Answer to Question #346636 in General Chemistry for Sage

Question

Calculate ∆H for the reaction CH4 (g) + NH3 (g) → HCN (g) + 3 H2 (g), from the reactions.

N2 (g) + 3 H2 (g) → 2 NH3 (g) ∆H = -91.8 kJ

C (s, graphite) + 2 H2 (g) → CH4 (g) ∆H = -74.9 kJ/mole

H2 (g) + 2 C (s, graphite) + N2 (g) → 2 HCN (g) ∆H = +270.3 kJ

Answer 1

eq - equation

(1^st eq): N₂(g) + 3H₂(g) → 2NH₃(g), ∆H₁ = −91.8 kJ

(2^nd eq): C(s, graphite) + 2H₂(g) → CH₄(g), ∆H₂ = −74.9 kJ/mol

(3^d eq): H₂(g) + 2C(s, graphite) + N₂(g) → 2HCN(g), ∆H₃ = +270.3 kJ

CH₄(g) + NH₃(g) → HCN(g) + 3H₂(g), ∆H_x = ???

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:

(1^st eq): flip it over and divide it by 2. We need 1 mol of NH₃(g) on the reactant side.

(2^nd eq): flip it over. We need 1 mol of CH₄(g) on the reactant side.

(3^d eq): divide it by 2. We need 1 mol of HCN(g) on the product side.

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

(1*^st eq): NH₃(g) → 0.5N₂(g) + 1.5H₂(g), ∆H₁^* = (−91.8 kJ) × (−1/2) = +45.9 kJ

(2*^nd eq): CH₄(g) → C(s, graphite) + 2H₂(g), ∆H₂^* = (−74.9 kJ) × (−1) = +74.9 kJ

(3*^d eq): 0.5H₂(g) + C(s, graphite) + 0.5N₂(g) → HCN(g), ∆H₃^* = (+270.3 kJ) × (1/2) = +135.15 kJ

3) Cancel out the common species on both sides:

0.5N₂(g) ⇒ (3*^d equation) & (1*^st equation)

C(s, graphite) ⇒ (3*^d equation) & (2*^nd equation)

0.5H₂(g) ⇒ (3*^d equation) & (sum of 1*^st and 2*^nd equations)

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

CH₄(g) + NH₃(g) → HCN(g) + 3H₂(g)

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

∆H_x = ∆H₁^* + ∆H₂^* + ∆H₃^*

∆H_x = (+45.9 kJ) + (+74.9 kJ) + (+135.15 kJ) = +255.95 kJ

∆H_x = +255.95 kJ

Answer: ∆H for the reaction is +255.95 kJ