Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution

    1. Select the most apropriate option.

    x. Bond enthalpies of H-H, Cl-Cl and
    H-Cl bonds are 434 kJ mol-1, 242 kJ
    mol-1 and 431 kJ mol-1, respectively.
    Enthalpy of formation of HCl is
    a. 245 kJ mol-1

    b. -93 kJmol-1
    c. -245 kJ mol-1

    d. 93 kJ mol-1

    Answer:- Bond enthalpies of H-H, Cl-Cl and H-Cl bonds are 434 kJ mol-1, 242 kJ mol-1 and 431 kJ mol-1, respectively. Enthalpy of formation of HCl is. -93 kJmol-1

    Explanation:- p>ΔrH° = Σ ΔH° (reactant bonds) - Σ ΔH° (product bonds)

    \[ \ce{H2_{(g)} + Cl2_{(g)} -> 2HCl_{(g)}} \]

    Therefore, ΔrH° = [1 mol × 434 kJ mol−1 + 1 mol × 242 kJ mol−1] - [2 mol × 431 kJ mol−1]

    = -186 kJ

    Therefore, \[ \ce{H2_{(g)} + Cl2_{(g)} -> 2HCl_{(g)}, ΔrH° = -186 kJ} \]

    For the enthalpy of formation of HCl, the reaction is

    \[ \ce{\frac{1}{2} H2_{(g)} + \frac{1}{2} Cl2_{(g)} -> HCl_{(g)}} \]

    ΔrH° = \(\frac{-186\, \text{kJ}}{2\, \text{mol}}\) = -93 kJ mol−1

    Q1.9
    Q 1.10
    Q 2.1

    Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution