Question

What is ΔHo in kJ for the reaction below?

Given the following enthalpies of formation, ΔHfo, values:

CH4(g) = -74.8 kJ/mol;

H2O(g) = -242 kJ/mol;

CO2(g) = -394 kJ/mol;

O3(g) = 143 kJ/mol;

What is ΔHo in kJ for the reaction:

3CH4(g) + 4O3(g) → 3CO2(g) + 6H2O(g)

A) -3x394 - 6x242 + 4(143) - 3x74.8
B) -3x394 - 6x242 - 4(143) - 3x74.8
C) -3x394 - 6x242 - 4(143) + 3x74.8
D) 3x394 + 6x242 + 4(143) + 3x74.8
E) 3x394 + 6x242 - 4(143) + 3x74.8

Answer 1

`("C")`

Explanation:

`"3CH"_4 + "4O"_3 -> "3CO"_2 + "6H"_2"O"`

`Δ"H"^@ = SigmaΔ"Hf"_"products"^@ - SigmaΔ"Hf"_"reactants"^@`

`Δ"H"^@ = (3 Δ"Hf"_("CO"_2)^@ + 6 Δ"Hf"_("H"_2"O")^@) - (3Δ"Hf"_("CH"_4)^@ + 4Δ"Hf"_("O"_3)^@)`

`Δ"H"^@ = [(3 × (-394)) + (6 × (-242)] - [(3 × (-74.8)) + (4 × 143)]`

`Δ"H"^@ = -(3 × 394) - (6 × 242) + (3 × 74.8) - (4 × 143)`

`("C")` is the correct option.