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 2 ZnO_((s)) , Delta H = -693.8 "kJ mol"^(-1)

Consider the following reaction between zinc and oxygen and choose the

1.	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 2 ZnO_((s)) , Delta H = -693.8 "kJ mol"^(-1)
Answer» 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. 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. `693.8 "kJ mol"^(-1)` energy is evolved in the reaction. `693.9 "kJ mol"^(-1)` energy is absorbed in the reaction. SOLUTION : For the given reaction, `2Zn_((s)) + O_(2(g)) to 2ZnO_((g)) , Delta H = - 693.8 "kJ mol"^(-1) Delta H = H_(P) - H_(R) ` A -ve value of `Delta H` shows that `H_R gt H_P "or" H_P lt H_R` i.e., enthalpy of two moles of ZnO is less than the enthalpy of two moles of zinc and one mole of oxygen by 693.8kJ. As `H_R gt H_P, 693.8 "kJ mol"^(-1)` of energy is evolved in the reaction.