Question

The enthalpy of dissociation of $P{H}_3$ is $954$ kJ/mol and that of $P_2{H}_4$ is $1.485$ MJ/mol. What is the bond enthalpy of $P-P$ bond?

• A. 213 kJ/mol
• B. 413 kJ/mol
• C. 200 kJ/mol
• D. Given data is incorrect

Answer 1

Answer: (A)

213 kJ/mol

$P{H}_3⟶P+3H\Delta H=954KJmo{l}^{-1}$

$\therefore 3\Delta H_{P-H}={\Delta }^{r}H\Rightarrow \Delta H_{P-H}=\frac{954}{3}KJmo{l}^{-1}$

Also given,

Enthalpy of dissociation of $P_2{H}_4$ is $1.485MJ/mol=1485KJmo{l}^{-1}$

$\therefore P_2{H}_4⟶2P+4H{\Delta }^{r}H=1485KJmo{l}^{-1}$

$\Rightarrow 4\Delta H_{P-H}+\Delta H_{P-P}={\Delta }^{r}H$

$\Rightarrow \Delta H_{P-P}={\Delta }^{r}H-4\Delta H_{P-H}=1485-4\times \frac{954}{3}=213KJmo{l}^{-1}$

$\Rightarrow \Delta H_{P-P}=213KJmo{l}^{-1}$

Thus bond enthalpy of $P-P$ bond $=213KJmo{l}^{-1}$

Hence, the correct option is $\text{A}$