Reaction 1: 4 NH3(g) + 5 O2(g) -> 4 NO(g) + 6 H2O(l) dG° = -1010 kJ/mol_rxn Reaction 2: 2 NO2(g) -> — Thermodynamics Chemistry Question

STEPS:

1. Identify the core chemical concept tested:
This question tests the application of Hess's Law to Gibbs free energy ($\Delta G^\circ$). Because Gibbs free energy is a thermodynamic state function, the overall free energy change of a multi-step process is independent of the pathway. Therefore, if a target chemical equation can be written as the sum of a series of individual reaction steps, the overall $\Delta G^\circ$ is simply the sum of the free energy changes of those steps.

2. Analyze the target reaction and align reactants/products:
Our goal is to manipulate the three given reactions so that they add up to the target overall reaction:
$$4\text{ NH}_3(g) + 8\text{ O}_2(g) \rightarrow 4\text{ HNO}_3(aq) + 4\text{ H}_2\text{O}(l) \quad \text{}$$
* reactant alignment ($\text{NH}_3$): The target reaction has $4\text{ NH}_3(g)$ on the reactant side. Reaction 1 also contains $4\text{ NH}_3(g)$ on the reactant side, so we keep Reaction 1 exactly as written:
$$\text{Reaction 1: } 4\text{ NH}_3(g) + 5\text{ O}_2(g) \rightarrow 4\text{ NO}(g) + 6\text{ H}_2\text{O}(l) \quad \Delta G^\circ_1 = -1010\text{ kJ/mol}_{\text{rxn}} \quad \text{}$$
* product alignment ($\text{HNO}_3$): The target reaction has $4\text{ HNO}_3(aq)$ on the product side. Reaction 3 also contains $4\text{ HNO}_3(aq)$ on the product side, so we keep Reaction 3 exactly as written:
$$\text{Reaction 3: } 4\text{ NO}_2(g) + \text{O}_2(g) + 2\text{ H}_2\text{O}(l) \rightarrow 4\text{ HNO}_3(aq) \quad \Delta G^\circ_3 = -170\text{ kJ/mol}_{\text{rxn}} \quad \text{}$$

3. Align intermediate species for cancellation ($\text{NO}$ and $\text{NO}_2$):
* The intermediate species $\text{NO}(g)$ and $\text{NO}_2(g)$ do not appear in the target overall reaction. Thus, they must cancel out when the steps are added.
* To consume the $4\text{ NO}(g)$ produced in Reaction 1 and supply the $4\text{ NO}_2(g)$ required in Reaction 3, we must manipulate Reaction 2.
* Reaction 2 is written as:
$$\text{Reaction 2: } 2\text{ NO}_2(g) \rightarrow 2\text{ NO}(g) + \text{O}_2(g) \quad \Delta G^\circ_2 = 70\text{ kJ/mol}_{\text{rxn}} \quad \text{}$$
* We must reverse Reaction 2 and multiply it by a factor of 2:
$$\text{Modified Reaction 2: } 4\text{ NO}(g) + 2\text{ O}_2(g) \rightarrow 4\text{ NO}_2(g)$$
* According to thermodynamic rules, when a reaction is reversed, the sign of $\Delta G^\circ$ is flipped. When it is multiplied by a coefficient, its $\Delta G^\circ$ is multiplied by that same factor.
$$\Delta G^\circ_{\text{modified 2}} = -2 \times (70\text{ kJ/mol}_{\text{rxn}}) = -140\text{ kJ/mol}_{\text{rxn}}$$

4. Sum the equations to verify the target reaction:
Add the three mathematical steps together:
$$\begin{array}{rll}\text{Reaction 1:}& 4{\text{ NH}}_3(g)+5{\text{ O}}_2(g)\to \cancel{4\text{ NO}(g)}+6{\text{ H}}_2\text{O}(l)& \mathrm{\Delta }{G}_1^{\circ }=-1010{\text{ kJ/mol}}_{\text{rxn}}\\ \text{Modified Reaction 2:}& \cancel{4\text{ NO}(g)}+2{\text{ O}}_2(g)\to \cancel{4{\text{ NO}}_2(g)}& \mathrm{\Delta }{G}_{\text{modified 2}}^{\circ }=-140{\text{ kJ/mol}}_{\text{rxn}}\\ \text{Reaction 3:}& \cancel{4{\text{ NO}}_2(g)}+{\text{O}}_2(g)+2{\text{ H}}_2\text{O}(l)\to 4{\text{ HNO}}_3(aq)& \mathrm{\Delta }{G}_3^{\circ }=-170{\text{ kJ/mol}}_{\text{rxn}}\\ \text{Summed Reaction:}& 4{\text{ NH}}_3(g)+8{\text{ O}}_2(g)\to 4{\text{ HNO}}_3(aq)+4{\text{ H}}_2\text{O}(l)& \end{array}$$
\begin{array}{rll}
\text{Reaction 1:} & 4\text{ NH}_3(g) + 5\text{ O}_2(g) \rightarrow \cancel{4\text{ NO}(g)} + 6\text{ H}_2\text{O}(l) & \Delta G^\circ_1 = -1010\text{ kJ/mol}_{\text{rxn}} \\
\text{Modified Reaction 2:} & \cancel{4\text{ NO}(g)} + 2\text{ O}_2(g) \rightarrow \cancel{4\text{ NO}_2(g)} & \Delta G^\circ_{\text{modified 2}} = -140\text{ kJ/mol}_{\text{rxn}} \\
\text{Reaction 3:} & \cancel{4\text{ NO}_2(g)} + \text{O}_2(g) + 2\text{ H}_2\text{O}(l) \rightarrow 4\text{ HNO}_3(aq) & \Delta G^\circ_3 = -170\text{ kJ/mol}_{\text{rxn}} \\
\hline
\text{Summed Reaction:} & 4\text{ NH}_3(g) + 8\text{ O}_2(g) \rightarrow 4\text{ HNO}_3(aq) + 4\text{ H}_2\text{O}(l) &
\end{array}

*(Note: The $2\text{ H}_2\text{O}$ on the reactant side cancels out with two of the $6\text{ H}_2\text{O}$ on the product side to leave net $4\text{ H}_2\text{O}$. The reactant $\text{O}_2$ moles sum to $5 + 2 + 1 = 8\text{ O}_2$).*

5. Calculate the final free energy change ($\Delta G^\circ_{\text{overall}}$):
* Sum the individual free energy changes:
$$\Delta G^\circ_{\text{overall}} = \Delta G^\circ_1 + \Delta G^\circ_{\text{modified 2}} + \Delta G^\circ_3$$
$$\Delta G^\circ_{\text{overall}} = (-1010) + (-140) + (-170) = \mathbf{-1320\text{ kJ/mol}_{\text{rxn}}}$$
* This confirms Option D is the correct answer.

*

WHY_OTHERS_WRONG:

• Option A is incorrect ($-1040\text{ kJ/mol}_{\text{rxn}}$): A student would arrive at this value if they multiplied Reaction 2 by 2 to get a magnitude of $140\text{ kJ/mol}_{\text{rxn}}$ but forgot to reverse its sign from positive to negative. This yields $-1010 + 140 - 170 = -1040\text{ kJ/mol}_{\text{rxn}}$.
• Option B is incorrect ($-1110\text{ kJ/mol}_{\text{rxn}}$): This represents a direct sum of the three given values without performing any of the necessary reversals or stoichiometric modifications for Reaction 2. The calculation is $-1010 + 70 - 170 = -1110\text{ kJ/mol}_{\text{rxn}}$.
• Option C is incorrect ($-1250\text{ kJ/mol}_{\text{rxn}}$): A student would obtain this value if they correctly reversed the sign of Reaction 2 to $-70\text{ kJ/mol}_{\text{rxn}}$ but forgot to multiply it by the coefficient of 2 to balance and cancel the intermediates. The calculation is $-1010 - 70 - 170 = -1250\text{ kJ/mol}_{\text{rxn}}$.