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A student carried out an experiment to find the temperature rise for a reaction between hydrochloric acid and sodium hydroxide solution - AQA - A-Level Chemistry - Question 4 - 2018 - Paper 3
Question 4
A student carried out an experiment to find the temperature rise for a reaction between hydrochloric acid and sodium hydroxide solution. - The student used a measur... show full transcript
Worked Solution & Example Answer:A student carried out an experiment to find the temperature rise for a reaction between hydrochloric acid and sodium hydroxide solution - AQA - A-Level Chemistry - Question 4 - 2018 - Paper 3
Step 1
0 4.1 -1 Plot a graph of temperature against time
Answer
To plot the graph, use the given time (min) on the x-axis and temperature (°C) on the y-axis. Each point from Table 3 will be plotted: (0, 19.8), (1, 19.8), (2, 19.8), (3, 19.8), (4, 21.1), (5, 21.6), (6, 21.5), (7, 21.4), (8, 21.3), (9, 21.2), (10, 21.1), (11, 21.0), (12, 21.1). Draw the lines of best fit between the first four points (0 to 3 min) and then again through the points from (4 to 12 min). The temperature rise, ΔT, at the fourth minute is calculated as follows: ΔT = Temperature at 4 min - Temperature at 3 min = 21.1 °C - 19.8 °C = 1.3 °C.
Step 2
0 4.1 -2 The uncertainty in each of the temperature readings
Answer
To calculate the percentage uncertainty in the temperature rise, first note that the uncertainty in the thermometer readings is ±0.1 °C. The formula for percentage uncertainty is:
ext{Percentage Uncertainty} = rac{ ext{Uncertainty}}{ ext{Value}} imes 100
In this case:
ext{Percentage Uncertainty} = rac{0.1}{1.3} imes 100 ext{ = 7.69 ext{ extperthousand} }
Step 3
Step 4
0 4.1 -4 Suggest and explain another change
Answer
Another change to decrease the percentage uncertainty could be to increase the magnitude of the temperature change being measured. A larger temperature change results in a higher signal-to-noise ratio, leading to more precise measurements.
Step 5
0 4.1 -5 Give an equation for the reaction
Answer
The balanced equation for the reaction between ethanoic acid (HOOC(CHOH)) and potassium hydroxide (KOH) is:
To calculate the enthalpy change (ΔH) per mole of water formed, first calculate the heat (q) absorbed using the formula:
Where:
- m = mass of solution (assuming the density is 1.00 g/cm³, the mass of 100 cm³ is approximately 100 g)
- c = specific heat capacity (4.2 J K⁻¹ g⁻¹)
- ΔT = change in temperature (3.2 °C)
Thus, q = 100 g imes 4.2 rac{J}{g imes K} imes 3.2 K = 1344 J
Now, for 0.080 mol of ethanoic acid used: The enthalpy change per mole of water formed is given by
Where n = 0.025 mol (since the reaction produces 1 mole of water for every mole of acid used), leading to:
Step 6
0 4.1 -6 Suggest an explanation for the difference
Answer
The difference in enthalpy values can be explained by the strength of the acids involved. Strong acids like sulfuric acid fully dissociate in solution, releasing more energy during the neutralization process compared to weaker acids like ethanoic acid, which do not fully dissociate. Therefore, different acid strengths lead to varying enthalpy changes during neutralization.