Question

Dissociation of sodium azide is given by,
$Na{N}_3\to Na+3/2N_2$
$\Delta H$ for this is__________.

• A. $3/2\Delta H_{f\left(N_2\right)}^o-\Delta H_{f\left(Na{N}_3\right)}^o$
• B. $-\Delta H_{f\left(Na{N}_3\right)}^o$
• C. $\Delta H_{f\left(Na{N}_3\right)}^o-[\Delta H_{f\left(Na\right)}^o+3/2\Delta H_{f\left(N_2\right)}^o]$
• D. $[\Delta H_{f\left(Na\right)}^o+3/2\Delta H_{f\left(N_2\right)}^o]-\Delta H_{f\left(Na{N}_3\right)}^o$

Answer 1

Answer: (B)

$-\Delta H_{f\left(Na{N}_3\right)}^o$

$Na{N}_3⟶Na+3/2N_2$

$\Delta H=\sum {\Delta H}_f^0\left(products\right)-\sum {\Delta H}_f^0\left(reactants\right)$

$={\Delta H}_f^0\left(Na\right)+\frac{3}{2}{\Delta H}_f^0\left(N_2\right)-{\Delta H}_f^0\left(Na{N}_3\right)$

But $Na$ and $N_2$ are in their standard states.

$\therefore {\Delta H}_f^0\left(Na\right)={\Delta H}_f^0\left(N_2\right)=0$

$\therefore \Delta H=-{\Delta H}_f^0\left({NaN}_3\right)$