11th Class Chemistry Thermodynamics

  • question_answer 27) Calculate the standard enthalpy of formation of\[C{{H}_{3}}OH(l)\]from the following data: \[C{{H}_{3}}OH(l)+\frac{3}{2}{{O}_{2}}(g)\to C{{O}_{2}}(g)+2{{H}_{2}}O(l);\] \[{{\Delta }_{r}}{{H}^{{}^\circ }}=-726kJmo{{l}^{-1}}\] \[C(s)+{{O}_{2}}(g)\to C{{O}_{2}}(g);\] \[{{\Delta }_{c}}{{H}^{{}^\circ }}=-393kJ\,mo{{l}^{-1}}\] \[{{H}_{2}}(s)+\frac{1}{2}{{O}_{2}}(g)\to H{{O}_{2}}(l);\] \[{{\Delta }_{f}}{{H}^{{}^\circ }}=-286kJ\,mo{{l}^{-1}}\]  

    Answer:

    The required equation is: \[C(s)+2{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\to C{{H}_{3}}OH(l)\] The given equations can be rearranged as: (i) \[C{{O}_{2}}(g)+2{{H}_{2}}O(l)\to C{{H}_{3}}OH(l)+\frac{3}{2}{{O}_{2}}(g);\] \[{{\Delta }_{r}}{{H}^{{}^\circ }}=726kJmo{{l}^{-1}}\] (ii) \[C(s)+{{O}_{2}}(g)\to C{{O}_{2}}(g);\] \[{{\Delta }_{c}}{{H}^{{}^\circ }}=-393kJ\,mo{{l}^{-1}}\] (iii) \[2{{H}_{2}}(g)+{{O}_{2}}(g)\to 2{{H}_{2}}O(l);\] \[{{\Delta }_{f}}{{H}^{{}^\circ }}=-286\times 2kJmo{{l}^{-1}}\] \[\frac{On\,adding,}{\frac{C(s)+2{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\to C{{H}_{3}}OH(l);\Delta H=-239kJmo{{l}^{-1}}}{{}}}\]  


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