Question 1:

The correct answer is equal to.

Explanation:
In a dynamic equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction.

This does not mean the concentrations of reactants and products are equal—rather, they are constant because the two reactions are occurring at the same rate in opposite directions. The system appears static at the macroscopic level, but at the molecular level, both reactions continue to happen simultaneously (hence "dynamic").

So the completed sentence is:
"When a reaction is in dynamic equilibrium the rate of the forward reaction is equal to the rate of the reverse reaction."

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 2:

The correct answer is D. constant.

Explanation:
In a dynamic equilibrium, the concentrations of reactants and products are constant (they do not change over time). This is because the forward and reverse reactions are occurring at the same rate, so any molecules of reactants that are converted into products are balanced by an equal number of product molecules converting back into reactants.

However, constant does NOT mean equal. The actual concentrations of reactants and products at equilibrium depend on the equilibrium position (which is determined by the equilibrium constant, Kc, and the reaction conditions). They could be equal in a rare case, but in general, they are not—they simply remain steady.

Here is why the other options are incorrect:

• A. Equal – Incorrect. The concentrations of reactants and products are not necessarily equal at equilibrium. They are constant but usually have different values.

• B. Unequal with the reactant concentration greater than the product concentration – Incorrect. While this can be true for some reactions (where the equilibrium lies to the left), it is not always true. It depends on the specific reaction and conditions.

• C. Unequal with the product concentration greater than the reactant concentration – Incorrect. This can be true for some reactions (equilibrium lies to the right), but it is not always true for all equilibria.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 3:

The correct answer is A. solid turning from white to blue.

Explanation:
The equation shows:

• CuSO₄ (s) – anhydrous copper(II) sulfate, which is white.

• CuSO₄·5H₂O (s) – hydrated copper(II) sulfate, which is blue.

If the reaction moves in the forward direction (left to right), anhydrous copper(II) sulfate reacts with water to form hydrated copper(II) sulfate. Therefore, you would observe the white solid turning blue.

Here is why the other options are incorrect:

• B. Solid dissolving – Incorrect. While copper(II) sulfate can dissolve in water, this reversible reaction specifically involves the formation of a hydrated solid, not dissolving in solution.

• C. Solid turning from blue to white – Incorrect. This would be the reverse reaction (heating the blue hydrated salt to drive off water, leaving white anhydrous salt).

• D. Condensation forming on the sides of the test tube – Incorrect. Condensation would occur in the reverse reaction, where water vapour is driven off and condenses on the cooler sides of the test tube when heating the blue solid.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 4:

The correct answer is D. decreases because the reverse reaction is endothermic.

Explanation:
The forward reaction (N₂ + 3H₂ → 2NH₃) is exothermic (ΔH = -92.4 kJ, meaning heat is released).
Therefore, the reverse reaction (2NH₃ → N₂ + 3H₂) is endothermic (absorbs heat).

According to Le Chatelier's principle, if you increase the temperature, the system will shift to oppose the change by favouring the endothermic direction (the reverse reaction), because that absorbs the excess heat. This means the equilibrium shifts to the left, converting ammonia back into nitrogen and hydrogen, which decreases the yield of ammonia.

Here is why the other options are incorrect:

• A. increases because the forward reaction is endothermic – Incorrect. The forward reaction is exothermic, not endothermic. Also, increasing temperature would not increase yield for an exothermic forward reaction.

• B. increases because the reverse reaction is endothermic – Incorrect. While it correctly states that the reverse reaction is endothermic, increasing temperature favours the endothermic reverse reaction, which decreases (not increases) ammonia yield.

• C. decreases because the forward reaction is endothermic – Incorrect. The forward reaction is exothermic, not endothermic. The reason for the decrease is that the reverse reaction is endothermic, not the forward.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 5:

The correct answer is A. increases because there are less moles of gas on the ‘products’ side.

Explanation:
For the Haber process:

• Reactants side: 1 mole of N₂ + 3 moles of H₂ = 4 moles of gas

• Products side: 2 moles of NH₃ = 2 moles of gas

According to Le Chatelier's principle, increasing the pressure favours the side of the reaction with fewer moles of gas (because this reduces the overall pressure). Since the products side has fewer moles of gas (2 moles vs 4 moles), the equilibrium shifts to the right (forward direction), which increases the yield of ammonia.

Here is why the other options are incorrect:

• B. increases because there are less moles of gas on the ‘reactants’ side – Incorrect. The reactants side has more moles of gas (4 moles), not fewer. If there were fewer on the reactants side, increasing pressure would favour the reactants and decrease ammonia yield.

• C. decreases because there are less moles of gas on the ‘products’ side – Incorrect. While it correctly states that there are fewer moles on the products side, increasing pressure would favour that side, increasing (not decreasing) the yield.

• D. decreases because there are less moles of gas on the ‘reactants’ side – Incorrect. The reactants side has more moles of gas, not fewer, and increasing pressure does not decrease the yield in this case.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 6:

Explanation:

1. Increasing the temperature:
   The reaction is endothermic in the forward direction (bonds are broken to form NO₂). According to Le Chatelier's principle, increasing the temperature favors the endothermic direction to absorb the added heat. Therefore, the equilibrium moves to the right (toward more NO₂).

2. Increasing the pressure:
   The reaction involves a change in the number of gas moles:

   • Left side: 1 mole of gas (N₂O₄)

   • Right side: 2 moles of gas (NO₂)
     Increasing the pressure favors the side with fewer gas moles to reduce the pressure. Thus, the equilibrium moves to the left (toward N₂O₄).

So, the effect is: Moves to the right for temperature, Moves to the left for pressure.

Answer: C

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 7:

For this reaction:

H₂(g) + Cl₂(g) ⇌ 2HCl(g)

If the pressure is increased, the equilibrium position will not change (it will remain constant).

Explanation:

• Count the number of moles of gas on each side of the equation:

  • Left side (reactants): 1 mole of H₂ + 1 mole of Cl₂ = 2 moles of gas.

  • Right side (products): 2 moles of HCl = 2 moles of gas.

• Since there are equal numbers of gas moles on both sides (2 ⇌ 2), changing the pressure does not favor either the forward or the reverse reaction.

• According to Le Chatelier's principle, increasing pressure shifts the equilibrium toward the side with fewer gas moles. Because both sides have the same number, there is no shift—the equilibrium position remains unchanged.

Final answer: The equilibrium position does not move (remains constant).

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 8:

The correct answer is B: low temperature and high pressure.

Explanation:

1. Temperature (Low):
   The reaction has a negative ΔH (-385 kJ), meaning the forward reaction (producing SO₃) is exothermic. According to Le Chatelier's principle, lowering the temperature favors the exothermic direction to release heat. Therefore, a low temperature shifts the equilibrium to the right, increasing the yield of SO₃.

2. Pressure (High):
   Count the moles of gas on each side:

   • Left side (reactants): 2 moles of SO₂ + 1 mole of O₂ = 3 moles of gas.

   • Right side (products): 2 moles of SO₃ = 2 moles of gas.
     Increasing the pressure favors the side with fewer gas moles to reduce the pressure. Since the right side has 2 moles compared to 3 on the left, high pressure shifts the equilibrium to the right, increasing the yield of SO₃.

So, to achieve a high yield of SO₃, you would use low temperature and high pressure.

Answer: B

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 9:

Answer: C
Both the carbon monoxide concentration and hydrogen concentration decrease.

Reason:
The forward reaction is endothermic (+210 kJ). Lowering the temperature from 1000 °C to 500 °C favors the exothermic reverse reaction, shifting equilibrium to the left. This consumes CO and H₂, so their concentrations both decrease.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 10:

The correct answer is C: decreases (for both Carbon Monoxide and Hydrogen).

Reason:
Count the moles of gas on each side:

• Left (reactants): 1 (CH₄) + 1 (H₂O) = 2 moles

• Right (products): 1 (CO) + 3 (H₂) = 4 moles

Increasing the pressure favors the side with fewer gas moles to reduce the pressure. That is the left (reactant) side.

So the equilibrium shifts to the left, consuming CO and H₂. Therefore, both concentrations decrease.

Answer: C

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.