📋 Structured Questions
These questions test key concepts from Section 3. Attempt each question on paper, then click "Show Mark Scheme" to check your answer.
Question 1: Calorimetry Experiment
4 marksA student carries out a calorimetry experiment to measure the temperature change when solid lithium chloride (LiCl) is dissolved in water. 2.0 g of lithium chloride is dissolved in 50.0 g of water. The temperature of the water rises by 6.5 degrees Celsius.
(a) Calculate the heat energy change (Q), in Joules, for this dissolving process.
(Specific heat capacity of water = 4.18 J/g/°C. Assume the specific heat capacity of the solution is the same as water.) [2]
(b) Calculate the molar enthalpy change (delta H), in kJ/mol, for dissolving lithium chloride. Include a sign in your answer.
(Relative formula mass: LiCl = 42.5) [2]
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(a)
- Calculation: Q = 50.0 g x 4.18 J/g/°C x 6.5°C [1]
- Q = 1358.5 Joules (or 1360 J) [1]
(b)
- Calculate moles: 2.0 g / 42.5 = 0.0471 moles [1]
- Calculate delta H: - (1358.5 / 1000) / 0.0471 = -28.9 kJ/mol (or -28.8 kJ/mol depending on rounding) [1]
Question 2: Reaction Profiles
4 marks(a) Describe the key features you would draw on a reaction profile diagram for an endothermic reaction. Explain where the reactants and products are placed relative to one another, and how you would represent the activation energy (Ea) and the enthalpy change (delta H). [4]
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- The reactants must be placed on a horizontal line at a lower energy level than the products [1]
- A curve rises from the reactants energy level to a peak and then drops to the products energy level [1]
- The activation energy (Ea) is represented by an arrow pointing upwards from the reactants energy level to the peak of the curve [1]
- The enthalpy change (delta H) is represented by a vertical arrow pointing upwards from the reactants energy level to the products energy level [1]
Question 3: Bond Energy Calculations
3 marksThe equation for the combustion of methane is:
CH4 + 2O2 → CO2 + 2H2O
(a) Calculate the enthalpy change (delta H), in kJ/mol, for this reaction using the following bond energies:
C-H = 413 kJ/mol, O=O = 498 kJ/mol, C=O = 805 kJ/mol, O-H = 464 kJ/mol. [3]
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- Energy required to break reactant bonds: (4 x C-H) + (2 x O=O) = (4 x 413) + (2 x 498) = 1652 + 996 = 2648 kJ/mol [1]
- Energy released in making product bonds: (2 x C=O) + (4 x O-H) = (2 x 805) + (4 x 464) = 1610 + 1856 = 3466 kJ/mol [1]
- delta H = 2648 - 3466 = -818 kJ/mol [1]
Question 4: Collision Theory & Temperature
4 marks(a) Explain, in terms of collision theory, why increasing the temperature of a reaction mixture increases the rate of reaction. [4]
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- Reactant particles gain kinetic energy and move faster [1]
- This increases the frequency of collisions / particles collide more often [1]
- A higher proportion of particles possess energy greater than or equal to the activation energy [1]
- This increases the rate / frequency of successful collisions per second [1]
Question 5: Compromise Conditions in the Haber Process
4 marksThe industrial production of ammonia by the Haber process uses a compromise temperature of 450 degrees Celsius and a pressure of 200 atmospheres:
N2(g) + 3H2(g) ⇌ 2NH3(g) delta H = -92 kJ/mol
(a) Explain why these specific conditions are used in industry. [4]
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- The forward reaction is exothermic, so a lower temperature would shift the equilibrium to the right to increase the yield of ammonia [1]
- However, a temperature of 450°C is used because lower temperatures would make the reaction rate too slow [1]
- A higher pressure shifts the equilibrium to the right (towards the side with fewer gas molecules: 2 moles vs 4 moles) to increase the yield and rate [1]
- A pressure of 200 atmospheres is used because higher pressures are extremely expensive to generate safely and require specialized equipment [1]
Question 6: Rates and Catalysts
6 marks(a) A student investigates the decomposition of hydrogen peroxide using manganese(IV) oxide as a catalyst. Compare the rate of reaction using a catalyst to the rate without a catalyst. Explain how a catalyst works, describe the shape of the volume of gas vs time graph, and explain why the reaction eventually stops. [6]
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Indicative content:
- How a catalyst works: A catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy. It remains chemically unchanged at the end of the reaction.
- Graph comparison and shape:
- With the catalyst, the reaction is faster, meaning the curve on the volume of gas vs time graph is steeper at the start.
- Both curves are steepest at the beginning (highest rate), then become less steep as the rate decreases, and eventually plateau (slope = 0).
- Since the same amount of hydrogen peroxide is used, both reactions will eventually produce the exact same final volume of oxygen gas (the plateaus will be at the same height).
- Why it stops: The reaction rate decreases because the concentration of hydrogen peroxide decreases as it is consumed. The reaction stops when all the hydrogen peroxide has reacted.
Marking guidance:
- 5 to 6 marks: Detailed explanation of how a catalyst works (alternative pathway, lower Ea). Compares the slopes of the graphs (steeper with catalyst) and explains that the final gas volume remains the same. Explains that the rate decreases and stops due to reactants being used up.
- 3 to 4 marks: Explains how a catalyst works and describes the graph shape, but with less detail on why it slows down or why the final volume is identical.
- 1 to 2 marks: Simple statement that a catalyst makes the reaction faster or that the graph levels off when finished.