Question

1. use the standard enthalpies of formation to determine the standard enthalpy change ($delta_{r}h^{circ}$) in the following reaction

$2 nh_{3}(g)\to n_{2}(g)+3 h_{2}(g)$
$delta_{f}h^{circ} nh_{3}(g)= - 46.11 kj mol^{-1}$
a) $-46.11 kj mol^{-1}$
b) $-92.22 kj mol^{-1}$
c) $92.22 kj mol^{-1}$
d) $5.42 kj mol^{-1}$
e) $46.11 kj mol^{-1}$

Explanation:

Step1: Recall enthalpy - change formula

The formula for the standard enthalpy change of a reaction $\Delta_{r}H^{\circ}=\sum n\Delta_{f}H^{\circ}_{products}-\sum m\Delta_{f}H^{\circ}_{reactants}$, where $n$ and $m$ are the stoichiometric coefficients. For the reaction $2NH_{3}(g)
ightarrow N_{2}(g)+3H_{2}(g)$, the standard enthalpy of formation of $N_{2}(g)$ and $H_{2}(g)$ are $\Delta_{f}H^{\circ}(N_{2},g) = 0\ kJ/mol$ and $\Delta_{f}H^{\circ}(H_{2},g)=0\ kJ/mol$ (by definition, the standard enthalpy of formation of an element in its standard state is zero), and $\Delta_{f}H^{\circ}(NH_{3},g)= - 46.11\ kJ/mol$.

Step2: Calculate $\Delta_{r}H^{\circ}$

$\Delta_{r}H^{\circ}=[1\times\Delta_{f}H^{\circ}(N_{2},g)+3\times\Delta_{f}H^{\circ}(H_{2},g)]-[2\times\Delta_{f}H^{\circ}(NH_{3},g)]$. Substitute the values: $\Delta_{r}H^{\circ}=[1\times0 + 3\times0]-[2\times(-46.11\ kJ/mol)]$. First, $1\times0 + 3\times0 = 0$. Then, $2\times(-46.11\ kJ/mol)=-92.22\ kJ/mol$. So, $\Delta_{r}H^{\circ}=0-(-92.22\ kJ/mol)=92.22\ kJ/mol$.

Answer:

C. $92.22\ kJ\ mol^{-1}$