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 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. `693.8 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

For the reaction,
`2 Zn(s) + O_(2) (g) rarr 2 Zn O (s), Delta H = - 693.8 kJ mol^(-1)`
As we know that, `Delta H = H_(p) - H_(R)`
A negative 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 oxgyen by `693.8 kJ`. A `H_(R) gt H_(P), 693.8 kJ mol^(-1)` of energy is evolved in the reaction