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NEET-XI-Chemistry

06: Chemical Thermodynamics

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  • #8
    The reaction of cyanamide, NH2CN(s),with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be –742.7 kJ mol–1at 298 K. Calculate enthalpy change for the reaction at 298 K.

    $$\ce{NH2CN(s) + \frac32 O2(g) -> N2(g) + CO2(g) + H2O(l) } $$
    Ans : Enthalpy change for a reaction (ΔH) is given by the expression,

    ΔH = ΔU + ΔngRT

    Where,

    ΔU = change in internal energy

    Δng = change in number of moles

    For the given reaction,

    Δng = ∑ng (products) – ∑ng (reactants)

    = (2 – 1.5) moles

    Δng = 0.5 moles

    And,

    ΔU = –742.7 kJ mol–1

    T = 298 K

    R = 8.314 × 10–3 kJ mol–1 K–1

    Substituting the values in the expression of ΔH:

    ΔH = (–742.7 kJ mol–1) + (0.5 mol) (298 K) (8.314 × 10–3 kJ mol–1 K–1)

    = –742.7 + 1.2

    ΔH = –741.5 kJ mol–1