1. Octane (C 8 H 18 ) undergoes combustion according to the following thermochem

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Question

1. Octane (C8H18) undergoes combustion according to the following thermochemical equation:

2C8H18(l) + 25O2(g) 16CO2(g) + 18H2O(l) H°rxn = –11,020 kJ/mol.

Given that DH°f[CO2(g)] = –393.5 kJ/mol and DH°f[H2O(l)] = –285.8 kJ/mol, calculate the standard enthalpy of formation of octane.

2. During volcanic eruptions, hydrogen sulfide gas is given off and oxidized by air according to the following chemical equation: (4points)

2H2S(g) + 3O2(g) ® 2SO2(g) + 2H2O(g)

Calculate the standard enthalpy change for the above reaction given:

3S(s) + 2H2O(g) 2H2S(g) + SO2(g) H° = 146.9 kJ/mol

S(s) + O2(g) SO2(g) H° = –296.4 kJ/mol

Given

Hof(CO2(g)) = -393.5 KJ/mol

Hof(H2O(l)) = -285.8 KJ/mol

we know,

Hof(O2(g)) = 0.0 KJ/mol. since O2 is in standard elemental state

Balanced chemical equation is:

2C8H18(l) + 25O2(g) ---> 16CO2(g) + 18H2O(l)

delta Ho rxn = 16*Hof(CO2(g)) + 18*Hof(H2O(l)) - 2*Hof( C8H18(l)) - 25*Hof(O2(g))

- 11,020 = 16*(-393.5) + 18*(-285.8) - 2*Hof( C8H18(l) - 25*0.0

2*Hof( C8H18(l) = 16*(-393.5) + 18*(-285.8) + 11,020

2*Hof( C8H18(l) = -420.4 KJ/mol

Hof( C8H18(l) = -210.2 KJ/mol

Answer: -210.2 KJ/mol

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