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Natural gas is a source of methane - Scottish Highers Chemistry - Question 7 - 2022
Question 7
Natural gas is a source of methane. (a) Methane, CH₄, can be used as a fuel. In an experiment, methane was burned to raise the temperature of 100 cm³ of water by 27... show full transcript
Worked Solution & Example Answer:Natural gas is a source of methane - Scottish Highers Chemistry - Question 7 - 2022
Step 1
Calculate the mass of methane burned in this experiment.
Answer
To find the mass of methane burned, we apply the formula: ( q = m \cdot c \cdot \Delta T ) where:
- ( q ) is the heat absorbed by the water,
- ( m ) is the mass of water (100 cm³, which is 100 g),
- ( c ) is the specific heat capacity of water (approximately 4.18 J g⁻¹ °C⁻¹),
- ( \Delta T ) is the temperature change (27 °C).
Calculating ( q ): [ q = 100 ext{ g} imes 4.18 ext{ J g}^{-1} ext{ °C}^{-1} imes 27 ext{ °C} = 11286 ext{ J} = 11.286 ext{ kJ} ]
Next, using the enthalpy of combustion of methane: [ ext{Moles of methane} = \frac{q}{\Delta H} = \frac{11.286 ext{ kJ}}{891 ext{ kJ mol}^{-1}} = 0.0127 ext{ mol} ]
Finally, converting moles to mass using the molar mass of methane (16 g mol⁻¹): [ \text{Mass} = 0.0127 ext{ mol} \times 16 ext{ g mol}^{-1} = 0.203 ext{ g} ]
Step 2
Calculate the enthalpy change for the combustion of methane.
Answer
Using bond enthalpies to find the theoretical enthalpy change:
- Breaking bonds:
- C-H bonds: 4 (412 kJ mol⁻¹) = 1648 kJ mol⁻¹
- O=O bonds: 2 (498 kJ mol⁻¹) = 996 kJ mol⁻¹
Total energy required to break bonds = 1648 + 996 = 2644 kJ mol⁻¹.
- Forming bonds:
- C=O bonds: 2 (804 kJ mol⁻¹) = 1608 kJ mol⁻¹
- O-H bonds: 4 (463 kJ mol⁻¹) = 1852 kJ mol⁻¹
Total energy released from forming bonds = 1608 + 1852 = 3460 kJ mol⁻¹.
Thus, the enthalpy change ( \Delta H ): [ \Delta H = \text{Energy required} - \text{Energy released} = 2644 - 3460 = -816 ext{ kJ mol}^{-1} ]
Step 3
Calculate the atom economy for the formation of hydrogen.
Answer
The atom economy is calculated using the formula: [ \text{Atom economy} = \left( \frac{\text{Molar mass of desired product}}{\text{Molar mass of all reactants}} \right) \times 100 ]
For the reaction ( CH₄(g) + H₂O(g) ⇌ CO(g) + 3 H₂(g) ):
- Desired product: Hydrogen (H₂), molar mass = 2 g mol⁻¹.
- Total molar mass of reactants: ( CH₄ = 16 ext{ g mol}^{-1} + H₂O = 18 ext{ g mol}^{-1} = 34 ext{ g mol}^{-1} ).
Thus, [ \text{Atom economy} = \left( \frac{2}{34} \right) \times 100 = 5.88% ]
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