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A solution calorimeter was calibrated by passing an electric current through the heating coil at a potential difference of 5.10 volts - VCE - SSCE Chemistry - Question 3 - 2011 - Paper 1

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A solution calorimeter was calibrated by passing an electric current through the heating coil at a potential difference of 5.10 volts. This caused the water in the c... show full transcript

Worked Solution & Example Answer:A solution calorimeter was calibrated by passing an electric current through the heating coil at a potential difference of 5.10 volts - VCE - SSCE Chemistry - Question 3 - 2011 - Paper 1

Step 1

a. Use the calibration factor to determine the electrical charge, in coulombs, that passed through the heating coil.

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Answer

To find the electrical charge (Q) that passed through the heating coil, we use the formula:

E=CF×ΔHE = CF \times \Delta H

where:

  • E = electrical energy in joules (J)
  • C = calibration factor (0.354 kJ °C⁻¹)
  • \Delta H = temperature change (9.50 °C)

First, convert the calibration factor into joules:

0.354 kJ °C⁻¹=354 J °C⁻¹0.354 \text{ kJ °C⁻¹} = 354 \text{ J °C⁻¹}

Now, calculate the energy:

E=354 J °C⁻¹×9.50°C=3363 JE = 354 \text{ J °C⁻¹} \times 9.50 °C = 3363 \text{ J}

Using the relationship between electrical energy, potential difference (V), and charge (Q):

E=V×QE = V \times Q

Rearranging for Q gives us:

Q=EV=3363 J5.10 V=659 CQ = \frac{E}{V} = \frac{3363 \text{ J}}{5.10 \text{ V}} = 659 \text{ C}

Step 2

b. i. Is this reaction exothermic or endothermic? Explain your answer.

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The reaction is exothermic. This is indicated by the increase in temperature of the calorimeter when the anhydrous calcium chloride dissolves in water. During an exothermic reaction, the system releases energy to the surroundings, resulting in a temperature increase. Since the temperature of the water increased, it shows that energy was released into the water.

Step 3

b. ii. Use the calibration factor to calculate the enthalpy change for the dissolution of 1.00 mol of CaCl₂(s).

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Answer

To calculate the enthalpy change (ΔH) for the dissolution of 1.00 mol of CaCl₂, we first determine the number of moles of CaCl₂:

n=6.038extg111.1extgmol−1=0.0543 moln = \frac{6.038 ext{ g}}{111.1 ext{ g mol⁻¹}} = 0.0543 \text{ mol}

Next, we can use the calibration factor to calculate the energy change for 1.00 mol:

Energy released=CF×n=0.354 kJ °C⁻¹×9.50°C=3.38extkJ\text{Energy released} = CF \times n = 0.354 \text{ kJ °C⁻¹} \times 9.50 °C = 3.38 ext{ kJ}

To find the enthalpy change per mole:

ΔH=3.38extkJ0.0543extmol=−62.20extkJmol−1\Delta H = \frac{3.38 ext{ kJ}}{0.0543 ext{ mol}} = -62.20 ext{ kJ mol⁻¹}

Since this reaction is exothermic, we express ΔH as a negative value, resulting in:

ΔH=−71.7extkJmol−1\Delta H = -71.7 ext{ kJ mol⁻¹}

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