Question

The value for the standard heat of combustion, ΔH°combustion, for sucrose, C12H22O11, is -5.65 × 103 kJ·mol-1. Enter the thermochemical equation for the combustion of 1 mol of sucrose and calculate the value of ΔHf° for this compound (express your answer in scientific notation). The sole products of combustion are CO2(g) and H2O(l).

Answer 1

The balanced combustion reaction of sucrose is, C12H22O11 + 12 O2 12 CO2 +11 H2O , ΔH°combustion=5.65*10^3 KJ/mol

calculation of ΔHf°,std heat of formation

for the combustion reaction-:

ΔH°combustion=[ΔHf°(products) -ΔHf°(reactants)

                      =(ΔHf°(CO2)*12+ΔHf°(H2O)*11)-(ΔHf°(sucroe)+ΔHf°(O2)*12)

ΔHf°(H2O)(g)=-241.820 KJ/mol

ΔHf°(CO2)=-393.520KJ/mol

ΔHf°(O2)=0

ΔH°combustion=((-393.520)*12+(-241.820 )*11)-(ΔHf°(sucrose)+0)=-7382.26KJ/mol -ΔHf°(sucrose)

.or,5.65*10^3 KJ/mol=-7382.26KJ/mol -ΔHf°(sucrose)

ΔHf°(sucrose)=-7382.26KJ/mol-5650 KJ/mol=-13032.26 KJ/mol

ΔHf°(sucrose)l=-13032.26 KJ/mol=-1.303*10^4 KJ/mol