When styrene C8H8 burns in oxygen to form carbon dioxide and liquid water under standard-state conditions at 25°C, 42.62 kJ are released per gram of styrene. Find the standard enthalpy of formation of styrene at 25°C. (Given: ΔH°f[CO2(g)] = –393.5 kJ/mol, ΔH°f[H2O(l)] = –285.8 kJ/mol)
Answer:-
Given:-
standard enthalpy of formation of CO2(g) i.e H0f [CO2(g)) ] = - 393.5 kJ/mol
standard enthalpy of formation of H2O(l) i.e H0f [H2O(l) ] = - 285.8 kJ/mol
heat energy of 1 g styrene (C8H8) = - 42.62 kJ ( since heat energy released i.e exothermic reaction)
standard enthalpy of formation of styrene (C8H8) i.e H0f [C8H8 ] = ?
As we know that
molar mass of styrene (C8H8) = 8 molar mass of C + 8 molar mass of H
molar mass of styrene (C8H8) = 8 12.0107 + 8 1.008
molar mass of styrene (C8H8) = 96.0856 + 8.064
molar mass of styrene (C8H8) = 104.15 g / mol
Also we know
1 mole styrene (C8H8) = 104.15 g
So
heat energy of 1 g styrene (C8H8) = -42.62 kJ
then
heat energy of 104.15 g styrene (C8H8) = 104.15 - 42.62 kJ
heat energy of 104.15 g styrene (C8H8) = - 4438.9 kJ
therefore when 1 mole styrene (C8H8) burns in oxygen to form carbon dioxide and liquid water to produce - 4438.9 kJ heat energy which is standard heat of reaction i.e H0rxn
So
standard heat of reaction (H0rxn ) = - 4438.9 kJ / mol
Since we know that
C8H8 + 10O2(g) 8CO2(g) + 4H2O(l)
So
standard heat of reaction (H0rxn ) = standard heat of products (H0products ) - standard heat of reactants (H0reactants )
standard heat of reaction (H0rxn ) = (8 H0f [CO2(g)) ] + 4 H0f [H2O(l) ] ) - ( H0f [C8H8 ] )
- 4438.9 kJ / mol = (8 - 393.5 kJ/mol + 4 - 285.8 kJ/mol] ) - ( H0f [C8H8 ] )
- 4438.9 kJ / mol = ( - 3148.0 kJ/mol + - 1143.2 kJ/mol] ) - ( H0f [C8H8 ] )
- 4438.9 kJ / mol + ( H0f [C8H8 ] ) = ( - 3148.0 kJ/mol + - 1143.2 kJ/mol] )
- 4438.9 kJ / mol + H0f [C8H8 ] = - 4291.2 kJ/mol]
H0f [C8H8 ] = - 4291.2 kJ/mol] + 4438.9 kJ / mol
standard enthalpy of formation of styrene (C8H8) i.e H0f [C8H8 ] = 147.7 kJ / mol (i.e the answer)
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