Question

What is the heat of formation of $$C_{6}H_{6}$$, give that the heat of combustion of Benzene, carbon and Hydrogen are $$-782,-94$$ and $$-68\ K$$ respectvely

A
+14 K.Cal
B
14 K.Cal
C
+28 K.Cal
D
28 K.Cal

Solution

The correct option is A $$+14\ K.Cal$$$$6C+3{ H }_{ 2 }\rightarrow { C }_{ 6 }{ H }_{ 6 }\quad \longrightarrow \left( i \right)$$   ;   $${ \Delta H }_{ f }=?$$$${ C }_{ 6 }{ H }_{ 6 }+\frac { 15 }{ 2 } { O }_{ 2 }\longrightarrow { 6CO }_{ 2 }+3{ H }_{ 2 }O\quad \longrightarrow \left( ii \right)$$   ;     $${ \Delta H }_{ h }=-782K$$ $$Cal$$given, that heat of combustion of benzene $$=-782K$$ $$Cal$$                                                            carbon $$=-94K$$ $$Cal$$                                                       Hydrogen $$=-68K$$ $$Cal$$$$C+{ O }_{ 2 }\rightarrow { CO }_{ 2 }\quad \longrightarrow \left( iii \right)$$     $${ \Delta H }_{ { f }_{ 2 } }=-94K$$ $$Cal$$$${ H }_{ 2 }+\dfrac { 1 }{ 2 } { O }_{ 2 }\rightarrow { H }_{ 2 }O\quad \longrightarrow \left( iv \right)$$      $${ \Delta H }_{ { f }_{ 3 } }=-68K$$ $$Cal$$$$(i)=6(ii)+3(iv)-ii$$$$\therefore \quad { \Delta H }_{ f }=\left( { 6\Delta H }_{ 2 }+3{ \Delta H }_{ 3 } \right) -{ \Delta H }_{ 1 }$$$${ \Delta H }_{ f }=\left( 6\times \left( -94 \right) +3\times \left( -68 \right) \right) -\left( -782 \right)$$$${ \Delta H }_{ f }=14K.Cal$$$$\Rightarrow$$ Heat of formation of Benzene is $$+14$$ $$K.Cal$$Chemistry

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