Energy required to dissociate $16 \mathrm{~g} \mathrm{O}_{2(\mathrm{~g})}$ into free atoms is $x…
Energy required to dissociate $16 \mathrm{~g} \mathrm{O}_{2(\mathrm{~g})}$ into free atoms is $x \mathrm{~kJ}$. The value of bond enthalpy of $\mathrm{O}=O$ bond is
$2 x KJ$
$\frac{x}{2} \mathbf{KJ}$
$4 x kJ$
16xKJ
Solution
To dissociate 1 mole of O 2 into free atoms, we need to break the $\mathrm{O}=\mathrm{O}$ bond. The bond enthalpy of the $\mathrm{O}=\mathrm{O}$ bond is the energy required to break this bond. Given that 16 g of O 2 corresponds to 0.5 moles (since the molar mass of O 2 is $32 \mathrm{~g} /$ $\mathrm{mol})$, the energy required to dissociate 0.5 moles of O 2 is xJ . Therefore, the energy required to dissociate 1 mole of O 2 would be $2 x \mathrm{~kJ}$. Thus, the bond enthalpy of the $\mathrm{O}=\mathrm{O}$ bond is 2 kJ .
Step by Step Solution:
Step 1
Calculate the number of moles in 16 g of $02: 16 \mathrm{~g} / 32 \mathrm{~g} / \mathrm{mol}=0.5$ moles.
Step 2
The energy required to dissociate 0.5 moles of O 2 is given as $\times \mathrm{kJ}$.
Step 3
To find the energy for 1 mole of O , multiply the energy for 0.5 moles by $2: 2$ * $(\mathrm{xkJ})=2 \mathrm{xkJ}$.
Step 4
The bond enthalpy of the $\mathrm{O}=\mathrm{O}$ bond is therefore 2 xkJ .
Final Answer:
$2 x \mathrm{~kJ}$