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Step-by-Step Solution
Step 1: Identify the given data
• Standard free energy of formation of NO(g), $ \Delta G^\circ_{NO} $, is given as:
$$
\Delta G^\circ_{NO} = 86.6 \, \text{kJ/mol} = 86600 \, \text{J/mol}.
$$
• Let $ \Delta G^\circ_{NO_2} = x\, (\text{J/mol}) $ be the standard free energy of formation of NO2(g).
• The reaction under consideration is:
$$
2\,\text{NO}(g) + \text{O}_2(g) \rightleftharpoons 2\,\text{NO}_2(g).
$$
• The temperature is $ T = 298 \,\text{K} $.
• The equilibrium constant is $ K_p = 1.6 \times 10^{12}. $
Step 2: Write the standard free energy change for the reaction
The standard Gibbs free energy change for the reaction, $ \Delta G^\circ_{\text{rxn}} $, is related to the equilibrium constant by the formula:
$$
\Delta G^\circ_{\text{rxn}} = - R T \,\ln(K_p),
$$
where $ R $ is the gas constant and $ T $ is the absolute temperature (in kelvin).
Step 3: Express $ \Delta G^\circ_{\text{rxn}} $ in terms of formation free energies
For the balanced reaction
$$
2\,\text{NO}(g) + \text{O}_2(g) \rightarrow 2\,\text{NO}_2(g),
$$
the standard Gibbs free energy change of the reaction can also be written using the standard free energies of formation:
$$
\Delta G^\circ_{\text{rxn}}
= 2\,\Delta G^\circ_{NO_2} - \Bigl[2\,\Delta G^\circ_{NO} + \Delta G^\circ_{O_2}\Bigr].
$$
Since the standard free energy of formation of $ \text{O}_2(g) $ in its elemental form is zero, this reduces to:
$$
\Delta G^\circ_{\text{rxn}}
= 2\,\Delta G^\circ_{NO_2} - 2\,\Delta G^\circ_{NO}.
$$
Step 4: Substitute known values and solve for $ \Delta G^\circ_{NO_2} $
According to step 2, we have:
$$
\Delta G^\circ_{\text{rxn}} = -R \times 298 \times \ln\bigl(1.6 \times 10^{12}\bigr).
$$
According to the expression in step 3:
$$
2\,\Delta G^\circ_{NO_2} - 2 \times 86600
= -R \times 298 \times \ln\bigl(1.6 \times 10^{12}\bigr).
$$
Rearrange to solve for $ 2\,\Delta G^\circ_{NO_2} $:
$$
2\,\Delta G^\circ_{NO_2}
= 2 \times 86600 \;-\; R \times 298 \;\ln\bigl(1.6 \times 10^{12}\bigr).
$$
Finally, divide both sides by 2 to find $ \Delta G^\circ_{NO_2} $:
$$
\Delta G^\circ_{NO_2}
= \frac{1}{2}\Bigl[\,2 \times 86600 \;-\; R \times 298 \;\ln\bigl(1.6 \times 10^{12}\bigr)\Bigr].
$$
Hence,
$$
\Delta G^\circ_{NO_2}
= 0.5\,\Bigl[\,2 \times 86600 \;-\; R \times 298 \;\ln\bigl(1.6 \times 10^{12}\bigr)\Bigr].
$$
Step 5: Identify the correct option
Comparing with the given options, the correct answer matches:
0.5[2×86,600 – R(298) ln(1.6×1012)].
