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7 December, 07:18

How much heat is produced by combustion 125 g of methanol under standard state condaitions?

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  1. 7 December, 07:27
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    For the reaction,

    2CH₃OH + 3O₂ → 2CO₂ + 4 H₂O

    For the above reaction,

    the change in enthalphy is calculated as

    Δ Hrxn = Δ H°f (products) - Δ H°f (reactants)

    In case the compound is in its standard state, enthalphy of formation is zero

    Hence,

    for the above reaction,

    ΔHrxn = (2 * Δ H° (CO₂) + 4 * Δ H° (H₂O)) - [ (2 * Δ H°CH₃OH) + (3 * Δ H° O₂) ]

    Δ H° (CO₂) = - 393.5kJ / mol

    Δ H° (H₂O) = - 241.8 kJ / mol

    Δ H°CH₃OH = - 239.2kJ / mol

    Δ H° O₂ = 0

    putting the corresponding values,

    ΔHrxn = (2 * - 393.5kJ / mol + 4 * - 241.8kJ / mol) - [ (2 * -239.2kJ / mol) + (3 * 0)

    ΔHrxn = - 1275.8 kJ / mol

    Moles of methanol,

    Moles is denoted by given mass divided by the molecular mass,

    Hence,

    n = w / m

    n = moles,

    w = given mass,

    m = molecular mass.

    From the question,

    w = 125 g

    as we know,

    m = 32 g / mol

    n = 125 g / 32 g / mol = 3.906 mol

    From, the reaction, 2 mol produces - 1275.8 kJ / mol heat,

    Now using unitary method,

    1 mol produces = - 1275.8 kJ / mol / 2 heat,

    3.906 mol produces = - 1275.8 kJ / mol / 2 * 3.906 heat

    3.906 mol produces = 249.7 kJ
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