• # 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

(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${{O}_{2}}$.