Question

The enthalpy of atomization of $P{H_3}$ is $954kJmo{l^{ - 1}}$and that of ${P_2}{H_4}$ is $1485kJmo{l^{ - 1}}$. What is the bond enthalpy of the $P - P$ bond ?

Answer 1

Hint: This question can be solved by the concept that enthalpy of atomisation is the total sum of bond enthalpy of all the bonds in that compound. By finding bond enthalpy from one compound we can use this bond enthalpy in another given compound.

Complete step by step answer:
In the question, by enthalpy of atomisation we mean that the amount of energy changes when a compound’s all bonds are broken to obtain atoms and bond enthalpy means the energy required to break a bond. Thus, we can say that enthalpy of atomisation is the sum of individual enthalpy that releases when each bond breaks.
The formula of enthalpy of atomisation can be given as:
Enthalpy of atomisation $ = $ Sum of bond enthalpy of each bond -(1)
Step $1$:
As given in question, we have
Enthalpy of atomisation of $P{H_3}$ $ = $ $954kJmo{l^{ - 1}}$
Now as we know $P{H_3}$ has three $P - H$bonds in it.
So, we can say from equation –(1)
Enthalpy of atomisation of $P{H_3} = $$3 \times $ (Bond enthalpy of $P - H$)
Now, $954 = 3 \times $(Bond Enthalpy of $P - H$ )
By solving the equation we get,
(Bond enthalpy of $P - H$)$ = 954 \div 3 = 318kJmo{l^{ - 1}}$ $ - (2)$
Step $2$:
Now, as given in question,
Enthalpy of atomisation of ${P_2}{H_4} = 1485kJmo{l^{ - 1}}$
As we know that in ${P_2}{H_4}$ there are four $P - H$ bonds and one $P - P$ bond.
So, from equation $ - (1)$ we can say that
Enthalpy of atomisation of ${P_2}{H_4} = 4 \times $(bond enthalpy of $P - H$)$ + 1 \times $(bond enthalpy of $P - P$) $ - (3)$
As we know the bond enthalpy of $P - H$ from equation $ - (2)$
So equation $ - (3)$ becomes
$1485 = 4 \times (318) + $(bond enthalpy of $P - P$)
$1485 = 1272 + $(bond enthalpy of $P - P$)
Bond enthalpy of $P - P = 213kJmo{l^{ - 1}}$
Hence, the required bond enthalpy of $P - P$ is $213kJmo{l^{ - 1}}$.

Note:
We may often get confused about the concept when bond energy is written in place of bond enthalpy but we need to remember that both of them are the same . Also we can use the bond energy of any bond in a compound in other compounds having the same bonds.