Consider the following reaction:
`2NaN_(3)(s) rarr 2Na(s) +3N_(2)(g)`
The enthalpy change for this reaction at `1atm` is equal to:

Options

A. `-2Delta_(f)H^(Theta) [NaN_(3)(s)]`
B. `+2Delta_(f)H^(Theta)[NaN_(3)(s)]`
C. `3Delta_(f)H^(Theta)[N_(2)(s)+2Delta_(f)H^(Theta)(Na(s)]`
D. `-Delta_(f)H^(Theta)[NaN_(3)(s)]`

Explanation

To find the enthalpy change (ΔH) for the reaction:

\[ 2 \text{NaN}_3(s) \rightarrow 2 \text{Na}(s) + 3 \text{N}_2(g) \]

we will use the concept of standard enthalpy of formation (ΔH_f). The enthalpy change for the reaction can be calculated using the following formula:

\[
\Delta H = \Delta H_f (\text{products}) - \Delta H_f (\text{reactants})
...