Sn (s) + 2Cl2(g) --> SnCl4(l).Sn(s) + Cl2(g) --> Sn Cl2(l) DH = -186.2 kJSnCl2(s) + Cl2(g) --> SnCl4(l) DH = -325.1 kJWhat is the change in enthalpy?

Answer 1

Sn (s) + 2Cl2(g) --> SnCl4(l)
Change in enthalpy = DH1 + DH2 + DH3

Sn(s) + Cl2(g) --> Sn Cl2(l)
DH1 = -186.2 kJ

Sn Cl2(l) --> SnCl2(s)
DH2 = ? <-- find this either in a table in your book or online

SnCl2(s) + Cl2(g) --> SnCl4(l)
DH3 = -325.1 kJ