Home / H2 Chemistry 9476 / H2 Chemistry (9476): Chemical Equilibria H2 Chemistry (9476): Chemical Equilibria Dynamic equilibrium, Le Chatelier’s principle, Kc, Kp, equilibrium composition and the Haber process.
Learning goals Dynamic Equilibrium and Le Chatelier Equilibrium Constants (Kc and Kp) Equilibrium Composition Calculations Haber Process (Case Study) H2 Chemical Equilibria: Learning outcomes This hub combines the explanation questions (Le Chatelier) and the calculation questions (Kc/Kp + equilibrium composition).
Before you start Beginner path: follow the Lessons (Recommended Order) from top to bottom.
Revision: Jump to: Quick Reference · What You Must Memorise · Common Exam Traps · Hub Quiz .
Prerequisites: skim the Prerequisites list first if you’re rusty.
Useful links: A Level portal · Exam Skills
What You’ll Learn
Use dynamic equilibrium language correctly and predict shifts using Le Chatelier’s principle.
Write correct K_c Kₚ
Combine calculation skill with exam explanation phrasing (not “it shifts right” without a reason).
Syllabus statements covered explain, in terms of rates of the forward and reverse reactions, what is meant by a reversible reaction and dynamic equilibrium state Le Chatelier’s Principle and apply it to deduce qualitatively (from appropriate information) the effects of changes in concentration, pressure or temperature, on a system at equilibrium deduce whether changes in concentration, pressure or temperature or the presence of a catalyst affect the value of the equilibrium constant for a reaction deduce expressions for equilibrium constants in terms of concentrations, Kc, and partial pressures, Kp [treatment of the relationship between Kp and Kc is not required] calculate the values of equilibrium constants in terms of concentrations or partial pressures from appropriate data calculate the quantities present at equilibrium, given appropriate data (such calculations will not require the solving of quadratic equations) describe and explain the conditions used in the Haber process, as an example of the importance of an understanding of chemical equilibrium in the chemical industry Quick Reference
What You Must Memorise
Dynamic equilibrium : forward rate = reverse rate (concentrations constant, not necessarily equal).Le Chatelier : equilibrium shifts to oppose a disturbance (disturbance → shift → named species increases/decreases).Only temperature changes K (concentration/pressure/catalyst do not change K Writing K_c/Kₚ : products over reactants, powers from coefficients, exclude pure solids/liquids.ICE tables : use coefficients (e.g. -2x K_c Haber trade-off trio : equilibrium yield vs rate vs economics/safety. Lessons (Recommended Order)
Dynamic Equilibrium and Le Chatelier Predict shifts (concentration/pressure/temperature) correctly.
Equilibrium Constants (Kc, Kp) Write K expressions and interpret magnitude.
Equilibrium Composition Calculations ICE tables, approximations, sanity checks.
Haber Process (Case Study) Yield vs rate vs economics trade-offs.
Common Exam Traps
Writing K_c
Changing K_c K
Mixing “rate” language with “equilibrium position” language.
Pressure questions: forgetting to count total moles of gas on each side first.
ICE tables: forgetting coefficient changes (e.g. -2x K_c
Kₚ Kₚ Vague Le Chatelier: writing “shifts right” with no named species increasing/decreasing.
Hub Quiz and Check Your Understanding
Use the quiz without notes first. Classify errors as equilibrium language, disturbance logic, K expression, ICE-table stoichiometry or industrial compromise.
H2 Chemical Equilibria Quiz Check Le Chatelier reasoning, Kc/Kp, equilibrium composition and Haber-process decisions.
H2 Chemical Equilibria Knowledge Check Find the reasoning step that needs more work, use the feedback, then try a fresh question independently.
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Hydrogen and iodine only are sealed in a flask at 700 K and react: H₂(g) + I₂(g) ⇌ 2HI(g). Which statement describes the rates as the mixture approaches equilibrium?
The forward rate falls and the reverse rate rises from zero until the two rates are equal; both reactions then continue. The forward rate falls to zero at equilibrium, and the reverse rate stays at zero. Both rates rise until the concentrations of H₂, I₂ and HI are equal. The reverse reaction begins only after all the H₂ and I₂ has reacted. N₂O₄(g) ⇌ 2NO₂(g), ΔH = +58 kJ mol⁻¹. The equilibrium mixture is compressed to a smaller volume at constant temperature. How does the position of equilibrium change?
It shifts to the left, towards the side with fewer gas molecules, which partly opposes the pressure increase. It shifts to the right, because the higher pressure makes collisions more frequent. It does not change, because the temperature is constant. It shifts to the right, because the forward reaction is endothermic. 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −197 kJ mol⁻¹. Which change alters the value of Kc?
Raising the temperature, which decreases Kc. Raising the pressure, which increases Kc. Adding a catalyst, which increases Kc. Removing SO₃ as it forms, which decreases Kc. 2NO(g) + O₂(g) ⇌ 2NO₂(g). At equilibrium a 1.00 dm³ vessel contains 0.20 mol NO, 0.10 mol O₂ and 0.40 mol NO₂. What is the value of Kc?
40 mol⁻¹ dm³ 20 mol⁻¹ dm³ 0.025 mol dm⁻³ 40 mol dm⁻³ 1.00 mol CO and 3.00 mol H₂ are sealed in a vessel: CO(g) + 2H₂(g) ⇌ CH₃OH(g). At equilibrium the vessel contains 0.40 mol CH₃OH. What amounts of CO and H₂ are present at equilibrium?
0.60 mol CO and 2.20 mol H₂ 0.60 mol CO and 2.60 mol H₂ 0.60 mol CO and 2.80 mol H₂ 0.40 mol CO and 0.80 mol H₂ N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹. Why is the Haber process operated at about 450 °C rather than a much lower temperature?
A lower temperature would give a higher equilibrium yield, but equilibrium would be reached too slowly even with the iron catalyst; 450 °C is a compromise between yield and rate. A higher temperature gives a higher equilibrium yield of ammonia. The iron catalyst increases the equilibrium yield only when it is hot. Below 450 °C the forward reaction becomes endothermic.
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A little radioactive iodine is added to an equilibrium mixture of H₂(g), I₂(g) and HI(g) at constant temperature. Later, radioactivity is found in both I₂ and HI, but the concentrations of all three substances are unchanged. What does this show?
The forward and reverse reactions are both still happening, at equal rates. The position of equilibrium has shifted to the right. The mixture had not yet reached equilibrium. The radioactive iodine acts as a catalyst for the reaction. CO(g) + 2H₂(g) ⇌ CH₃OH(g), ΔH = −91 kJ mol⁻¹. The temperature of the equilibrium mixture is raised at constant pressure. What happens to the equilibrium amount of CH₃OH?
It decreases, because the position shifts in the endothermic (reverse) direction to absorb the added heat. It increases, because both reactions become faster. It is unchanged, because temperature affects only the rate. It increases, because the position shifts to the side with fewer gas molecules. N₂O₄(g) ⇌ 2NO₂(g), ΔH = +58 kJ mol⁻¹. Which change increases the value of Kc?
Raising the temperature. Lowering the pressure at constant temperature. Removing NO₂ as it forms at constant temperature. Adding a catalyst. 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). The equilibrium concentrations are [SO₂] = 0.20, [O₂] = 0.10 and [SO₃] = 0.60 mol dm⁻³. What is the value of Kc?
90 mol⁻¹ dm³ 30 mol⁻¹ dm³ 90 mol dm⁻³ 0.011 mol dm⁻³ 0.80 mol H₂S is heated in a sealed vessel: 2H₂S(g) ⇌ 2H₂(g) + S₂(g). At equilibrium 25% of the H₂S has dissociated. What amount of S₂ is present at equilibrium?
0.10 mol 0.20 mol 0.40 mol 0.60 mol In the Haber process a pressure of about 200 atm is used. A higher pressure would give a higher equilibrium yield of ammonia. Why is a much higher pressure not used?
The extra yield does not justify the higher cost of stronger plant and of the energy for compression, and the greater safety risk. Higher pressure would decrease Kc, lowering the yield. Higher pressure would shift the position to the left, towards more gas molecules. Higher pressure would poison the iron catalyst. CaCO₃(s) ⇌ CaO(s) + CO₂(g). At 1100 K the equilibrium partial pressure of carbon dioxide is 0.45 atm. What is Kp at this temperature?
0.45 atm 2.2 atm⁻¹ It cannot be found without the amounts of CaCO₃ and CaO. 0.45, with no units
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A little radioactive iodine is added to an equilibrium mixture of H₂(g), I₂(g) and HI(g) at constant temperature. Later, radioactivity is found in both I₂ and HI, but the concentrations of all three substances are unchanged. What does this show?
The forward and reverse reactions are both still happening, at equal rates. The position of equilibrium has shifted to the right. The mixture had not yet reached equilibrium. The radioactive iodine acts as a catalyst for the reaction. CO(g) + 2H₂(g) ⇌ CH₃OH(g), ΔH = −91 kJ mol⁻¹. The temperature of the equilibrium mixture is raised at constant pressure. What happens to the equilibrium amount of CH₃OH?
It decreases, because the position shifts in the endothermic (reverse) direction to absorb the added heat. It increases, because both reactions become faster. It is unchanged, because temperature affects only the rate. It increases, because the position shifts to the side with fewer gas molecules. N₂O₄(g) ⇌ 2NO₂(g), ΔH = +58 kJ mol⁻¹. Which change increases the value of Kc?
Raising the temperature. Lowering the pressure at constant temperature. Removing NO₂ as it forms at constant temperature. Adding a catalyst. 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). The equilibrium concentrations are [SO₂] = 0.20, [O₂] = 0.10 and [SO₃] = 0.60 mol dm⁻³. What is the value of Kc?
90 mol⁻¹ dm³ 30 mol⁻¹ dm³ 90 mol dm⁻³ 0.011 mol dm⁻³ 0.80 mol H₂S is heated in a sealed vessel: 2H₂S(g) ⇌ 2H₂(g) + S₂(g). At equilibrium 25% of the H₂S has dissociated. What amount of S₂ is present at equilibrium?
0.10 mol 0.20 mol 0.40 mol 0.60 mol In the Haber process a pressure of about 200 atm is used. A higher pressure would give a higher equilibrium yield of ammonia. Why is a much higher pressure not used?
The extra yield does not justify the higher cost of stronger plant and of the energy for compression, and the greater safety risk. Higher pressure would decrease Kc, lowering the yield. Higher pressure would shift the position to the left, towards more gas molecules. Higher pressure would poison the iron catalyst. CaCO₃(s) ⇌ CaO(s) + CO₂(g). At 1100 K the equilibrium partial pressure of carbon dioxide is 0.45 atm. What is Kp at this temperature?
0.45 atm 2.2 atm⁻¹ It cannot be found without the amounts of CaCO₃ and CaO. 0.45, with no units
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