Consider the following reaction between zinc and oxygen and choose the correct options out of the options given below:
$2 Z n (s) + O_{2} (g) \to 2 Z n O (s)$ ; $Δ H = - 693.8 k J m o l^{- 1}$

The enthalpy of two moles of

Z n O

is less than the total enthalpy of two moles of

Z n

and one mole of oxygen by 693.8 kJ.

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

The enthalpy of two moles of

Z n O

is more than the total enthalpy of two moles of

Z n

and one mole of oxygen by 693.8 kJ.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

693.8 k J m o l^{- 1}

energy is evolved in the reaction

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

693.8 k J m o l^{- 1}

energy is absorbed in the reaction.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $693.8 k J m o l^{- 1}$ energy is evolved in the reaction

we know :
$Δ H$ = $H_{p r o d u c t s} - H_{r e a c t a n t}$
A negative value of $Δ H$ shows that $H_{P} < H_{R}$ , i.e.,enthalpy of two moles of $Z n O$ is less than the enthalpy of two moles of zinc and one mole of oxygen by $693.8 k J$ .
As $H_{P} < H_{R}$ , the reaction is exothermic and $693.8 k J m o l^{- 1}$ of energy is evolved in the reaction.

Suggest Corrections

Similar questions

Consider the following reaction between zinc and oxygen and choose the correct options out of the options given below
$2 Z n (s) + O_{2} (g) \to 2 Z n O (s); Δ H = - 693.8 k J m o l^{- 1}$
(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 $m o l^{- 1}$ energy is evolved in the reaction
(d) 693.8 kJ $m o l^{- 1}$ energy is absorbed in the reaction