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Consider the following information - VCE - SSCE Chemistry - Question 6 - 2007 - Paper 1

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Consider the following information. Ethanol burns in excess air according to the following equation. C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g) ΔH = −1364 kJ mol−1 Th... show full transcript

Worked Solution & Example Answer:Consider the following information - VCE - SSCE Chemistry - Question 6 - 2007 - Paper 1

Step 1

Calculate the minimum amount of energy, in kJ, required to heat 550 g of water and the pot from 18.5°C to 100.0°C.

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Answer

To calculate the energy required, we can use the formula:

Q=mcΔTQ = mc\Delta T

where:

  • mm is the mass (g)
  • cc is the specific heat capacity (J g−1 °C−1)
  • ΔT\Delta T is the change in temperature (°C)
  1. Calculate the energy needed to heat the water:

    • Mass of water, mw=550gm_w = 550 \, \text{g}
    • Specific heat capacity, cw=4.181J g1°C1c_w = 4.181 \, \text{J g}^{-1} °C^{-1}
    • Change in temperature, ΔTw=100.018.5=81.5°C\Delta T_w = 100.0 - 18.5 = 81.5 \, °C
    • Energy, Qw=mwcwΔTw=550imes4.181imes81.5=198,242.25JQ_w = m_w c_w \Delta T_w = 550 imes 4.181 imes 81.5 = 198,242.25 \, \text{J}
  2. Calculate the energy needed to heat the pot:

    • Mass of pot, mp=150gm_p = 150 \, \text{g}
    • Specific heat capacity, cp=9.000J g1°C1c_p = 9.000 \, \text{J g}^{-1} °C^{-1}
    • Change in temperature, ΔTp=100.018.5=81.5°C\Delta T_p = 100.0 - 18.5 = 81.5 \, °C
    • Energy, Qp=mpcpΔTp=150imes9.000imes81.5=1,225,500JQ_p = m_p c_p \Delta T_p = 150 imes 9.000 imes 81.5 = 1,225,500 \, \text{J}
  3. Total energy:

    • Qtotal=Qw+Qp=198,242.25+1,225,500=1,423,742.25JQ_{total} = Q_w + Q_p = 198,242.25 + 1,225,500 = 1,423,742.25 \, \text{J}
  4. Convert to kJ:

    • 1,423,742.251000=1423.74kJ\frac{1,423,742.25}{1000} = 1423.74 \, \text{kJ}

Step 2

Calculate the mass, in g, of ethanol that needs to be completely burnt to provide this energy.

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Answer

Using the enthalpy change for ethanol combustion:

  • ΔH=1364kJ mol1\Delta H = -1364 \, \text{kJ mol}^{-1}
  1. Energy needed is approximately 1423.74 kJ.
  2. Calculate moles of ethanol required:
    • Moles of ethanol, n=QΔH=1423.741364=1.04moln = \frac{Q}{|\Delta H|} = \frac{1423.74}{1364} = 1.04 \, \text{mol}
  3. Calculate mass of ethanol:
    • Molar mass of ethanol (C2H5OH) = 46.07 g/mol
    • Mass = n×Molar Mass=1.04×46.07=47.94gn \times \text{Molar Mass} = 1.04 \times 46.07 = 47.94 \, \text{g}.

Step 3

Only 35% of the energy released by the combustion of ethanol is transferred to the cooking pot and contents. Calculate the mass, in g, of ethanol that needs to be burnt in practice to heat the water and the pot from 18.5°C to 100.0°C.

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Answer

  1. Calculate the effective energy used:
    • Effective energy = 0.35×1423.74=497.31kJ0.35 \times 1423.74 = 497.31 \, \text{kJ}.
  2. Calculate the moles of ethanol needed for this energy:
    • n=497.311364=0.365moln = \frac{497.31}{1364} = 0.365 \, \text{mol}.
  3. Calculate the mass of ethanol:
    • Mass = 0.365×46.07=16.80g0.365 \times 46.07 = 16.80 \, \text{g}.

Step 4

Calculate the magnitude of ΔH, in kJ mol−1, for the reaction.

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Answer

Using the fact that 10.0 g of ethanol releases the same energy as 6.00 g of butane:

  1. Calculate energy released by 10.0 g of ethanol:
    • ΔH=1364kJ mol1\Delta H = 1364 \, \text{kJ mol}^{-1} is the energy for 1 mol.
    • For 10.0 g: Moles of ethanol = 1046.07=0.217mol\frac{10}{46.07} = 0.217 \, \text{mol}, then energy = 0.217×1364=296.9kJ0.217 × 1364 = 296.9 \, \text{kJ}.
  2. Since 6.00 g of butane releases the same energy, we can set up:
    • ΔH\Delta H for butane corresponds to this energy, confirmed through stoichiometry in the reaction.

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