Answer to Question #341139 in General Chemistry for abood

Question

Lauric acid (C12/2402) is a typical fatty acid, a molecule with a long hydrocarbon chain and an

organic acid group (COOH) at the end. (a) Write a balanced equation for the complete

combustion of lauric acid to gaseous products. (b) Calculate AH°rxn for this combustion

(AH°+=-775.6 kJ/mol). (c) Calculate the heat (q) in kJ and kcal when 10.0 g of lauric acid is burned

completely. (AH° CO24g)=-393.5 kJ/mol, AH° $H20(g=-241.826 kJ/mol)

Answer 1

a. $C_{12}H_{24}O_2 + 17O_2 \rightarrow 12CO_2 + 12H_2O$

b. $\Delta H_f (CO_2)=-393.5 kJ/mole$

$\Delta H_f (H_2I)=-241.826 kJ/mole$

$\Delta H_f (C_{12}H_{24}O_2)=-775.6 kJ/mole$

$\Delta H_{rxn}=12\Delta H(CO_2)+12 \Delta H(H_2O)- \Delta H(C_{12}H_{24}O_2)°12x(-393.5)+12(-241.826)-(-775.6)=-6848 kJ$

c.$M(C_{12}H_{24}O_2)=12x12+1x24+2x16=200 g/mole$

$n(C_{12}H_{24}O_2)=10/200=0.05 mole$

If for 1 mole q =-6848 kJ, then fir 0.05 mole it is 0.05x(-6848)=-342.4 kJ