Given the following (hypothetical) thermochemical equations:
A + B → 2C; ∆H = –447 kJ
A + 3D → 2E; ∆H = –484 kJ
2D + B → 2F; ∆H = –429 kJ
Calculate ∆H, in kJ, for the equation
4E + 5B → 4C + 6F