The following heats of formation are used to find the change in enthalpy for the formation of tin chloride in the following reaction Sn (s) + 2Cl2(g)Ã*SnCl4(l).
Sn(s) + Cl2(g) Ã* Sn Cl2(l)
DH = -186.2 kJ
SnCl2(s) + Cl2(g)Ã* SnCl4(l)
DH = -325.1 kJ
What is the change in enthalpy?
Give examples of situations in which the entropy is low, and examples of situations in which the entropy is high.
Sn(s) + Cl2(g) Ã* Sn Cl2(l)
DH = -186.2 kJ
SnCl2(s) + Cl2(g)Ã* SnCl4(l)
DH = -325.1 kJ
What is the change in enthalpy?
Give examples of situations in which the entropy is low, and examples of situations in which the entropy is high.