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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 calculate the electrical charge (Q) that passed through the heating coil, we use the relationship between electrical energy, potential difference, charge, and the calibration factor:

  1. Calculate Energy (E):

    E=CF×ΔTE = CF \times \Delta T

    Where:

    • C = Calibration factor = 0.354 J°C⁻¹

    • (\Delta T = 9.50 °C)

    Plugging in the values:

    E=0.354×9.50=3.36JE = 0.354 \times 9.50 = 3.36 J

  2. Find Charge (Q):

    We also know that:

    E=V×QE = V \times Q Where:

    • V = Potential difference = 5.10 V

    Rearrange the equation to solve for Q:

    Q=EV=3.365.10≈0.659 CQ = \frac{E}{V} = \frac{3.36}{5.10} \approx 0.659 \text{ C}

Step 2

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

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Answer

The reaction is exothermic.

As the calcium chloride (CaCl₂) dissolves in water, it releases heat, demonstrated by the observed increase in the temperature of the solution. The temperature of the water increased, indicating that energy was released during the dissolution process.

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

First, we need to convert the mass of CaCl₂ to moles:

  1. Calculate Moles of CaCl₂:

    n=massmolar mass=6.038g111.1g/mol≈0.0543moln = \frac{\text{mass}}{\text{molar mass}} = \frac{6.038 g}{111.1 g/mol} \approx 0.0543 mol

  2. Calculate Energy Released per mole:

    Using the calibration factor:

    Energy=En=3.36 J0.0543 mol≈61.98J/mol\text{Energy} = \frac{E}{n} = \frac{3.36 \text{ J}}{0.0543 \text{ mol}} \approx 61.98 J/mol

    To convert to kJ:

    ΔH=61.981000≈0.062kJ/mol\Delta H = \frac{61.98}{1000} \approx 0.062 kJ/mol

  3. Final Enthalpy Change:

    Thus, the enthalpy change for the dissolution of 1.00 mol of CaCl₂ is approximately:

    ΔH=−0.062kJ/mol\Delta H = -0.062 kJ/mol (the negative sign indicates that it is an exothermic process).

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