Question

Consider the following reaction between zinc and oxygen and choose the correct option out of the options given below:

$2Zn\left(s\right)+O_2\left(g\right)\to 2ZnO\left(s\right);△H=-693.8kjmo{l}^{-1}$

• A. $693.8kjmo{l}^{-1}$ energy is evolved in the reaction
• B. The enthalpy of two moles of ZnO is less than the total enthalpy of two moles of Zn and one mole of oxygen by $693.8kj.$
• C. The enthalpy of two mole of ZnO is more than the total enthalpy of two moles of Zn and one mole of oxygen by $693.8kj.$
• D. $693.8kjmo{l}^{-1}$ energy is absorbed in the reaction

Answer 1

Answer: (A), (B)

The enthalpy of two moles of ZnO is less than the total enthalpy of two moles of Zn and one mole of oxygen by $693.8kj.$

$2Zn\left(s\right)+O_2\left(g\right)\to 2ZnO\left(s\right);△H=-693.8kjmo{l}^{-1}$

Since the change is enthalpy is negative
$\therefore$ enthalpy of product will be less than the sum of both the reactions by$693.8kj/mol$ because the amount of energy released is $693.8kj/mol$

Hence, option (A) and (C) are correct.