Yohan C. answered • 11/22/14

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Math Tutor (up to Calculus) (not Statistics and Finite)

Hi Alex,

From the equation above (CH

_{4}(g) + 2O_{2}(g) → 2H_{2}O (l) + CO_{2}), Δ H = -802.4 kJ, if we show water vapor rather than liquid water as a product.Reversing the melting of ice and the combustion of methane, the thermochemical equation

(2H

_{2}O (l) + CO_{2}→ CH_{4}(g) + 2O_{2}(g) ) Δ H = 890.4 kJand what was an endothermic process becomes exothermic, and vice versa.

2H

_{2}O (s) → 2H_{2}O (l) Δ H = 2(6.01 kJ) = 12.0 kJthe enthalpy change is -802.4 kJ rather than -890.4 kJ because 88.0 kJ are needed to convert 2 moles of liquid water to vapor

2H

_{2}O (l) → 2H_{2}O (g) Δ H = 88.0 kJRemember this? Δ H = H (products) - H (reactants)

CH

_{4}(g) + 2O_{2}(g) → 2H_{2}O (l) + CO_{2}, Δ H = -802.4 kJWith these being said, where did you get

**4.1798 J /g C**?I'm asking you this because specific heat of water is

**4.184 J /g C**.You must find this from your Chemistry Textbook to do this

Δ H

_{f}(kJ /mol) standard enthalpies of Formation of Inorganic Substance at 25 C.By the way, molar mass of methane (CH

_{4}) is 16.01 g /mol. And there are 2 mol of H_{2}O in this equation. And molar mass of water is 18.0 g /mol. You might need these so you can do the stoichiometry to solve this problem.Good luck to you.