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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 determine the electrical charge passed through the heating coil, we can use the relationship between energy, voltage, and charge:

E=V⋅QE = V \cdot Q

Where:

  • EE is the electrical energy in joules,
  • VV is the potential difference in volts,
  • QQ is the electrical charge in coulombs.

Given:

  • Voltage, V=5.10extVV = 5.10 ext{ V}
  • Temperature change, ΔT=9.50ext°C\Delta T = 9.50 ext{ °C}
  • Calibration factor, C=0.354 kJ/°C=354extJ/°CC = 0.354 \text{ kJ/°C} = 354 ext{ J/°C}

Calculating the energy:

E=C⋅ΔT=354extJ/°C⋅9.50ext°C=3363extJE = C \cdot \Delta T = 354 ext{ J/°C} \cdot 9.50 ext{ °C} = 3363 ext{ J}

Now using this energy to find the charge:

Q=EV=3363extJ5.10extV≈658.43extCQ = \frac{E}{V} = \frac{3363 ext{ J}}{5.10 ext{ V}} \approx 658.43 ext{ C}

Therefore, the charge that passed through the heating coil is approximately 659 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 water after adding CaCl₂ to it. In exothermic reactions, energy is released into the surroundings, which causes the temperature of the surroundings (i.e., the water) to rise. As a result, the dissolution of CaCl₂ releases heat energy, leading to a temperature increase, confirming that the reaction is exothermic.

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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First, we need to calculate the number of moles of CaCl₂:

n=massmolar mass=6.038extg111.1extg/mol≈0.0543extmoln = \frac{\text{mass}}{\text{molar mass}} = \frac{6.038 ext{ g}}{111.1 ext{ g/mol}} \approx 0.0543 ext{ mol}

Using the calibration factor to find the energy change:

Knowledge from part 'a' gives us an energy change of 3.39 kJ for the dissolution of 0.0543 mol of CaCl₂. Thus, to find the enthalpy change per mole:

ΔH=3.39extkJ0.0543extmol≈62.4extkJ/mol\Delta H = \frac{3.39 ext{ kJ}}{0.0543 ext{ mol}} \approx 62.4 ext{ kJ/mol}

Hence, the enthalpy change for the dissolution of 1.00 mol of CaCl₂(s) is approximately -62.4 kJ/mol.

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