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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
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B
14 K.Cal
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C
+28 K.Cal
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D
28 K.Cal
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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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