To find the standard enthalpy change, Δ_rH°, for the reaction C_(graphite) + 2H₂(g) → CH₄(g), we can use Hess's Law. Hess's Law states that the total enthalpy change for a reaction is the same, regardless of the number of steps taken, as long as the initial and final conditions are identical. This means we can manipulate the given thermochemical equations algebraically to obtain the target reaction.

Step 1: List the given equations with their enthalpy changes:

(1) C_(graphite) + O₂(g) → CO₂(g); Δ_rH° = –393.5 kJ mol^–1

(2) H₂(g) + $\frac{1}{2}$ O₂(g) → H₂O(l); Δ_rH° = –285.8 kJ mol^–1

(3) CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g); Δ_rH° = +890.3 kJ mol^–1

Step 2: Identify the target reaction:

C_(graphite) + 2H₂(g) → CH₄(g)

Step 3: Manipulate the given equations to match the target:

Notice that in the target, we have C_(graphite) and 2H₂(g) as reactants, and CH₄(g) as the product. We need to eliminate CO₂, O₂, and H₂O, which are not in the target.

• Equation (1) already has C_(graphite) as a reactant and produces CO₂. We can use it as is.
• Equation (2) produces H₂O, but we need 2H₂(g) as reactants. So, multiply equation (2) by 2:

  2 × [H₂(g) + $\frac{1}{2}$ O₂(g) → H₂O(l)] gives:

  2H₂(g) + O₂(g) → 2H₂O(l); Δ_rH° = 2 × (–285.8) = –571.6 kJ mol^–1

• Equation (3) has CH₄ as a product, which we need, but it is written in reverse compared to our needs. To use it properly, reverse equation (3):

  Reverse of (3): CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l); Δ_rH° = –890.3 kJ mol^–1 (since reversing changes the sign)

Step 4: Add the manipulated equations together:

(1) C_(graphite) + O₂(g) → CO₂(g); ΔH = –393.5 kJ

(2 multiplied) 2H₂(g) + O₂(g) → 2H₂O(l); ΔH = –571.6 kJ

(3 reversed) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l); ΔH = –890.3 kJ

Now, add these three equations:

C_(graphite) + O₂(g) + 2H₂(g) + O₂(g) + CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) + CO₂(g) + 2H₂O(l)

Simplify by canceling species that appear on both sides:

• CO₂(g): one on left and one on right? Wait, careful: Actually, after adding, we have:

  Left: C_(graphite) + 2H₂(g) + CH₄(g) + 4O₂(g)

  Right: 2CO₂(g) + 4H₂O(l)

• But we need to get the target reaction. Notice that CH₄(g) is on the left in the reversed equation, which is not what we want. We actually need to have CH₄ as a product. So, let's rearrange the addition properly.

Better Approach: Directly combine to cancel intermediates:

We want: C_(graphite) + 2H₂(g) → CH₄(g)

From the equations, we see that if we take:
(1) C_(graphite) + O₂(g) → CO₂(g); ΔH₁ = –393.5
(2 multiplied) 2H₂(g) + O₂(g) → 2H₂O(l); ΔH₂ = –571.6
(3) CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g); ΔH₃ = +890.3

Now, add these three equations:
Left: C_(graphite) + O₂(g) + 2H₂(g) + O₂(g) + CO₂(g) + 2H₂O(l)
Right: CO₂(g) + 2H₂O(l) + CH₄(g) + 2O₂(g)

Cancel common terms on both sides:
- CO₂(g) cancels.
- 2H₂O(l) cancels.
- O₂(g): on left, 2O₂; on right, 2O₂; they cancel.

We are left with: C_(graphite) + 2H₂(g) → CH₄(g)

This is exactly our target reaction.

Step 5: Calculate the total enthalpy change:

Δ_rH° = ΔH₁ + ΔH₂ + ΔH₃ = (–393.5) + (–571.6) + (+890.3) kJ

Compute step by step:
–393.5 – 571.6 = –965.1
–965.1 + 890.3 = –74.8 kJ mol^–1

Final Answer: The value of Δ_rH° for C_(graphite) + 2H₂(g) → CH₄(g) is $–74.8\mathrm{kJ}{\mathrm{mol}}^{–1}$

Related Topics and Formulae

Hess's Law: The total enthalpy change for a reaction is independent of the pathway, and can be calculated by the algebraic sum of enthalpy changes of individual steps that lead from reactants to products.

Standard Enthalpy of Formation (Δ_fH°): The enthalpy change when one mole of a compound is formed from its elements in their standard states. The reaction we solved is actually the formation reaction for CH₄(g), so Δ_rH° = Δ_fH°(CH₄, g).

Manipulating Thermochemical Equations: When reversing an equation, the sign of ΔH is changed. When multiplying an equation by a factor, ΔH is multiplied by that factor.