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): DeltaH=-693.8 kJ mol^(-1)`

Options

A. The enthalpy of two moles 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 enthalpyof two moles of Zn and one mole of oxygen by 693.8kJ
C. `693.8 kJ mol^(-1)` energy is evolved in the reaction
D. `693.8 kJ mol^(-1)` energy is absorbed in the reaction

Answer

A, C

Explanation

In the reaction:
`2Zn(s)+O_(2)(g) rarr 2ZnO(s), Delta_(r)H = -693.8 kJ mol^(-1)`
`Delta_(r)H = sum` Enthalpy of formation of products `-sum` Enthalpy of formation of reactants
`=[2Delta_(f)H_(ZnO)]-[2Delta_(Zn)+Delta_(f)H_(O_(2))]`
For exothermic reactions. `Delta_(r)H` is negative, i.e., 693.8 kJ heat is evolved
Enthalpy of 2 moles of ZnO will less than the enthalpy of two moles of zinc and one mole of `O_(2)`