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Question: The heat of formation of \(PC{l_5}(s)\) from the following data will be- \(2P(s) + 3C{l_2}(g) \to ...

The heat of formation of PCl5(s)PC{l_5}(s) from the following data will be-
2P(s)+3Cl2(g)→2PCl3(l);ΔH=−151.8kcal2P(s) + 3C{l_2}(g) \to 2PC{l_3}(l);\Delta H = - 151.8kcal
PCl3(l)+Cl2(g)→PCl5(s);ΔH=−32.8kcalPC{l_3}(l) + C{l_2}(g) \to PC{l_5}(s);\Delta H = - 32.8kcal
A.−108.7kcal - 108.7kcal
B.+108.7kcal + 108.7kcal
C.−184.6kcal - 184.6kcal
D.+184.6kcal + 184.6kcal

Explanation

Solution

Standard enthalpy of formation is also called as standard heat of formation. First we need to find the reaction for formation of PCl5(s)PC{l_5}(s) from P(s)P(s) and Cl2(g)C{l_2}(g) from the given two equations. Enthalpy of formation can be positive or negative.

Complete step by step answer:
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 ΔfHΘ{\Delta _f}{H^\Theta } .
Now the reaction for formation of PCl5(s)PC{l_5}(s) can be given as-
P(s)+52Cl2(g)→PCl5(s)P(s) + {5}{2}C{l_2}(g) \to PC{l_5}(s) ……..(1)(1)
The given equations are-
2P(s)+3Cl2(g)→2PCl3(l);ΔH=−151.8kcal2P(s) + 3C{l_2}(g) \to 2PC{l_3}(l);\Delta H = - 151.8kcal……..(2)(2)
PCl3(l)+Cl2(g)→PCl5(s);ΔH=−32.8kcalPC{l_3}(l) + C{l_2}(g) \to PC{l_5}(s);\Delta H = - 32.8kcal ……(3)(3)
The equation (1)(1) can be obtained by dividing equation (2)(2) with two and then adding equation (2)(2) and (3)(3) .
Equation (2)(2) on dividing by two becomes-
P(s)+32Cl2(g)→PCl3(l);ΔH=−151.82=−75.9kcalP(s) + {3}{2}C{l_2}(g) \to PC{l_3}(l);\Delta H = {{ - 151.8}}{2} = - 75.9kcal
Now adding equation (3)(3) and above equation-
PCl3(l)+Cl2(g)+P(s)+32Cl2(g)→PCl5(s)+PCl3(l)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)+52Cl2(g)→PCl5(s)P(s) + {5}{2}C{l_2}(g) \to PC{l_5}(s)
So, ΔfHΘ{\Delta _f}{H^\Theta } of PCl5(s)PC{l_5}(s) will be
Adding the enthalpies of both the equations we get :
ΔfHΘ(PCl5)=−75.9−32.8=−108.7kcal{\Delta _f}{H^\Theta }(PC{l_5}) = - 75.9 - 32.8 = - 108.7kcal
The correct option is A.

Additional information:
-The compounds which have positive enthalpies of formation are endothermic compounds and are less stable than reactants.
-The compounds which have negative enthalpies of formation are known as exothermic compounds and are more stable than the reactants.

Note:
-Energy is released when a bond is broken and energy is needed to make a bond.
-Enthalpy is a form of energy. In simple words it is the amount of internal energy that is present in any system.
-1calorie=4.184joules1calorie = 4.184joules .
-Enthalpy is an extensive property. Its value depends on the mass of the system.