Calc Heat of Vaporization TiCl4: -804.2 + -763.2 KJ/mole

[chemister]

How would you calculate the heat of vaporization of titanium (IV) chloride using the following data:

Ti (s) + 2Cl2 (g)-----TiCl4 (l) Standard Molar Enthalpy of formation= -804.2 KJ/mole

Ti (s) + 2Cl2 (g)-----TiCl4 (g) Standard Molar Enthalpy of formation= -763.2 KJ/mole



Is it something as simple as subtracting or adding the two numbers?

 

[so-crates]

What would be your reaction for the heat ot vaporization? How would that relate to the two equations you have written?

 

[DH]

How would you calculate the heat of vaporization of titanium (IV) chloride using the following data:

Ti (s) + 2Cl2 (g)-----TiCl4 (l) Standard Molar Enthalpy of formation= -804.2 KJ/mole

Ti (s) + 2Cl2 (g)-----TiCl4 (g) Standard Molar Enthalpy of formation= -763.2 KJ/mole



Is it something as simple as subtracting or adding the two numbers?

this is very easy problem!
Heat of vaporization of Titanium 4 cloride: heat required to vaporized Tian...
--> from liquid to gas
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Solve: you reverse the first equation:
TiCl4 (l) --->Ti (s) + 2Cl2 (g) Standard Molar Enthalpy of formation= +804.2 KJ/mole
Then you combine with the second

Ti (s) + 2Cl2 (g)-----TiCl4 (g) Standard Molar Enthalpy of formation= -763.2 KJ/mole
You have:
Ti (s) + 2Cl2 (g) + TiCl4 (l)-----TiCl4 (g) + Ti (s) + 2Cl2 (g)
so, TiCl4(l) --> TiCl4(g) Enthalpy = -763.2 + 804.2 = 41 KJ/mole

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This may be the correct answer!
But this problem is not like that!