The standard enthalpy change for the following reaction is 511 kJ at 298 K. SnCl4(l) --> Sn(s, white) + 2 Cl2(g) What is the standard heat o?

The given thermochemical equation:
SnCl₄(l) → Sn(s, white) + 2 Cl₂(g) …… ΔH° = 511 kJ

Reverse the given thermochemical equation:
Sn(s, white) + 2 Cl₂(g) →SnCl₄(l) ……ΔH° = -511 kJ

The above equation represents that of the standard enthalpy heat of formation of SnCl₄(l).
Hence, ΔHf°[SnCl₄(l)] = -511 kJ/mol