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Using the following thermochemical data, calculate Hf° of Ti2O3(s). 2TiCl3(s) + 3H2O(l)  Ti2O3(s) + 6HCl(g)...

Using the following thermochemical data, calculate Hf° of Ti2O3(s). 2TiCl3(s) + 3H2O(l)  Ti2O3(s) + 6HCl(g) H° = 224.5 kJ/mol 2Ti(s) + 3Cl2(g)  2TiCl3(s) H° = –1441.8 kJ/mol 4HCl(g) + O2(g)  2Cl2(g) + 2H2O(l) H° = –202.4 kJ/mol A) –1520.9 kJ/mol B) –1463.9 kJ/mol C) 1868.7 kJ/mol D) –1419.7 kJ/mol E) 1014.9 kJ/mol

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Answer #1

2TiCl3(s) + 3H2O(l) ---> Ti2O3(s) + 6HCl(g) DH° = 224.5 kJ/mol

2Ti(s) + 3Cl2(g) ---> 2TiCl3(s) dH° = –1441.8 kJ/mol.

4HCl(g) + O2(g) ---> 2Cl2(g) + 2H2O(l) dH° = –202.4 kJ/mol.

from the above data.

2TiCl3(s) + 3H2O(l) ---> Ti2O3(s) + 6HCl(g)     DH° = 224.5 kJ/mol

2Ti(s) + 3Cl2(g) ---> 2TiCl3(s)               dH° = –1441.8 kJ/mol.

(4HCl(g) + O2(g) ---> 2Cl2(g) + 2H2O(l)        dH° = –202.4 kJ/mol.)*1.5

--------------------------------------------------------------------------

2Ti(s) + 1.5O2(g)   -----------> Ti2O3(s)   DHrxn =

--------------------------------------------------------------------

DHrxn = 224.5 +(-1441.8)+(-202.4*1.5) = -1520.9 kj/mol


answer: A

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