: The values of heat of combustion of graphite and \[{H_2}\] are -395 and -269 KJ/mole respectively. If heat of formation of glucose is -1169 KJ/mole, then the heat of combustion of glucose is:
A.-2815 KJ/mole
B.-1169 KJ/mole
C.1169 KJ/mole
D.2815 KJ/mole

Answer Verified Verified
The combustion and formation both are opposite processes. Combustion reaction is always exothermic i.e. heat of that reaction is negative then other one can be exothermic or endothermic i.e. heat of reaction can be negative or positive.

Complete step by step answer:
Heat of combustion and heat of formation are basically two kinds of heat of reaction. Heat of reaction is the energy which is released or absorbed during the reaction. This can be calculated from the change of energy between all reactants and all products.
The definition of heat of formation is, the amount of energy change due to the formation of 1 mole of a compound from its constituent elements.
The definition of heat of combustion is, the amount of heat or energy developed due to the total combustion of 1mole of a compound.
Now heat of the combustion reaction of glucose can be calculated on the basis of Hess’ law. Using formula,
\[\Delta H = \sum {\Delta {H_f}(product)} - \sum {\Delta {H_f}({\text{reactant}})} \]
The equation of combustion of graphite is,
\[C + {O_2}\xrightarrow{\Delta }C{O_2},\Delta H = - 395KJ/mole\] ……(1)
This reaction can also be considered as a formation reaction of \[C{O_{2.}}\]
 The equation of combustion of \[{{\text{H}}_{\text{2}}}\] is,
\[H{}_2 + \dfrac{1}{2}O{}_2\xrightarrow{\Delta }H{}_2H = - 269KJ/MOLE\] ……(2)
This reaction can also be considered as a formation reaction of \[{H_2}O\]
 The equation of formation of glucose is,
\[6C + 6{H_2} + 3{O_2} \to {C_6}{H_{12}}{O_6},\Delta H = - 1169{\text{ }}KJ/mole\] ……(3)
 The equation of combustion of glucose is,
\[C{}_6H{}_{12}O{}_6 + 6O{}_2\xrightarrow{\Delta }6CO{}_2 + 6H{}_2O,\Delta H\]
Now according to the Hess’ law, the heat of reaction is,
  \Delta H = \sum {\Delta {H_f}(product)} - \sum {\Delta {H_f}({\text{reactant}})} \\
   = \sum {6 \times \Delta {H_f}(C{O_2}) + 6 \times } \Delta {H_f}({H_2}O) - \sum {\Delta {H_f}({{\text{C}}_6}{H_{12}}{O_6})} + 6 \times \Delta {H_f}({O_2}) \\
   = \sum {6 \times \left( { - 395} \right) + 6 \times } \left( { - 269} \right) - \sum { - 1169} + 0 \\
   = \sum { - 2370 - } 1614 + 1169 \\
   = - 2815KJ/mole \\

The correct answer is A.

Remember the formula \[\Delta H = \sum {\Delta {H_f}(product)} - \sum {\Delta {H_f}({\text{reactant}})} \]. The molar heat of combustion is the amount of heat released when one mole of substance is completely burnt.

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