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NEET Chemistry Thermodynamics / रासायनिक उष्मागतिकी NEET PYQ-Thermodynamics

  • question_answer
    The bond dissociation energies of \[{{\text{X}}_{\text{2}}}\text{,}{{\text{Y}}_{\text{2}}}\] and \[\text{XY}\] are in the ratio of 1 : 0.5 : 1. \[\Delta \text{H}\] for the formation of XY is \[\text{--200 kJ mo}{{\text{l}}^{\text{--1}}}\]. The bond dissociation energy of \[{{\text{X}}_{\text{2}}}\] will be                         [NEET - 2018]

    A)  \[\text{800 kJ mo}{{\text{l}}^{\text{--1}}}\]

    B)       \[\text{100 kJ mo}{{\text{l}}^{\text{--1}}}\]

    C)  \[\text{200 kJ mo}{{\text{l}}^{\text{--1}}}\]

    D)       \[\text{400 kJ mo}{{\text{l}}^{\text{--1}}}\]

    Correct Answer: A

    Solution :

    [a] The reaction for \[{{\Delta }_{f}}\text{H }\!\!{}^\circ\!\!\text{ (XY)}\]
    \[\frac{1}{2}{{X}_{2}}(g)+\frac{1}{2}{{Y}_{2}}(g)\xrightarrow{{}}XY(g)\]
    Bond energies of \[{{X}_{2}},{{Y}_{2}}\] and \[XY\] are \[X,\frac{X}{2},X\] respectively
    \[\therefore \]      \[\Delta H=\left( \frac{X}{2}+\frac{X}{4} \right)-x=-200\]
    On solving, we get
    \[\Rightarrow -\frac{X}{2}+\frac{X}{4}=-200\]
    \[\Rightarrow \text{X=800}\,\,\text{kJ/mole}\]


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