Answer to Question #256759 in Chemistry for levi

Question

Grasping the Concept

Use the following thermochemical equations as needed to find the heat of formation of diamond:

C(diamond) + O2(g) → CO2(g) ΔHo = -395.3 KJ

2 CO2(g) → 2 CO(g) + O2(g) ΔHo= 566.0 KJ

C(graphite) + O2(g) → CO2(g) ΔHo = -395.3 KJ

2 CO(g) → C(graphite) + CO2(g) ΔHo= -172.5 KJ

Answer 1

eq - equation

(1^st eq): C(diamond) + O₂(g) → CO₂(g), ΔH_o = −395.3 kJ

(2^nd eq): 2CO₂(g) → 2CO(g) + O₂(g), ΔH_o = 566.0 kJ

(3^d eq): C(graphite) + O₂(g) → CO₂(g), ΔH_o = −393.5 kJ (NB! −393.5 kJ not −395.3 kJ)

(4^th eq): 2CO(g) → C(graphite) + CO₂(g), ΔH_o = −172.5 kJ

C(graphite) → C(diamond), ΔH_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 four given equations to get the target equation:

(1^st eq): multiply by −1

(2^nd eq): do nothing

(3^d eq): multiply by 2

(4^th eq): do nothing

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

(1*^st eq): CO₂(g) → C(diamond) + O₂(g), ΔH₁ = (−1) × (−395.3 kJ) = 395.3 kJ

(2^nd eq): 2CO₂(g) → 2CO(g) + O₂(g), ΔH₂ = 566.0 kJ

(3*^d eq): 2C(graphite) + 2O₂(g) → 2CO₂(g), ΔH₃ = (2) × (−393.5 kJ) = −787.0 kJ

(4^th eq): 2CO(g) → C(graphite) + CO₂(g), ΔH₄ = −172.5 kJ

3) Cancel out the common species on both sides:

CO₂(g) ⇒ the sum of 1*^st and 2^nd equations versus the sum of 3*^d and 4^th equations

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

CO(g) ⇒ 4^th equation versus 2^nd equation

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

C(graphite) → C(diamond)

4) Add the ΔH values of (1*^st), (2^nd), (3*^d) and (4^th) equations to get your answer:

ΔH_x = ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄

ΔH_x = (395.3 kJ) + (566.0 kJ) + (−787.0 kJ) + (−172.5 kJ) = 1.8 kJ

ΔH_x = 1.8 kJ

C(graphite) → C(diamond), ΔH_o = 1.8 kJ

Answer: ΔH_o = 1.8 kJ