Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The heat of formation of $PC{l_5}(s)$ from the following data will be-$2P(s) + 3C{l_2}(g) \to 2PC{l_3}(l);\Delta H = - 151.8kcal$$PC{l_3}(l) + C{l_2}(g) \to PC{l_5}(s);\Delta H = - 32.8kcal$A.$- 108.7kcal$B.$+ 108.7kcal$C.$- 184.6kcal$D.$+ 184.6kcal$

Last updated date: 19th Jun 2024
Total views: 394.2k
Views today: 8.94k
Verified
394.2k+ views
Hint:Standard enthalpy of formation is also called as standard heat of formation. First we need to find the reaction for formation of $PC{l_5}(s)$ from $P(s)$ and $C{l_2}(g)$ from the given two equations. Enthalpy of formation can be positive or negative.

The process in which a compound is formed from its constituent elements in their standard state is called formation.
Definition: The amount of heat released or absorbed when one mole of substance is formed from its constituent elements is called an enthalpy of formation. If all substances of the chemical reaction are in their standard states then the enthalpy of formation is called as standard enthalpy of formation. It is denoted by ${\Delta _f}{H^\Theta }$ .
Now the reaction for formation of $PC{l_5}(s)$ can be given as-
$P(s) + {5}{2}C{l_2}(g) \to PC{l_5}(s)$ ……..$(1)$
The given equations are-
$2P(s) + 3C{l_2}(g) \to 2PC{l_3}(l);\Delta H = - 151.8kcal$……..$(2)$
$PC{l_3}(l) + C{l_2}(g) \to PC{l_5}(s);\Delta H = - 32.8kcal$ ……$(3)$
The equation $(1)$ can be obtained by dividing equation $(2)$ with two and then adding equation $(2)$ and $(3)$ .
Equation $(2)$ on dividing by two becomes-
$P(s) + {3}{2}C{l_2}(g) \to PC{l_3}(l);\Delta H = {{ - 151.8}}{2} = - 75.9kcal$
Now adding equation $(3)$ and above equation-
$PC{l_3}(l) + C{l_2}(g) + P(s) + {3}{2}C{l_2}(g) \to PC{l_5}(s) + PC{l_3}(l)$
$P(s) + {5}{2}C{l_2}(g) \to PC{l_5}(s)$
So, ${\Delta _f}{H^\Theta }$ of $PC{l_5}(s)$ will be
Adding the enthalpies of both the equations we get :
${\Delta _f}{H^\Theta }(PC{l_5}) = - 75.9 - 32.8 = - 108.7kcal$
The correct option is A.

-$1calorie = 4.184joules$ .