Answer:
(a, c) In the reaction:
\[2Zn(s)+{{O}_{2}}(g)\to
2ZnO(s);\,\,\,\,{{\Delta }_{r}}H=-693.8kJ\,,mo{{l}^{-1}}\]
\[{{\Delta }_{r}}H=\Sigma \]Enthalpy
of formation of products
\[-\Sigma \] Enthalpy of formation of
reactants
\[=[2{{\Delta
}_{f}}{{H}_{ZnO}}]-[2{{\Delta }_{f}}{{H}_{Zn}}+{{\Delta }_{f}}{{H}_{{{O}_{2}}}}]\]
For exothermic reaction: \[{{\Delta
}_{r}}H\] is negative, i.e., \[693.8\text{ kJ}\]heat is evolved.
\[\therefore \] Enthalpy of 2
moles of \[ZnO\]will less than the enthalpy of two moles of zinc and one mole
of\[{{O}_{2}}\].
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