Answer to Question #159080 in Chemistry for Arvin Shopon

Question

Solve the enthalpy of methane formation using the following reaction equations

1. $2H₂(g)+O₂(g) → 2H₂O(g)$ $∆H = -571,6 kJ$
2. $C(s)+O₂(g) → CO₂(g)$ $∆H = -393,5 kJ$
3. CH₄ (g) + 2 O₂ (g) → CO₂ (g) + 2 H₂O (g) $∆H = -889,8 kJ$

Answer 1

eq - equation

(1^st eq): 2H₂(g) + O₂(g) → 2H₂O(g), ∆H = -571.6 kJ/mol

(2^nd eq): C(s) + O₂(g) → CO₂(g), ∆H = -393.5 kJ/mol

(3^d eq): CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g), ∆H = -889.8 kJ/mol

C(s) + 2H₂(g) → CH₄(g), ∆H = ???

Solution:

According to Hess's law, the heat of reaction depends upon Initial and final conditions of reactants and does not depend of the intermediate path of the reaction.

1) Modify the three given equations to get the target equation:

a) (1^st eq): do nothing. We need 2H₂(g) on the reactant side and that's what we have.

b) (2^nd eq): do nothing. We need one C(s) on the reactant side and that's what we have.

c) (3^d eq): flip it so as to put CH₄(g) on the product side.

2) Rewrite all three equations with the changes made (including changes in their enthalpy):

(1^st eq): 2H₂(g) + O₂(g) → 2H₂O(g), ∆H = -571.6 kJ/mol

(2^nd eq): C(s) + O₂(g) → CO₂(g), ∆H = -393.5 kJ/mol

(3*^deq): CO₂(g) + 2H₂O(g) → CH₄(g) + 2O₂(g), ∆H = +889.8 kJ/mol

3) Cancel out the common species on both sides:

2O₂(g) ⇒ sum of 1^st and 2^nd equation & 3*^d equation

2H₂O(g) ⇒ 3*^d & 1^st equation

CO₂(g) ⇒ 3*^d & 2^nd equation

Thus, adding modified equations and canceling out the common species on both sides, we get:

C(s) + 2H₂(g) → CH₄(g)

5) Add the ΔH values of (1^st), (2^nd) and (3*^d) equations to get your answer:

∆H_x = (-571.6 kJ/mol) + (-393.5 kJ/mol) + 889.9 kJ/mol = -75.3 kJ/mol

Hence the enthalpy of formation of methane (CH₄) is -75.3 kJ/mol.

Answer: The enthalpy of formation of methane (CH₄) is -75.3 kJ/mol.