Answer
$Ti(s) + 2Cl_2(g) -- \gt TiCl_4(l)$ $\Delta_fH^{\circ}\{TiCl_4(l)\} = -804.2kJ/mol$
Work Step by Step
1. Write the separated atoms, and their coefficients, to form a single $NH_3$
$Ti + 4Cl -- \gt TiCl_4$
2. Analyze the appendix J, and find the standard state for each element:
- Hint: It is the compound with $\Delta_fH^{\circ}$ = 0.
- Divide the coefficient by the number of atoms in each compound in standard state:
$\frac{1}{1} Ti(s) + \frac{4}{2}Cl_2(g) -- \gt TiCl_4(l)$
Just add the $\Delta_fH^{\circ}$** after the reaction.
** Appendix J.
$Ti(s) + 2Cl_2(g) -- \gt TiCl_4(l)$ $\Delta_fH^{\circ}\{TiCl_4(l)\} = -804.2kJ/mol$
