SCH4U - Chemistry 12 (2024-25) - A

Hess's Law Solution

Table of Standard Enthalpies of Formation

Solve for ΔH:

CH_4(g) + 2O_2(g)  →  CO_2(g) + 2H₂O_(g)    ΔH = ?

Step 1. Find an equation on the table that has CH₄ in it:

C_(s) + 2H_2(g) → CH_4(g)    ΔH₁ = -75.18 kJ

We need to have CH_4(g) on the reactants side though, so we need to reverse this reaction, therefore we must reverse the sign of ΔH.

CH_4(g) → C_(s) + 2H₂(g)   ΔH₁ = +75.18 kJ

Step 2. Find an equation with CO₂ as a product that has preferably has either C_(s) or H₂(g) as a reactant.

C_(s) + O_2(g) → CO_2(g)  ΔH₂ = -395.28 kJ

Step 3. Find an equation with H₂(g) as a reactant.

H_2(g) + 1/2O_2(g) → H₂O_(g)     ΔH₃ = -242.76 kJ

Step 4. Add up the equations

1. CH_4(g) → C_(s) + 2H₂(g)         ΔH₁ = +75.18 kJ

2. C_(s) + O_2(g) → CO_2(g)               ΔH₂ = -395.28 kJ

3. H_2(g) + 1/2O_2(g) → H₂O_(g)     ΔH₃ = -242.76 kJ

We have a problem here however, because we have only 1 mole of H₂(g) on the reactant side (equation #3) and 2 moles on the product side (equation #1), therefore we need to multiply equation #3 by a factor of 2: (remember that this means we must multiply the ΔH by 2 as well).

2H_2(g) + O_2(g) → 2H₂O_(g)     ΔH₃ = -485.52 kJ

So lets add up our equations now:

1. CH_4(g)           → C_(s) + 2H₂(g)         ΔH₁ = +75.18 kJ

2. C_(s) + O_2(g)    → CO_2(g)                        ΔH₂ = -395.28 kJ

3. 2H_2(g) + O_2(g) → 2H₂O_(g)                ΔH₃ = -485.52 kJ

CH_4(g)  + O_2(g)   → CO_2(g) + 2H₂O_(g)    ΔH = -805.62 kJ

Therefore, the ΔH for the combustion of methane (CH₄), using Hess's Law is -805.62 kJ.