Question

standard enthalpies of formation
substance\tδhf (kj/mol)
c₂h₂ (g)\t−26.7
nh₃ (g)\t−46.19
hbr (g)\t236.23
hcl (g)\t−92.30
hf (g)\t−268.6
hi (g)\t25.9
nacl (s)\t−411.0

based on the equation and the information in the table, what is the enthalpy of the reaction?
○ δhₓₙ = 46.19 kj
○ δhₓₙ = −46.19 kj
○ δhₓₙ = 92.38 kj
○ δhₓₙ = −92.38 kj

Explanation:

Step1: Recall Enthalpy Formula

The formula for the enthalpy of a reaction ($\Delta H_{rxn}$) using standard enthalpies of formation ($\Delta H_f^\circ$) is:
$\Delta H_{rxn} = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$

Step2: Identify Reaction (Assumed: $\text{N}_2 + 3\text{H}_2

ightarrow 2\text{NH}_3$)
For the reaction forming $\text{NH}_3$:

• Reactants: $\text{N}_2$ (element, $\Delta H_f^\circ = 0$) and $\text{H}_2$ (element, $\Delta H_f^\circ = 0$)
• Products: $2\,\text{NH}_3(g)$, $\Delta H_f^\circ(\text{NH}_3) = -46.19\,\text{kJ/mol}$

Step3: Calculate $\sum \Delta H_f^\circ(\text{products})$

$\sum \Delta H_f^\circ(\text{products}) = 2 \times (-46.19) = -92.38\,\text{kJ}$

Step4: Calculate $\sum \Delta H_f^\circ(\text{reactants})$

$\sum \Delta H_f^\circ(\text{reactants}) = 0 + 3 \times 0 = 0\,\text{kJ}$

Step5: Compute $\Delta H_{rxn}$

$\Delta H_{rxn} = -92.38 - 0 = -92.38\,\text{kJ}$

Answer:

$\boldsymbol{\Delta H_{rxn} = -92.38\,\text{kJ}}$ (Option: $\Delta H_{rxn} = -92.38\,\text{kJ}$)