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27 May, 06:09

Be sure to answer all parts. Coal gasification is a multistep process to convert coal into cleaner-burning fuels. In one step, a coal sample reacts with superheated steam: C (s) + H2O (g) → CO (g) + H2 (g) ΔH o rxn = 129.7 kJ (a) Combine the reaction above with the following two to write an overall reaction for the production of methane: CO (g) + H2O (g) → CO2 (g) + H2 (g) ΔH o rxn = - 41 kJ CO (g) + 3H2 (g) → CH4 (g) + H2O (g) ΔH o rxn = - 206 kJ In the overall reaction, include the physical states of each product and reactant. (b) Calculate ΔH o rxn for this overall change. 12.03 kJ (c) Using the value in (b) and calculating ΔH o rxn for the combustion of methane, find the total heat for gasifying 6.28 kg of coal and burning the methane formed. Assume water forms as a gas and the molar mass of coal is 12.00 g/mol.

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  1. 27 May, 10:03
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    A.

    C (s) + H2O (g) - -> CO (g) + H2 (g) (x2)

    CO (g) + H2O (g) - -> CO2 (g) + H2 (g)

    CO (g) + 3 H2 (g) - -> CH4 (g) + H2O (g)

    Adding the three equations up:

    2C (s) + 2H2O (g) - -> 2CO (g) + 2H2 (g)

    CO (g) + H2O (g) - -> CO2 (g) + H2 (g)

    CO (g) + 3 H2 (g) - -> CH4 (g) + H2O (g)

    Overall equation:

    2C (s) + 2H2O (g) - -> CO2 (g) + CH4 (g)

    B.

    2C (s) + 2H2O (g) - -> 2CO (g) + 2H2 (g)

    = 2x 129.7 kJ/mol = 259.4 kJ/mol

    CO (g) + H2O (g) - -> CO2 (g) + H2 (g) ΔH°rxn = - 41.2 kJ/mol

    Using their individual heat of formation;

    ΔH°rxn = H products - H reactants

    = (-393.5 + 0) - (-110.5 + - 241.8)

    = - 41.2 kJ/mol

    CO (g) + 3H2 (g) - -> CH4 (g) + H2O (g) ΔH°rxn = - 206.17 kJ/mol

    Using their individual heat of formation;

    ΔH°rxn = H products - H reactants

    = (-74.87 + - 241.8) - (-110.5 + 0)

    = - 206.17 kJ

    ΔH°overall rxn = sum of energies for the 3 reactions above

    ΔH°overall reaction = 259.4 + - 41.2 + - 206.17

    = 12.03 kJ/mol

    Overall equation for the reaction: 2C (coal) + 2H2O (g) - -> CO2 (g) + CH4 (g)

    overall ΔH°rxn = 12.03 kJ/mol

    C.

    Number of moles of coal = mass/molar mass

    = 6.28/12

    = 0.523 mol

    Total heat = number of moles * enthalpy change of the reaction

    = 0.523 * 12.03

    = 6.30 kJ
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