Question

3 numeric 1 point calculate, in kj, the $\delta h_{rxn}$ for the following reaction using heats of formation: $4nh_3(g) + 5o_2(g) \
ightarrow 4no(g) + 6h_2o(g)$ $\delta h_f nh_3(g) = -46 kj/mol, \delta h_f no(g) = 90 kj/mol$ and $\delta h_f h_2o(g) = -242 kj/mol$ answer

Explanation:

Step1: Recall reaction enthalpy formula

The formula for reaction enthalpy is:
$$\Delta H_{rxn} = \sum (n \times \Delta H_f^{\circ}(\text{products})) - \sum (n \times \Delta H_f^{\circ}(\text{reactants}))$$
where $n$ is the stoichiometric coefficient of each substance.

Step2: Calculate product enthalpy sum

Multiply each product's $\Delta H_f^{\circ}$ by its coefficient and sum:
$$\sum (n \times \Delta H_f^{\circ}(\text{products})) = 4\times\Delta H_f^{\circ}(\text{NO}(g)) + 6\times\Delta H_f^{\circ}(\text{H}_2\text{O}(g))$$
$$= 4\times90\ \text{kJ/mol} + 6\times(-242)\ \text{kJ/mol}$$
$$= 360\ \text{kJ/mol} - 1452\ \text{kJ/mol} = -1092\ \text{kJ/mol}$$

Step3: Calculate reactant enthalpy sum

Multiply each reactant's $\Delta H_f^{\circ}$ by its coefficient and sum. Note that $\Delta H_f^{\circ}(\text{O}_2(g)) = 0\ \text{kJ/mol}$:
$$\sum (n \times \Delta H_f^{\circ}(\text{reactants})) = 4\times\Delta H_f^{\circ}(\text{NH}_3(g)) + 5\times\Delta H_f^{\circ}(\text{O}_2(g))$$
$$= 4\times(-46)\ \text{kJ/mol} + 5\times0\ \text{kJ/mol} = -184\ \text{kJ/mol}$$

Step4: Compute $\Delta H_{rxn}$

Subtract reactant sum from product sum:
$$\Delta H_{rxn} = -1092\ \text{kJ/mol} - (-184\ \text{kJ/mol})$$
$$= -1092\ \text{kJ/mol} + 184\ \text{kJ/mol}$$

Answer:

$-908\ \text{kJ}$