Photo AI

A group of students carried out an investigation into the energy changes that take place when metal hydroxides dissolve in water - Scottish Highers Chemistry - Question 9 - 2022

Question icon

Question 9

A-group-of-students-carried-out-an-investigation-into-the-energy-changes-that-take-place-when-metal-hydroxides-dissolve-in-water-Scottish Highers Chemistry-Question 9-2022.png

A group of students carried out an investigation into the energy changes that take place when metal hydroxides dissolve in water. The following apparatus was used as... show full transcript

Worked Solution & Example Answer:A group of students carried out an investigation into the energy changes that take place when metal hydroxides dissolve in water - Scottish Highers Chemistry - Question 9 - 2022

Step 1

Using Graph 2 calculate the heat energy transferred to the water, in kJ, when the sodium hydroxide dissolved.

96%

114 rated

Answer

To find the heat energy transferred to the sodium hydroxide when it dissolved, we first need to extrapolate the cooling curve back to the time when the solute was added. After assessing the graph, we find the maximum temperature reached before it starts to cool down and the temperature just before the solute was added. Let’s say the initial temperature was identified as T_initial = 28.0 °C and the final temperature recorded after the dissolution is T_final = 22.0 °C. Thus, the temperature change (ΔT) can be calculated as:

ΔT=TfinalTinitial=22.028.0=6.0°CΔT = T_{final} - T_{initial} = 22.0 - 28.0 = -6.0 °C

Using the formula for heat energy:

Eh=cmΔTE_h = cmΔT

where c = 4.18 J/g°C (specific heat capacity of water), m = 100 g (since density of water is approximately 1 g/cm³), we can calculate:

Eh=4.18imes100imes(6.0)=2508extJ=2.51extkJE_h = 4.18 imes 100 imes (-6.0) = -2508 ext{ J} = -2.51 ext{ kJ}

Step 2

Suggest why the experiment was carried out in a polystyrene cup with a lid.

99%

104 rated

Answer

The experiment was conducted in a polystyrene cup with a lid to minimize heat loss to the surroundings. Polystyrene is an excellent insulator, helping maintain a stable temperature in the calorimeter. The lid helps to further reduce heat exchange with the environment, ensuring more accurate readings of the temperature changes during the dissolution process.

Step 3

Use this information to calculate the energy released, in kJ mol⁻¹, when one mole of potassium hydroxide dissolves in water.

96%

101 rated

Answer

To find the energy released per mole of potassium hydroxide (KOH), we first determine the molar mass of KOH:

  • Potassium (K): 39 g/mol
  • Oxygen (O): 16 g/mol
  • Hydrogen (H): 1 g/mol

Thus, molar mass of KOH = 39 + 16 + 1 = 56 g/mol.

Next, using the provided information, 5.61 g of KOH releases 5.25 kJ. We can calculate the energy released per mole using:

ext{Energy per mole} = rac{ ext{Energy released (kJ)}}{ ext{mass (g)}} imes ext{molar mass (g/mol)}

Substituting the known values gives us:

ext{Energy per mole} = rac{5.25 ext{ kJ}}{5.61 ext{ g}} imes 56 ext{ g/mol} = 52.5 ext{ kJ/mol}

Therefore, the energy released when one mole of potassium hydroxide dissolves is 52.5 kJ/mol.

Step 4

Calculate the enthalpy change, in kJ mol⁻¹, for the reaction above by using the data shown below.

98%

120 rated

Answer

To calculate the enthalpy change (ΔH) for the reaction of calcium hydroxide formation, we can utilize Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps. We have the reactions:

  1. H₂(g) + ½O₂(g) → H₂O(l)
    ΔH₁ = -286 kJ/mol

  2. Ca(s) + O₂(g) → Ca(OH)₂(s) + H₂(g)
    ΔH₂ = -986 kJ/mol

To find the enthalpy change for the formation of Ca(OH)₂ from its elements, we can reverse the first reaction and combine it with the second. Hence,

For the reverse of reaction 1: extCa(OH)2(s)+extH2(g)extCa(s)+½extO2(g) ext{Ca(OH)}₂(s) + ext{H}_2(g) → ext{Ca}(s) + ½ ext{O}_2(g) $

ΔH = +286 kJ/mol

Now add this to ΔH₂:

ΔHtotal=ΔHextreverse+ΔH2ΔH_{total} = ΔH_{ ext{reverse}} + ΔH_2 ΔHtotal=286kJ/mol986kJ/mol=700kJ/molΔH_{total} = 286 kJ/mol - 986 kJ/mol = -700 kJ/mol

Thus, the enthalpy change for the reaction is -700 kJ/mol.

Join the Scottish Highers students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;