4 NH3(g) + 3 O2(g) → 2 N2(g) + 6 H2O(g) If the standard molar heats of formation of ammonia, NH3(g), — Thermodynamics Chemistry Question

STEPS:

1. Recall the standard enthalpy of reaction formula: The standard enthalpy of a reaction ($\Delta H^\circ$) can be calculated from the standard enthalpies of formation ($\Delta H^\circ_f$) of the reactants and products using Hess's Law:
$$\Delta H^\circ_{rxn} = \sum n \Delta H^\circ_f(\text{products}) - \sum m \Delta H^\circ_f(\text{reactants})$$
2. Identify the standard enthalpies of formation for elements: By definition, the standard enthalpy of formation of pure elements in their standard states is zero. Therefore:
* $\Delta H^\circ_f (\text{O}_2(g)) = 0\text{ kJ/mol}$
* $\Delta H^\circ_f (\text{N}_2(g)) = 0\text{ kJ/mol}$
3. Set up the summation with stoichiometric coefficients: Plug the coefficients from the balanced chemical equation ($4\text{ NH}_3(g) + 3\text{ O}_2(g) \rightarrow 2\text{ N2}(g) + 6\text{ H}_2\text{O}(g)$) into the Hess's Law expression:
$$\Delta H^\circ_{rxn} = [2 \times \Delta H^\circ_f(\text{N}_2) + 6 \times \Delta H^\circ_f(\text{H}_2\text{O})] - [4 \times \Delta H^\circ_f(\text{NH}_3) + 3 \times \Delta H^\circ_f(\text{O}_2)]$$
4. Substitute the given values into the equation:
$$\Delta H^\circ_{rxn} = [2(0) + 6(-242)] - [4(-46) + 3(0)]$$
5. Simplify the mathematical terms:
* Products: $6 \times (-242) = -1452\text{ kJ/mol}$
* Reactants: $4 \times (-46) = -184\text{ kJ/mol}$
6. Subtract the reactant total from the product total:
$$\Delta H^\circ_{rxn} = -1452 - (-184) = -1452 + 184 = \mathbf{-1268\text{ kJ/mol}_{rxn}}$$
Rounding to three significant figures gives $-1,270\text{ kJ/mol}_{rxn}$, which matches Option D.

*

WHY_OTHERS_WRONG:

• A is incorrect: This value (~$-190\text{ kJ/mol}$) is close to $-196\text{ kJ/mol}$, which is obtained if a student neglects the stoichiometric coefficients entirely and simply subtracts the two given formation values: $(-242) - (-46)$.
• B is incorrect: This value (~$-290\text{ kJ/mol}$) is close to $-288\text{ kJ/mol}$, which a student would get by adding the two given enthalpies of formation directly without accounting for stoichiometry or the subtraction required by the Hess's Law formula: $(-242) + (-46)$.
• C is incorrect: This value is approximately double the incorrect sum from option B, representing another failure to correctly apply stoichiometric balancing to the Hess's Law formula.
• E is incorrect: This value (~$-1,640\text{ kJ/mol}$) is close to $-1636\text{ kJ/mol}$, which is the result of adding the reactant term and product term magnitudes instead of subtracting them: $-1452 - 184$. This represents a sign error during subtraction.