Try these eight questions on paper before opening the answers. They cover the thermochemistry chapter and rise in difficulty. Use 4.2 J g⁻¹ °C⁻¹ for water and 1 g cm⁻³ for its density.
Questions
Question 1. When a solid dissolves, the temperature of the water rises. Is the dissolving exothermic or endothermic, and what is the sign of ΔH?
Answer
A rising temperature means heat is released to the water. The change is exothermic and ΔH is negative.
Question 2. An energy-level diagram shows reactants at 120 kJ, a peak at 180 kJ and products at 50 kJ. Find Ea and ΔH.
Answer
Ea = 180 − 120 = 60 kJ mol⁻¹.
ΔH = 50 − 120 = −70 kJ mol⁻¹. The reaction is exothermic.
Question 3. Another diagram shows reactants at 40 kJ, a peak at 110 kJ and products at 85 kJ. Find Ea and ΔH.
Answer
Ea = 110 − 40 = 70 kJ mol⁻¹.
ΔH = 85 − 40 = +45 kJ mol⁻¹. The reaction is endothermic.
Question 4. 25 cm³ of 2.0 mol dm⁻³ hydrochloric acid is mixed with 25 cm³ of 2.0 mol dm⁻³ sodium hydroxide. The temperature rises by 13.4 °C. Find ΔH.
Answer
Moles of each = 0.025 × 2.0 = 0.050 mol, so 0.050 mol of water forms.
Q = 50 × 4.2 × 13.4 = 2 814 J.
ΔH = −2 814 ÷ 0.050 = −56 280 J mol⁻¹ = −56.3 kJ mol⁻¹.
Question 5. A student records ΔH = +45 kJ mol⁻¹ for a reaction in which the temperature of the solution rose. Explain the error.
Answer
A temperature rise means heat was released, so the reaction is exothermic and ΔH must be negative.
The student should record a negative value, with the same size.
Question 6. Give two ways to reduce heat loss in a neutralisation experiment, and say how heat loss affects the result.
Answer
Use a polystyrene cup with a lid, and stir gently while recording the highest temperature.
Heat loss makes the temperature rise smaller, so the calculated heat and the size of ΔH are smaller than the true value.
Question 7. In Question 2, a catalyst lowers the peak to 155 kJ. Find the new Ea and ΔH.
Answer
New Ea = 155 − 120 = 35 kJ mol⁻¹.
ΔH is still −70 kJ mol⁻¹, because the reactants and products stay at the same levels.
Question 8. 0.020 mol of a salt dissolves in 100 cm³ of water and the temperature falls by 1.5 °C. Find ΔH.
Answer
Q = 100 × 4.2 × 1.5 = 630 J.
ΔH = +630 ÷ 0.020 = +31 500 J mol⁻¹ = +31.5 kJ mol⁻¹.
The sign is positive because the temperature fell.
If you got these wrong
- Questions 2, 3 and 7 test reading energy-level diagrams.
- Questions 1, 4, 5 and 8 test explaining sign conventions and sources of error.
- Question 6 also tests the error explanation in that lesson.
Log the step that broke in the mistake log and build a timed set with the timed original practice session builder. For a teacher to find where your reasoning stops, see online one-to-one Chemistry tuition.