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NEET-XI-Chemistry

06: Chemical Thermodynamics

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  • #14
    Calculate the standard enthalpy of formation of CH3OH(l) from the following data:

    CH3OH(l) + O2(g) CO2(g) + 2H2O(l) ; ΔrHθ = -726 kJ mol-1

    C(g) + O2(g) CO2(g) ; ΔcHθ = -393 kJ mol-1

    H2(g) +O2(g) H2O(l) ; ΔfHθ = -286 kJ mol-1.
    Ans : The reaction that takes place during the formation of CH3OH(l) can be written as:

    C(s) + 2H2O(g) + O2(g)

    CH3OH(l) (1)

    The reaction (1) can be obtained from the given reactions by following the algebraic calculations as:

    Equation (ii) + 2 × equation (iii) - equation (i)

    ΔfHθ [CH3OH(l)] = ΔcHθ + 2ΔfHθ [H2O(l)] - ΔrHθ

    = (-393 kJ mol-1) + 2(-286 kJ mol-1) - (-726 kJ mol-1)

    = (-393 - 572 + 726) kJ mol-1

    ΔfHθ [CH3OH(l)] = -239 kJ mol-1