Question

The standard enthalpy change for the following reaction is 305 kJ at 298 K.
NiCl2(s) ------> Ni(s) + Cl2(g)
What is the standard heat of formation of NiCl2(s)? ________ kJ/mol

Answer 1

$$NiCl_2(s)\rightarrow Ni(s) + Cl_2(g)$$

$$\Delta H_{rxn} = \sum \Delta H_f^0 (products) -\sum \Delta H_f^0(reactants) = [\Delta H_f^0(Ni(s)) + \Delta H_f^0 (Cl_2(g))] - \Delta H_f^0 (NiCl_2(s))$$

$\Delta H_f^0(Ni(s)) = 0 \frac{kJ}{mol}$

$\Delta H_f^0 (Cl_2(g)) = 0 \frac{kJ}{mol}$

$$305 \frac{kJ}{mol} = [0\frac{kJ}{mol}+ 0\frac{kJ}{mol}] - \Delta H_f^0(NiCl_2(S))$$ $$\Delta H_f^0 (NiCl_2(s)) = -305 \frac{kJ}{mol}$$