Free Chemical Equilibrium MCQs with Answers

1. A chemical system is said to be in dynamic equilibrium when

• A. both reactions have stopped completely
• B. the forward and reverse reactions continue at equal rates
• C. the amounts of reactants and products are equal
• D. all the reactants have been converted to products

Explanation: The word dynamic means both reactions are still running; they simply cancel out, so the concentrations stay constant while molecules continue to convert in both directions. Equal concentrations of reactant and product are not required and are usually not the case, which is the standard misconception. Equilibrium can only be reached in a closed system.

2. For the reaction N2 + 3H2 gives 2NH3, the equilibrium constant expression Kc is

• A. [N2][H2]^3 divided by [NH3]^2
• B. [NH3] divided by [N2][H2]
• C. [NH3]^2 divided by [N2][H2]^3
• D. 2[NH3] divided by [N2] + 3[H2]

Explanation: Kc is written as the product concentrations over the reactant concentrations, each raised to the power of its coefficient in the balanced equation. Inverting the expression gives the equilibrium constant for the reverse reaction, which is why the first option is offered. Coefficients become exponents, never multipliers, which rules out the last option.

3. A very large value of the equilibrium constant Kc indicates that

• A. the reaction reaches equilibrium very quickly
• B. the reaction is exothermic
• C. the reactants are favoured at equilibrium
• D. the products are strongly favoured at equilibrium

Explanation: Kc compares product to reactant concentrations, so a large value means the equilibrium position lies far to the right and the reaction goes almost to completion. Kc says nothing at all about how fast equilibrium is reached, which is a matter of kinetics, and this is the confusion the question targets. The Haber process is the classic case of a favourable equilibrium that is nevertheless slow without a catalyst.

4. Le Chatelier's principle states that when a stress is applied to a system at equilibrium, the system

• A. shifts in the direction that relieves the stress
• B. shifts in the direction that increases the stress
• C. stops reacting altogether
• D. always shifts towards the products

Explanation: The position of equilibrium moves so as to partly oppose the change imposed, whether that change is in concentration, pressure or temperature. Adding a reactant therefore drives the reaction forward, and removing a product does the same. The principle predicts direction only; it says nothing about how far or how fast the shift occurs.

5. For the equilibrium N2 + 3H2 gives 2NH3, which is exothermic, the yield of ammonia is increased by

• A. raising the temperature and lowering the pressure
• B. raising the pressure and lowering the temperature
• C. adding a catalyst
• D. removing nitrogen from the mixture

Explanation: There are four moles of gas on the left and two on the right, so high pressure pushes the equilibrium towards the smaller volume, and since the forward reaction releases heat a lower temperature also favours it. A catalyst speeds the approach to equilibrium without changing the yield at all, which is the standard trap. In practice a compromise temperature near 450 degrees Celsius is used because the reaction is too slow when cold.

6. Increasing the pressure has NO effect on the position of the equilibrium H2 + I2 gives 2HI because

• A. the reaction is exothermic
• B. hydrogen iodide is a gas
• C. the number of moles of gas is the same on both sides
• D. the reaction has no catalyst

Explanation: Pressure changes shift an equilibrium only when the two sides differ in the number of gas molecules, and here there are two moles on each side, so neither direction relieves the stress. The rate of reaching equilibrium does rise with pressure, but the position stays put. Temperature changes will still shift this equilibrium, since the two directions differ in enthalpy.

7. Adding a catalyst to a system at equilibrium

• A. shifts the equilibrium towards the products
• B. shifts the equilibrium towards the reactants
• C. increases the value of Kc
• D. speeds up the forward and reverse reactions equally, so equilibrium is reached sooner but is unchanged

Explanation: A catalyst lowers the activation energy of both directions by the same amount, so both rates rise in proportion and they still become equal at exactly the same position. What changes is the time taken to get there, which is commercially vital in a process such as the Haber synthesis. Only temperature changes the value of Kc itself.

8. In the Haber process for manufacturing ammonia, the catalyst used is

• A. finely divided iron with promoters
• B. platinum
• C. vanadium pentoxide
• D. nickel

Explanation: Iron with small amounts of potassium and aluminium oxide as promoters is used because it is effective and cheap enough for a process operating on this scale. Vanadium pentoxide is the catalyst of the Contact process for sulphuric acid and nickel is used in the hydrogenation of oils, both of which appear here as familiar but wrong answers. The gases are recycled repeatedly since a single pass converts only about fifteen per cent.

9. The conditions actually used in the Haber process are about

• A. 1000 degrees Celsius and 1 atmosphere
• B. 25 degrees Celsius and 1000 atmospheres
• C. 450 degrees Celsius and 200 atmospheres
• D. 450 degrees Celsius and 1 atmosphere

Explanation: A moderately high temperature is a compromise, since a low temperature would give a better yield but far too slowly for industry, while high pressure genuinely improves the yield and is limited only by the cost and strength of the plant. Ammonia is removed by liquefaction and the unreacted gases are recycled, so the overall conversion is high even though each pass is poor.

10. The solubility product Ksp of a sparingly soluble salt is

• A. the total mass of the salt dissolved
• B. the product of the concentrations of its ions in a saturated solution, each raised to the power of its coefficient
• C. the concentration of the undissolved solid
• D. the ratio of dissolved to undissolved salt

Explanation: For a saturated solution in contact with excess solid, an equilibrium exists between the solid and its ions, and Ksp is the equilibrium constant for that dissolving process. The concentration of the solid itself is constant and so is not included in the expression. If the ionic product exceeds Ksp a precipitate forms, which is the basis of qualitative analysis.