question_answer 54)

  Consider the following reaction between zinc and oxygen and choose the correct options out of the options given below: \[2Zn(s)+{{O}_{2}}(g)\to 2ZnO(s);\,\,\,\Delta H=-693.8kJ\,mo{{l}^{-1}}\](a) The enthalpy of two moles of \[ZnO\]is less than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ (b) The enthalpy of two moles of \[ZnO\]is more than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ (c) \[693.8kJ\,mo{{l}^{-1}}\]energy is evolved in the reaction (d) \[693.8kJ\,mo{{l}^{-1}}\]energy is absorbed in the reaction

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\[{{O}_{2}}\].