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QUESTION 3 (Start on a new page.) A learner investigates bond energies in two similar compounds: Chloromethane Bromomethane Refer to the table of bond energies below:

Chemistry

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2 parts
3.3

409.9 kJ⋅mol−1\text{409.9 kJ}\cdot\text{mol}^{-1}

3.4

1509.6 kJ⋅mol−1\text{1509.6 kJ}\cdot\text{mol}^{-1}

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Part 3.1

Bond energy is the amount of energy required to break one mole of a specific type of bond in the gaseous state.

Part 3.2

The C-Cl bond has a higher bond energy (340 kJ/mol) than the C-Br bond (280 kJ/mol). This difference is primarily due to the atomic size of chlorine versus bromine. Chlorine is a smaller atom than bromine, leading to a shorter and stronger C-Cl bond compared to the C-Br bond. Shorter bonds generally require more energy to break.

Part 3.3

409.9 kJ⋅mol−1\text{409.9 kJ}\cdot\text{mol}^{-1}

  1. Calculate the molar mass of chloromethane (CH3Cl\text{CH}_3\text{Cl}).

    M(CH3Cl)=(1×12.011)+(3×1.008)+(1×35.453)M(\text{CH}_3\text{Cl}) = (1 \times 12.011) + (3 \times 1.008) + (1 \times 35.453) M(CH3Cl)=12.011+3.024+35.453=50.488 g/molM(\text{CH}_3\text{Cl}) = 12.011 + 3.024 + 35.453 = 50.488 \text{ g/mol}

  2. Calculate the number of moles of chloromethane in the sample.

    n=mM=5.05 g50.488 g/mol=0.10002376 moln = \frac{m}{M} = \frac{5.05 \text{ g}}{50.488 \text{ g/mol}} = 0.10002376 \text{ mol}

  3. Calculate the total energy required to break all bonds in 1 mole of chloromethane.

    Energy per mole=Total energy in sampleNumber of moles in sample=157 kJ0.10002376 mol=1569.62 kJ/mol\text{Energy per mole} = \frac{\text{Total energy in sample}}{\text{Number of moles in sample}} = \frac{157 \text{ kJ}}{0.10002376 \text{ mol}} = 1569.62 \text{ kJ/mol}

  4. Set up an equation for the total bond energy in 1 mole of CH3Cl\text{CH}_3\text{Cl}.

    A molecule of CH3Cl\text{CH}_3\text{Cl} contains three C-H bonds and one C-Cl bond.

    Total energy per mole=(3×EC-H)+(1×EC-Cl)\text{Total energy per mole} = (3 \times E_{\text{C-H}}) + (1 \times E_{\text{C-Cl}})

    We know EC-Cl=340 kJ/molE_{\text{C-Cl}} = 340 \text{ kJ/mol} from the table.

    1569.62 kJ/mol=(3×EC-H)+340 kJ/mol1569.62 \text{ kJ/mol} = (3 \times E_{\text{C-H}}) + 340 \text{ kJ/mol}

  5. Solve for the bond energy of the C-H bond (EC-HE_{\text{C-H}}).

    3×EC-H=1569.62 kJ/mol−340 kJ/mol3 \times E_{\text{C-H}} = 1569.62 \text{ kJ/mol} - 340 \text{ kJ/mol} 3×EC-H=1229.62 kJ/mol3 \times E_{\text{C-H}} = 1229.62 \text{ kJ/mol} EC-H=1229.62 kJ/mol3=409.87 kJ/molE_{\text{C-H}} = \frac{1229.62 \text{ kJ/mol}}{3} = 409.87 \text{ kJ/mol}

    The bond energy of the C-H bond is 409.9 kJ⋅mol−1\boxed{\text{409.9 kJ}\cdot\text{mol}^{-1}}.

Part 3.4

1509.6 kJ⋅mol−1\text{1509.6 kJ}\cdot\text{mol}^{-1}

  1. Identify the bonds present in 1 mole of bromomethane (CH3Br\text{CH}_3\text{Br}) and their quantities.

    A molecule of CH3Br\text{CH}_3\text{Br} contains three C-H bonds and one C-Br bond.

  2. Use the calculated C-H bond energy from 3.3 and the given C-Br bond energy.

    EC-H=409.87 kJ/molE_{\text{C-H}} = 409.87 \text{ kJ/mol}
    EC-Br=280 kJ/molE_{\text{C-Br}} = 280 \text{ kJ/mol} (from the table)

  3. Calculate the total energy required to break all bonds in 1 mole of bromomethane.

    Total energy=(3×EC-H)+(1×EC-Br)\text{Total energy} = (3 \times E_{\text{C-H}}) + (1 \times E_{\text{C-Br}}) Total energy=(3×409.87 kJ/mol)+(1×280 kJ/mol)\text{Total energy} = (3 \times 409.87 \text{ kJ/mol}) + (1 \times 280 \text{ kJ/mol}) Total energy=1229.61 kJ/mol+280 kJ/mol\text{Total energy} = 1229.61 \text{ kJ/mol} + 280 \text{ kJ/mol} Total energy=1509.61 kJ/mol\text{Total energy} = 1509.61 \text{ kJ/mol}

    The energy required to break all bonds in 1 mole of bromomethane is 1509.6 kJ⋅mol−1\boxed{\text{1509.6 kJ}\cdot\text{mol}^{-1}}.

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