NEET Chemistry Thermodynamics / रासायनिक उष्मागतिकी Question Bank Self Evaluation Test - Thermodynamics

  • question_answer
    The standard enthalpy of formation \[({{\Delta }_{f}}{{H}^{o}}_{298})\] for methane, \[C{{H}_{4}}\] is\[-74.9\text{ }kJ\text{ }mo{{l}^{-1}}\]. In order to calculate the average energy given out in the formation of a \[C-H\] bond from this it is necessary to know which one of the following?

    A) The dissociation energy of the hydrogen molecule, \[{{H}_{2}}\].

    B) The first four ionisation energies of carbon.

    C) The dissociation energy of \[{{H}_{2}}\] and enthalpy of sublimation of carbon (graphite).

    D) The first four ionisation energies of carbon and electron affinity of hydrogen.

    Correct Answer: C

    Solution :

    [c] To calculate average enthalpy of \[C-H\] bond in methane following in formations are needed (i) (i) dissociation energy of \[{{H}_{2}}\] i.e. \[\frac{1}{2}{{H}_{2}}(g)\xrightarrow{{}}H(g);\Delta H=x(suppose)\] (ii) Sublimation energy of C(graphite) to C(g) \[C\left( graphite \right)\xrightarrow{{}}C\left( g \right);\text{ }\Delta H=y\left( Suppose \right)\] Given \[C(graphite)+2{{H}_{2}}(g)\xrightarrow{{}}C{{H}_{4}}(g);\] \[\Delta H=75\text{ }kJ\text{ }mo{{l}^{-1}}\]


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