Question

Calculate ?H (298 K) per gram of fuel units of kJ*g^-1 (exclude oxygen)

a.) H2(g) + 1/2 O2(g) = H2O(g)

b.) CH4(g) + 2O2(g) = CO2(g) + 2H2O(g)

c.) CH3OH(l) + 3/2 O2(g) = CO2(g) + 2H2O(g)

d.) C6H14(g) + 9 1/2 O2(g) = 6CO2(g) + 7H2O(g)

Answer 1

a.) H2(g) + 1/2 O2(g) = H2O(g)

delta H = sum of enthalpies of formation of products - sum of enthalpies of formation of reactants

delta H = delta H of H2O - ( delta H of H2 + 1/2 delta H of O2) = ?241.818 - (0+0) = -241.818 KJ / mol

so for one gram = -241.818 / 2 = -120.909 KJ / g

b.) CH4(g) + 2O2(g) = CO2(g) + 2H2O(g)

delta H = sum of enthalpies of formation of products - sum of enthalpies of formation of reactants

delta H = (H of CO2 + 2 X H H2O ) -( H of CH4) =( ?393.509 + 2X  -241.818 ) - (?74.87) = -802.275 KJ / mol

so for 1 g = -560.457 / 16 = -50.14 KJ / g

c.) CH3OH(l) + 3/2 O2(g) = CO2(g) + 2H2O(g)

delta H = sum of enthalpies of formation of products - sum of enthalpies of formation of reactants

delta H = (H of CO2 + 2 X H H2O ) -( H of CH3OH) =( ?393.509 + 2X  -241.818 ) - (?238.4) = -638.74 KJ / mol

so for one gram =  -638.74 KJ / 32 = -19.96 KJ / g

d.) C6H14(g) + 9 1/2 O2(g) = 6CO2(g) + 7H2O(g)

delta H = sum of enthalpies of formation of products - sum of enthalpies of formation of reactants

delta H = (6 X H of CO2 + 7 X H H2O ) -( H of C6H14) =( ?2361.054 + (- 1692.726) ) - (?167.4) = -3886.38 KJ / mol

For one g =  -3886.38 KJ / 86 = 45.19 KJ / g