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JEE Main & Advanced Chemistry Thermodynamics / रासायनिक उष्मागतिकी Sample Paper Topic Test - Thermo Chemistry (Thermodynamics)

  • question_answer
    The \[\Delta H_{F}^{O}\] for \[C{{O}_{2}}(g),\,\,CO(g)\] and  \[{{H}_{2}}O(g)\] are  \[-\,393.5,\,-110.5\] and \[-\,241.8\text{ }kJ\text{ }mol\],  respectively. The standard enthalpy change  (in kJ) for the reaction \[C{{O}_{2(g)}}+{{H}_{2}}_{(g)}\xrightarrow{{}}C{{O}_{(g)}}+{{H}_{2}}{{O}_{(g)}}\] is:

    A) + 524.1

    B) + 41.2  

    C) - 262.5  

    D) - 41.2

    Correct Answer: B

    Solution :

    [b]  (i) \[{{C}_{\left( s \right)}}+{{O}_{2(g)}}\to C{{O}_{2(g)}};\,\,\,\,\Delta H=-393.5kJ/mol\]
    (ii)  \[{{C}_{\left( s \right)}}+\frac{1}{2}{{O}_{2(g)}}\to C{{O}_{(g)}};\,\,\,\,\Delta H=-110.5kJ/mol\]
    (iii) \[{{H}_{2\left( g \right)}}+\frac{1}{2}{{O}_{2(g)}}\to {{H}_{2}}{{O}_{(g)}};\,\,\,\,\,\,\Delta H=-241.8kJ/mol\]
    For getting \[C{{O}_{2\left( g \right)}}+{{H}_{2(g)}}\to C{{O}_{(g)}}+{{H}_{2}}{{O}_{(g)}}\,\] add (ii) and (iii) and subtract (i).
    Thus,  - 110.5 - 241.8 + 393.5 = 395.3 = 41.2 kJ/mol


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