Question

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 1

Answer:-

Given:-

standard enthalpy of formation of CO_2(g) i.e $\Delta$H⁰f [CO_2(g)) ] = - 393.5 kJ/mol

standard enthalpy of formation of H₂O_(l) i.e $\Delta$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 $\Delta$H⁰f [C₈H₈ ] = ?

As we know that

molar mass of styrene (C₈H₈) = 8 $\times$ molar mass of C + 8 $\times$​​​​​​​ molar mass of H

molar mass of styrene (C₈H₈) = 8 $\times$​​​​​​​ 12.0107 + 8 $\times$​​​​​​​ 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 $\times$ - 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 $\Delta$H⁰_rxn

So

standard heat of reaction ($\Delta$H⁰_rxn ) =   - 4438.9 kJ / mol

Since we know that

C₈H₈ + 10O_2(g)  $\rightarrow$  8CO_2(g) + 4H₂O_(l)

So

standard heat of reaction ($\Delta$H⁰_rxn ) = standard heat of products ($\Delta$H⁰_products ) - standard heat of reactants ($\Delta$H⁰_reactants )

standard heat of reaction ($\Delta$H⁰_rxn ) = (8 $\times$$\Delta$H⁰f [CO_2(g)) ] + 4 $\times$$\Delta$H⁰f [H₂O_(l) ]   ) - ( $\Delta$​​​​​​​H⁰f [C₈H₈ ] )

- 4438.9 kJ / mol = (8 $\times$ - 393.5 kJ/mol + 4 $\times$- 285.8 kJ/mol]   ) - ( $\Delta$​​​​​​​H⁰f [C₈H₈ ] )

- 4438.9 kJ / mol = ( - 3148.0 kJ/mol + - 1143.2 kJ/mol]   ) - ( $\Delta$​​​​​​​H⁰f [C₈H₈ ] )

- 4438.9 kJ / mol +  ( $\Delta$​​​​​​​H⁰f [C₈H₈ ] ) = ( - 3148.0 kJ/mol + - 1143.2 kJ/mol] )

- 4438.9 kJ / mol + $\Delta$​​​​​​​H⁰f [C₈H₈ ] = - 4291.2 kJ/mol]

$\Delta$​​​​​​​H⁰f [C₈H₈ ] = - 4291.2 kJ/mol] + 4438.9 kJ / mol

standard enthalpy of formation of styrene (C₈H₈) i.e $\Delta$​​​​​​​H⁰f [C₈H₈ ] = 147.7 kJ / mol (i.e the answer)