11th Class Chemistry Thermodynamics

  • question_answer 18) The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are \[-890.3kJmo{{l}^{-1}},\]\[-393.5kJmo{{l}^{-1}}\]and\[-285.8kJmo{{l}^{-1}}\] respectively. Enthalpy of formation of \[C{{H}_{4}}(g)\] will be: (i) \[-74.8kJmo{{l}^{-1}}\] (ii)\[-52.27kJmo{{l}^{-1}}\] (iii)\[+\text{ }74.8kJmo{{l}^{-1}}\] (iv)\[+52.26kJ\text{ }mo{{l}^{-1}}\]  

    Answer:

    Required equation: \[C(s)+2{{H}_{2}}(g)\to C{{H}_{4}}(g)\] (i) \[C{{H}_{4}}(g)+2{{O}_{2}}(g)\to C{{O}_{2}}(g)+2{{H}_{2}}O(l)\] \[=\Delta {{H}^{{}^\circ }}=-890.3KJmo{{l}^{-1}}\] (ii) \[C(s)+{{O}_{2}}(g)\to C{{O}_{2}}(g)\Delta {{H}^{{}^\circ }}=-393.5kJ\,mo{{l}^{-1}}\] (iii) \[2{{H}_{2}}(g)+{{O}_{2}}(g)\to 2{{H}_{2}}O(l)\Delta {{H}^{{}^\circ }}=-2\times 285.8kJmo{{l}^{-1}}\] Adding (ii) and (iii), subtracting (i) we get \[C(s)+2{{H}_{2}}(g)\to C{{H}_{4}}(g)\] \[\Delta {{H}^{{}^\circ }}=(-393.5)+(-2\times 285.8)-(-890.3)\] \[=-74.8kJ\,mo{{l}^{-1}}\]  


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