Question

determine the standard enthalpy change for each of the following reactions. report your answers to the nearest 0.1 kj. appendix d
2nh₃(g)+3o₂(g)+2ch₄(g)→2hcn(g)+6h₂o(g)
etextbook and media
2al(s)+3cl₂(g)→2alcl₃(s)
3no₂(g)+h₂o(l)→2hno₃(g)+no(g)

Explanation:

Step1: Recall the formula for $\Delta H^{\circ}_{rxn}$

$\Delta H^{\circ}_{rxn}=\sum n\Delta H^{\circ}_{f}(products)-\sum m\Delta H^{\circ}_{f}(reactants)$, where $n$ and $m$ are the stoichiometric coefficients and $\Delta H^{\circ}_{f}$ are the standard - enthalpies of formation. We need to look up the standard - enthalpies of formation values from Appendix D.

Reaction 1: $2NH_{3}(g)+3O_{2}(g)+2CH_{4}(g)

ightarrow2HCN(g)+6H_{2}O(g)$
Let's assume the standard - enthalpies of formation values: $\Delta H^{\circ}_{f}(NH_{3}(g))=-46.1\text{ kJ/mol}$, $\Delta H^{\circ}_{f}(O_{2}(g)) = 0\text{ kJ/mol}$, $\Delta H^{\circ}_{f}(CH_{4}(g))=-74.6\text{ kJ/mol}$, $\Delta H^{\circ}_{f}(HCN(g))=135.1\text{ kJ/mol}$, $\Delta H^{\circ}_{f}(H_{2}O(g))=-241.8\text{ kJ/mol}$

Step2: Calculate $\sum n\Delta H^{\circ}_{f}(products)$

$\sum n\Delta H^{\circ}_{f}(products)=2\times\Delta H^{\circ}_{f}(HCN(g)) + 6\times\Delta H^{\circ}_{f}(H_{2}O(g))=2\times135.1+6\times(-241.8)=270.2-1450.8=- 1180.6\text{ kJ/mol}$

Step3: Calculate $\sum m\Delta H^{\circ}_{f}(reactants)$

$\sum m\Delta H^{\circ}_{f}(reactants)=2\times\Delta H^{\circ}_{f}(NH_{3}(g))+3\times\Delta H^{\circ}_{f}(O_{2}(g))+2\times\Delta H^{\circ}_{f}(CH_{4}(g))=2\times(-46.1)+3\times0 + 2\times(-74.6)=-92.2-149.2=-241.4\text{ kJ/mol}$

Step4: Calculate $\Delta H^{\circ}_{rxn}$

$\Delta H^{\circ}_{rxn}=\sum n\Delta H^{\circ}_{f}(products)-\sum m\Delta H^{\circ}_{f}(reactants)=-1180.6-(-241.4)=-939.2\text{ kJ}$

Reaction 2: $2Al(s)+3Cl_{2}(g)

ightarrow2AlCl_{3}(s)$
Assume $\Delta H^{\circ}_{f}(Al(s)) = 0\text{ kJ/mol}$, $\Delta H^{\circ}_{f}(Cl_{2}(g)) = 0\text{ kJ/mol}$, $\Delta H^{\circ}_{f}(AlCl_{3}(s))=-704.2\text{ kJ/mol}$

Step1: Calculate $\sum n\Delta H^{\circ}_{f}(products)$

$\sum n\Delta H^{\circ}_{f}(products)=2\times\Delta H^{\circ}_{f}(AlCl_{3}(s))=2\times(-704.2)=-1408.4\text{ kJ/mol}$

Step2: Calculate $\sum m\Delta H^{\circ}_{f}(reactants)$

$\sum m\Delta H^{\circ}_{f}(reactants)=2\times\Delta H^{\circ}_{f}(Al(s))+3\times\Delta H^{\circ}_{f}(Cl_{2}(g))=2\times0+3\times0 = 0\text{ kJ/mol}$

Step3: Calculate $\Delta H^{\circ}_{rxn}$

$\Delta H^{\circ}_{rxn}=\sum n\Delta H^{\circ}_{f}(products)-\sum m\Delta H^{\circ}_{f}(reactants)=-1408.4 - 0=-1408.4\text{ kJ}$

Reaction 3: $3NO_{2}(g)+H_{2}O(l)

ightarrow2HNO_{3}(g)+NO(g)$
Assume $\Delta H^{\circ}_{f}(NO_{2}(g)) = 33.2\text{ kJ/mol}$, $\Delta H^{\circ}_{f}(H_{2}O(l))=-285.8\text{ kJ/mol}$, $\Delta H^{\circ}_{f}(HNO_{3}(g))=-135.1\text{ kJ/mol}$, $\Delta H^{\circ}_{f}(NO(g)) = 91.3\text{ kJ/mol}$

Step1: Calculate $\sum n\Delta H^{\circ}_{f}(products)$

$\sum n\Delta H^{\circ}_{f}(products)=2\times\Delta H^{\circ}_{f}(HNO_{3}(g))+\Delta H^{\circ}_{f}(NO(g))=2\times(-135.1)+91.3=-270.2 + 91.3=-178.9\text{ kJ/mol}$

Step2: Calculate $\sum m\Delta H^{\circ}_{f}(reactants)$

$\sum m\Delta H^{\circ}_{f}(reactants)=3\times\Delta H^{\circ}_{f}(NO_{2}(g))+\Delta H^{\circ}_{f}(H_{2}O(l))=3\times33.2+(-285.8)=99.6-285.8=-186.2\text{ kJ/mol}$

Step3: Calculate $\Delta H^{\circ}_{rxn}$

$\Delta H^{\circ}_{rxn}=\sum n\Delta H^{\circ}_{f}(products)-\sum m\Delta H^{\circ}_{f}(reactants)=-178.9-(-186.2)=7.3\text{ kJ}$

Answer:

Reaction 1: $-939.2\text{ kJ}$
Reaction 2: $-1408.4\text{ kJ}$
Reaction 3: $7.3\text{ kJ}$