Given the following thermochemical equations:

C(graphite) + O₂(g) → CO₂(g); ΔrH° = -393.5 kJ/mol
H₂(g) + ½O₂(g) → H₂O(l); ΔrH° = -285.8 kJ/mol
CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g); ΔrH° = +890.3 kJ/mol

Based on the above thermochemical equations, the value of ΔrH° at 298K for the reaction C(graphite) + 2H₂(g) → CH₄(g) will be:

C(graphite) + O₂(g) → CO₂(g); ΔrH° = -393.5 kJ/mol (1)
H₂(g) + ½O₂(g) → H₂O(l); ΔrH° = -285.8 kJ/mol (2)
CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g); ΔrH° = +890.3 kJ/mol (3)

We need to find ΔrH° for the reaction: C(graphite) + 2H₂(g) → CH₄(g) (4)

Multiply equation (2) by 2:
2H₂(g) + O₂(g) → 2H₂O(l); ΔrH° = 2 × (-285.8 kJ/mol) = -571.6 kJ/mol (5)

Add equations (1), (5), and (3):
[C(graphite) + O₂(g) → CO₂(g)] + [2H₂(g) + O₂(g) → 2H₂O(l)] + [CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g)]

C(graphite) + 2H₂(g) + 2O₂(g) → CH₄(g) + 2O₂(g)

Simplifying, we get:
C(graphite) + 2H₂(g) → CH₄(g)

The ΔrH° for this reaction will be the sum of the ΔrH° values for equations (1), (5), and (3):
ΔrH° = -393.5 kJ/mol + (-571.6 kJ/mol) + 890.3 kJ/mol = -74.8 kJ/mol