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 \[{{H}_{2}}O=-286kJ\,mo{{l}^{-1}}\]
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