverse)
Remove product $[\text{C}]$ or $[\text{D}]$	Equilibrium shifts to replace removed product; makes more product	Shift right (forward)

Change 2: Temperature

Temperature changes are unique because they actually change the value of $K_c$ (unlike concentration and pressure, which only shift the position).

Treat heat as a “reactant” or “product”:

• For an exothermic forward reaction: heat is a product. Increasing temperature is like adding a product → shift left (reverse).
• For an endothermic forward reaction: heat is a reactant. Increasing temperature is like adding a reactant → shift right (forward).

Change	Exothermic forward ($\Delta H < 0$)	Endothermic forward ($\Delta H > 0$)
Increase temperature	Shift left; $K_c$ decreases	Shift right; $K_c$ increases
Decrease temperature	Shift right; $K_c$ increases	Shift left; $K_c$ decreases

Change 3: Pressure (gases only)

Changing pressure has an effect only if there are different numbers of moles of gas on each side of the equation.

• Increasing pressure → system shifts toward the side with fewer moles of gas (to reduce pressure)
• Decreasing pressure → system shifts toward the side with more moles of gas (to increase pressure)
• If moles of gas are equal on both sides → pressure change has no effect on equilibrium position

Change 4: Catalyst

Adding a catalyst speeds up both the forward and reverse reactions equally. Therefore:

• The equilibrium position does not shift (no change in concentrations at equilibrium)
• $K_c$ is unchanged
• The system reaches equilibrium faster — that is the only effect

Summary Table — Le Chatelier Disturbances

Disturbance	Direction of Shift	Effect on $K_c$
Add reactant	Right (forward)	No change
Remove reactant	Left (reverse)	No change
Add product	Left (reverse)	No change
Remove product	Right (forward)	No change
Increase temperature (exothermic forward)	Left (reverse)	Decreases
Increase temperature (endothermic forward)	Right (forward)	Increases
Decrease temperature (exothermic forward)	Right (forward)	Increases
Increase pressure (fewer gas moles on right)	Right (forward)	No change
Increase pressure (fewer gas moles on left)	Left (reverse)	No change
Increase pressure (equal gas moles both sides)	No shift	No change
Add catalyst	No shift	No change

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6. Equilibrium Constant $K_c$

Writing the $K_c$ Expression

For any reversible reaction at equilibrium:

$a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$

The equilibrium constant $K_c$ is defined as:

$K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$

Where:

• Square brackets $[ \, ]$ denote equilibrium concentration in mol dm$^{-3}$
• The products go in the numerator (top)
• The reactants go in the denominator (bottom)
• Each concentration is raised to the power of its stoichiometric coefficient from the balanced equation

Magnitude of $K_c$

Value of $K_c$	Meaning
$K_c \gg 1$ (e.g. $K_c = 10^{10}$)	Products strongly favoured; reaction goes nearly to completion
$K_c \approx 1$	Roughly equal amounts of reactants and products at equilibrium
$K_c \ll 1$ (e.g. $K_c = 10^{-10}$)	Reactants strongly favoured; very little product formed

What Changes $K_c$?

$K_c$ depends only on temperature. Changing concentration, pressure, or adding a catalyst does NOT change $K_c$ — only the position of equilibrium shifts.

Worked Example — Calculating $K_c$

Worked Example — ICE Table

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7. Industrial Application — The Haber Process

The Reaction

The Haber Process is the industrial synthesis of ammonia from nitrogen and hydrogen:

$\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g}) \quad \Delta H = -92 \text{ kJ mol}^{-1}$

Ammonia is one of the most important industrial chemicals — it is the starting material for fertilisers, explosives, cleaning products, and many other chemicals. Understanding the Haber Process means applying both Le Chatelier’s Principle and collision theory to choose the best industrial conditions.

Applying Le Chatelier’s Principle

Pressure

• Left side: $1 + 3 = \mathbf{4}$ moles of gas
• Right side: $\mathbf{2}$ moles of gas
• High pressure favours the right side (fewer moles of gas) → more $\text{NH}_3$ produced
• Industrial pressure used: ~200 atm

However, very high pressures are extremely expensive to maintain (requires thick-walled vessels, powerful compressors, high energy costs) and pose safety risks. ~200 atm is a compromise between yield and cost.

Temperature

• The forward reaction is exothermic ($\Delta H = -92$ kJ mol$^{-1}$)
• Low temperature shifts equilibrium right → higher yield of $\text{NH}_3$
• But low temperature means a very slow rate of reaction (molecules have less energy; fewer exceed $E_a$)
• High temperature means faster rate but lower equilibrium yield
• Industrial temperature used: 400–500 °C

This is a critical compromise: at 400–500 °C, the rate is fast enough to be economically viable, even though the equilibrium yield of $\text{NH}_3$ is only ~15%. The ammonia is continuously removed (liquefied and extracted), which shifts equilibrium right (Le Chatelier) and allows the unreacted $\text{N}_2$ and $\text{H}_2$ to be recycled.

Catalyst

• An iron catalyst (with potassium oxide and alumina as promoters) is used
• The catalyst increases the rate of reaction without affecting $K_c$ or the equilibrium position
• This allows the reaction to proceed at an acceptable rate at the lower 400–500 °C temperature

Summary of Haber Process Conditions

Condition	Value	Reason
Pressure	~200 atm	High pressure favours products (fewer gas moles); limited by cost/safety
Temperature	400–500 °C	Compromise: fast enough rate; acceptable (but not maximum) yield
Catalyst	Iron (+ promoters)	Increases rate; does not change equilibrium position or $K_c$
$\text{NH}_3$ removal	Continuous liquefaction	Removes product; shifts equilibrium right by Le Chatelier; unreacted gases recycled

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8. Practice MCQs

The following 10 questions are written in IB exam style. Try each question before revealing the answer.

MCQ 1. Which of the following changes will increase the rate of a reaction between zinc and dilute hydrochloric acid? I. Using more concentrated acid II. Increasing the surface area of the zinc III. Adding a catalyst A. I only B. I and II only C. II and III only D. I, II, and III

Answer: D — I, II, and III

All three changes increase the rate:

• I: More concentrated acid → more $\text{H}^+$ ions per unit volume → more frequent collisions with zinc
• II: More surface area → more zinc particles exposed → more collisions per second
• III: A catalyst provides a lower-$E_a$ pathway → greater fraction of collisions are effective

MCQ 2. A reaction is carried out at two temperatures: 30 °C and 50 °C. Using the Maxwell-Boltzmann distribution, which statement best explains the increased rate at 50 °C? A. The activation energy decreases at higher temperature B. More molecules have energy greater than or equal to the activation energy C. The molecules collide more frequently only D. The activation energy is exceeded by all molecules at 50 °C

Answer: B — More molecules have energy greater than or equal to the activation energy

At higher temperature, the Maxwell-Boltzmann distribution shifts right — a larger proportion of molecules now have kinetic energy ≥ $E_a$. The activation energy itself does not change (A is wrong). While collision frequency also increases (part of C), the dominant effect is the larger fraction exceeding $E_a$. Option D is incorrect — even at very high temperatures, only a fraction of molecules exceed $E_a$.

MCQ 3. For the equilibrium: $\text{PCl}_5(\text{g}) \rightleftharpoons \text{PCl}_3(\text{g}) + \text{Cl}_2(\text{g})$ What is the effect of increasing pressure on the equilibrium position? A. Shifts right; more $\text{PCl}_3$ and $\text{Cl}_2$ are produced B. Shifts left; more $\text{PCl}_5$ is produced C. No shift; pressure does not affect gas-phase equilibria D. Shifts right; $K_c$ increases

Answer: B — Shifts left; more $\text{PCl}_5$ is produced

Left side: 1 mole of gas. Right side: 1 + 1 = 2 moles of gas. Increasing pressure favours the side with fewer moles of gas — the left side. The equilibrium shifts left, producing more $\text{PCl}_5$. $K_c$ is unchanged by pressure (D is wrong — only temperature changes $K_c$).

MCQ 4. Which of the following will change the value of $K_c$ for a reaction at equilibrium? A. Adding more reactant B. Increasing the pressure C. Adding a catalyst D. Raising the temperature

Answer: D — Raising the temperature

$K_c$ depends only on temperature. Adding reactants, changing pressure, and adding catalysts all change the rate or shift the position of equilibrium, but leave $K_c$ unchanged. Only a temperature change alters $K_c$.

MCQ 5. Consider the reaction: $2\text{NO}_2(\text{g}) \rightleftharpoons \text{N}_2\text{O}_4(\text{g}) \quad \Delta H = -57 \text{ kJ mol}^{-1}$ What are the effects of simultaneously increasing temperature and decreasing pressure? A. Both changes shift the equilibrium to the right B. Temperature shifts left; pressure shifts right — net effect depends on magnitudes C. Temperature shifts right; pressure shifts left D. Temperature shifts left; pressure shifts left

Answer: B — Temperature shifts left; pressure shifts right — net effect depends on magnitudes

Temperature: forward reaction is exothermic ($\Delta H = -57$ kJ mol$^{-1}$), so increasing temperature shifts equilibrium left (reverse).

Pressure: left side has 2 moles of gas; right side has 1 mole. Decreasing pressure favours more moles of gas — shifts right (forward).

The two effects oppose each other. The net direction cannot be determined without knowing the magnitudes.

MCQ 6. At equilibrium, $K_c = 1.2 \times 10^{-5}$ for a reaction. Which statement is correct? A. The forward reaction is faster than the reverse reaction B. The reaction has not yet reached equilibrium C. Reactants are strongly favoured at equilibrium D. Products are strongly favoured at equilibrium

Answer: C — Reactants are strongly favoured at equilibrium

$K_c = 1.2 \times 10^{-5} \ll 1$, which means the equilibrium lies far to the left — the concentration of reactants greatly exceeds the concentration of products. At equilibrium, forward rate = reverse rate (A is wrong). B is wrong because $K_c$ is defined at equilibrium.

MCQ 7. Which expression correctly gives $K_c$ for: $\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g})$ A. $K_c = \dfrac{[\text{NH}_3]}{[\text{N}_2][\text{H}_2]}$ B. $K_c = \dfrac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$ C. $K_c = \dfrac{[\text{N}_2][\text{H}_2]^3}{[\text{NH}_3]^2}$ D. $K_c = \dfrac{2[\text{NH}_3]}{[\text{N}_2] + 3[\text{H}_2]}$

Answer: B

$K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$

Products over reactants; each raised to its stoichiometric coefficient. $\text{NH}_3$ has coefficient 2, so it is squared. $\text{H}_2$ has coefficient 3, so it is cubed. Option C is the inverse (reactants over products). Option D incorrectly uses addition instead of multiplication and linear coefficients.

MCQ 8. The Haber Process uses a temperature of 400–500 °C rather than a lower temperature such as 100 °C. Why? A. A higher temperature increases $K_c$, favouring product formation B. The forward reaction is endothermic, so higher temperature increases yield C. A lower temperature would give a higher yield but an unacceptably slow rate D. The catalyst only works above 400 °C

Answer: C — A lower temperature would give a higher yield but an unacceptably slow rate

The Haber Process forward reaction is exothermic, so lower temperature increases yield (shifts equilibrium right). However, the rate at 100 °C would be far too slow to be economically viable. 400–500 °C is a compromise: the rate is fast enough, even though yield is lower (~15%). Option A is wrong — increasing temperature for an exothermic reaction decreases $K_c$. Option B is wrong — the reaction is exothermic, not endothermic.

MCQ 9. Which of the following correctly describes a heterogeneous catalyst? A. The catalyst and reactants are in the same physical state B. The catalyst is consumed during the reaction and must be regenerated C. The catalyst changes the enthalpy change ($\Delta H$) of the reaction D. The catalyst is in a different physical state from the reactants

Answer: D — The catalyst is in a different physical state from the reactants

A heterogeneous catalyst is in a different phase from the reactants. Example: solid iron catalyst with gaseous $\text{N}_2$ and $\text{H}_2$ in the Haber process. Option A describes a homogeneous catalyst (same phase as reactants). B is wrong — catalysts are not consumed; they are regenerated. C is wrong — catalysts do not change $\Delta H$; they only lower the activation energy $E_a$.

MCQ 10. For the equilibrium: $\text{Fe}^{3+}(\text{aq}) + \text{SCN}^-(\text{aq}) \rightleftharpoons [\text{FeSCN}]^{2+}(\text{aq})$ The product $[\text{FeSCN}]^{2+}$ is deep red. A solution at equilibrium is pale orange. What would be observed if excess $\text{Fe}^{3+}$ ions are added? A. The solution becomes paler B. The solution becomes deeper red C. No change — the system is already at equilibrium D. The solution turns colourless

Answer: B — The solution becomes deeper red

Adding more $\text{Fe}^{3+}$ (a reactant) disturbs the equilibrium. By Le Chatelier’s Principle, the system shifts right (forward) to oppose the increase in $[\text{Fe}^{3+}]$, consuming $\text{Fe}^{3+}$ and $\text{SCN}^-$ to produce more $[\text{FeSCN}]^{2+}$. Since $[\text{FeSCN}]^{2+}$ is deep red, the solution darkens. This is a classic demonstration of Le Chatelier’s Principle in the IB lab.

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Summary — Key Formulas and Definitions

Concept	Formula / Key Statement
Rate of reaction	$\text{Rate} = \dfrac{\Delta[\text{concentration}]}{\Delta t}$ in mol dm$^{-3}$ s$^{-1}$
Effective collision	Requires sufficient energy (≥ $E_a$) AND correct orientation
Activation energy ($E_a$)	Minimum energy for reaction to occur
Catalyst effect	Provides lower-$E_a$ alternative pathway; does not change $\Delta H$ or $K_c$
Dynamic equilibrium	Forward rate = reverse rate; concentrations constant; system is closed
Le Chatelier’s Principle	System opposes any imposed change; shifts to re-establish equilibrium
$K_c$ expression	$K_c = \dfrac{[\text{products}]^{\text{coeff}}}{[\text{reactants}]^{\text{coeff}}}$ (exclude pure solids/liquids)
$K_c$ and temperature	Only temperature changes $K_c$
Haber Process	$\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$, $\Delta H = -92$ kJ mol$^{-1}$; 400–500 °C, ~200 atm, Fe catalyst

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May 2026 Prediction Questions

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