Ncert Exercises & Solutions

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

$2 Zn (s) + O_{2} (g) \to 2 ZnO (s); ∆ H = - 693.8 kJ {mol}^{- 1}$

1. 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

2. 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

3. $693.8 kJ {mol}^{- 1}$ energy is evolved in the reaction

4. $693.8 kJ {mol}^{- 1}$ energy is absorbed in the reaction

(a, c)

For the reaction,

2 Zn (s) + O_{2} (g) \to 2 ZnO (s); ∆ H = - 693.8 kJ {mol}^{- 1}

As we know that,

∆ H = H_{p} - H_{R}

A negative value of

∆ H

shows that

H_{R} > H_{P}

H_{P} < 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 693.8 kJ. As

H_{R} > H_{P}

693.8 kJ {mol}^{- 1}

of energy is evolved in the reaction.