Question

The average energy required to break a P - P bond in $P_4\left(s\right)$ into gaseous atoms is $53.2kcalmo{l}^{-1}.$ The bond dissociation energy of $H_2\left(g\right)$ is $104.2kcalmo{l}^{-1};{△}_f{H}^o$ of $P{H}_3\left(g\right)$ from $P_4\left(s\right)$ is $5.5kcalmo{l}^{-1}.$ The P - H bond energy in $kcalmo{l}^{-1}$ is [Neglect presence of van der Waals forces in $P_4\left(s\right)$]
Given: $P_4\left(s\right)$ contains $6P-P$ single bonds.

• A. 85.2 $kcalmo{l}^{-1}$
• B. 57.6 $kcalmo{l}^{-1}$
• C. 76.9 $kcalmo{l}^{-1}$
• D. 63.3 $kcalmo{l}^{-1}$

Answer 1

The correct option is C 76.9 $kcalmo{l}^{-1}$

First we write all the equations along with energy required for them
Since $P_4$ molecule has 6 $P-P$ bond
$P_4\left(s\right)\to 4P\left(g\right)△H_{P-P}^o=53.2\times 6kcalmo{l}^{-1}$
$H_2\left(g\right)\to 2H\left(g\right)△H_{H-H}^o=104.2kcalmo{l}^{-1}$
$\frac{1}{4}{P}_4\left(s\right)+\frac{3}{2}{H}_2\left(g\right)\to P{H}_3\left(g\right){△}_{f}H=5.5kcalmo{l}^{-1}.$
But on calculation $△H$ for this reaction will be;
$\frac{1}{4}\times 6\times 53.2+\frac{3}{2}\times 104.2-3\Delta H_{P-H}^o$
But also, given
${△}_{f}H=5.5kcalmo{l}^{-1}.$
Comparing both the $△H$'s we get,
$\frac{1}{4}\times 6\times 53.2+\frac{3}{2}\times 104.2-3\Delta H_{P-H}^o=5.5$
$\frac{1}{4}\times 6\times 53.2+\frac{3}{2}\times 104.2-5.5=3\Delta H_{P-H}^o$
$\Rightarrow \Delta H_{P-H}^o=76.866i.e.76.9kcalmo{l}^{-1}$
hence, the P - H bond energy is $76.9kcalmo{l}^{-1}$