NEET-XI-Chemistry

06: Chemical Thermodynamics

  • #4
    ΔUθof combustion of methane is - X kJ mol-1. The value of ΔHθ is
    (i) = ΔUθ
    (ii) > ΔUθ
    (iii) < ΔUθ
    (iv) = 0
    Ans : Since ΔHθ = ΔUθ + ΔngRT and ΔUθ = -X kJ mol-1,

    ΔHθ = (-X) + ΔngRT.

    ⇒ ΔHθ < ΔUθ

    Therefore, alternative (iii) is correct.