Students conducted an experiment to determine ΔH for the reaction between sodium hydroxide and hydrochloric acid. The data from one student are shown in the table below. Mass of 100.0 mL of 0.50 mol L⁻¹ HCl Mass of 100.0 mL of 0.50 mol L⁻¹ NaOH Initial temperature of HCl solution Initial temperature of NaOH solution Final temperature of mixture Assume that all the solutions have the same specific heat capacity as water. (a) Calculate the heat energy released in this experiment. (b) A second student using the same procedure obtained 2.6 × 10³ J for the heat energy released in their experiment. Use this value to determine the enthalpy of neutralisation, ΔH, in kJ mol⁻¹, for the reaction shown. NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

SSCE HSC Chemistry Energy Changes in Chemical Reactions: Students conducted an experiment

Students conducted an experiment to determine ΔH for the reaction between sodium hydroxide and hydrochloric acid - HSC - SSCE Chemistry - Question 26 - 2022 - Paper 1

Question 26

Students-conducted-an-experiment-to-determine-ΔH-for-the-reaction-between-sodium-hydroxide-and-hydrochloric-acid-HSC-SSCE Chemistry-Question 26-2022-Paper 1.png

Students conducted an experiment to determine ΔH for the reaction between sodium hydroxide and hydrochloric acid. The data from one student are shown in the table b... show full transcript

Worked Solution & Example Answer:Students conducted an experiment to determine ΔH for the reaction between sodium hydroxide and hydrochloric acid - HSC - SSCE Chemistry - Question 26 - 2022 - Paper 1

Step 1

Calculate the heat energy released in this experiment.

96%

114 rated

Answer

To calculate the heat energy released, we can use the formula:

$q = mc\Delta T$

Where:

$q$ = heat energy (in joules)
$m$ = mass of the solution (in kg)
$c$ = specific heat capacity of water, $4.18 \times 10^3 \text{ J K}^{-1} \text{ kg}^{-1}$
$\Delta T$ = change in temperature

First, we find the change in temperature:

$\Delta T = T_{final} - T_{initial} = 24.4°C - 21.1°C = 3.3 K$

Next, we calculate the total mass of the final solution:

$m = 100.7 g + 102.0 g = 202.7 g = 0.2027 kg$

Now, substituting into the formula:

$q = 0.2027 kg \times 4.18 \times 10^3 \text{ J K}^{-1} \text{ kg}^{-1} \times 3.3 K = 2796 J = 2.8 kJ$

Step 2

Use this value to determine the enthalpy of neutralisation, ΔH, in kJ mol⁻¹, for the reaction shown.

99%

104 rated

Answer

To calculate the enthalpy of neutralisation, we use the heat energy from the second student's experiment:

The formula for enthalpy is given by:

$\Delta H = \frac{q}{n}$

where $q$ is the heat energy and $n$ is the number of moles of HCl.

From the data provided:

Heat energy ( $q$ ) = 2.6 × 10³ J = 2.6 kJ
Moles of HCl reacted = 0.1000 L × 0.50 mol L⁻¹ = 0.0500 mol

Substituting these values into the equation:

$\Delta H = \frac{2.6 \text{ kJ}}{0.0500 \text{ mol}} = 52 \text{ kJ mol}^{-1}$

Students conducted an experiment to determine ΔH for the reaction between sodium hydroxide and hydrochloric acid - HSC - SSCE Chemistry - Question 26 - 2022 - Paper 1

Worked Solution & Example Answer:Students conducted an experiment to determine ΔH for the reaction between sodium hydroxide and hydrochloric acid - HSC - SSCE Chemistry - Question 26 - 2022 - Paper 1

Calculate the heat energy released in this experiment.

Use this value to determine the enthalpy of neutralisation, ΔH, in kJ mol⁻¹, for the reaction shown.

Join the SSCE students using SimpleStudy...