Question

For the decomposition of $1$ mole of sodium chlorate, determine $\Delta H_{reaction}$.
($\Delta H_f^0$ values: $NaCl{O}_3\left(s\right)=-85.7$ kcal$/$mol, $NaCl\left(s\right)=-98.2$ kcal$/$mol, $O_2\left(g\right)=0$ kcal$/$mol).

• A. $-183.9$ kcal
• B. $-91.9$ kcal
• C. $+45.3$ kcal
• D. $+22.5$ kcal
• E. $-12.5$ kcal

Answer 1

The correct option is E $-12.5$ kcal

The decomposition reaction of sodium chlorate is as shown:
$2NaCl{O}_3\to 2NaCl+3O_2$
$\Delta H_{reaction}=2\Delta H_{f,NaCl}+3\Delta H_{f,O2}-2\Delta H_{f,NaCl}$
$\Delta H_{reaction}=2\times (-98.2\text{kcal/mol})+3\times \left(0\text{kcal/mol}\right)-2\times (-85.7\text{kcal/mol})$
$\Delta H_{reaction}=-25\text{kcal/mol}$
This is the enthalpy change for decomposition of 2 moles of sodium chlorate.
The enthalpy change for the decomposition of 1 mole of sodium chlorate is $\frac{-25\text{kcal/mol}}{2}=-12.5\text{kcal/mol}$