question_answer 79)

  The enthalpy of reaction for the reaction: \[2{{H}_{2}}(g)+{{O}_{2}}(g)\to 2{{H}_{2}}O(l)\]is\[{{\Delta }_{r}}{{H}^{O-}}=-572kJ\,mo{{l}^{-1}}\] What will be standard enthalpy of formation of\[{{H}_{2}}O(l)\]?

Answer:

  \[2{{H}_{2}}(g)+{{O}_{2}}(g)\to 2{{H}_{2}}O(l){{\Delta }_{r}}{{H}^{{}^\circ }}=-572kJ\,mo{{l}^{-1}}\]\[{{\Delta }_{r}}H=\Sigma 2{{\Delta }_{f}}H_{{{H}_{2}}O}^{{}^\circ }-[\Sigma 2{{\Delta }_{f}}H_{{{H}_{2}}}^{{}^\circ }+{{\Delta }_{f}}H_{{{O}_{2}}}^{{}^\circ }]\] \[-572=2x-0+0\]                 \[x=-\frac{572}{2}=-286kJ\,mo{{l}^{-1}}\] \[\therefore \] Heat of formation of \[{{H}_{2}}O=-286kJ\,mo{{l}^{-1}}\]