Question

The following reaction is performed at $298\, K ?$ $2 NO ( g )+ O _{2}( g ) \rightleftharpoons 2 NO _{2}( g )$ The standard free energy of formation of $NO ( g )$ is $86.6 \,kJ /\, mol$ at $298 \,K .$ What is the standard free energy of formation of $NO _{2}( g )$ at $298 \,K ?\left( K _{n}=1.6 \times 10^{12}\right)$

• A. $R (298) \, in \, (1.6 \times 10^{12}) - 86600$
• B. $86600 + R(298) \, in \, (1.6 \times 10^{12})$
• C. $86600 - \frac{In \, (1.6 \times 10^{12})}{R(298)}$
• D. $0.5[2 \times 86600 - R(298) \, In \, (1.6 \times 10^{12})]$

Answer 1

The Correct Option is D

To determine the standard free energy of formation of \( \text{NO}_2(g) \) at 298 K, we can utilize the given chemical equilibrium and thermodynamic data. Let's break down the problem and solve it step by step.

1. The given chemical reaction is:

   2 \, \text{NO}(g) + \text{O}_2(g) \rightleftharpoons 2 \, \text{NO}_2(g)

2. The equilibrium constant \( K_n \) for this reaction at 298 K is given as \( 1.6 \times 10^{12} \).

3. We use the relation between the standard free energy change (\( \Delta G^\circ \)) and the equilibrium constant:

   \Delta G^\circ = -RT \ln K_n

   where \( R \) is the universal gas constant (8.314 J/mol·K) and \( T \) is the temperature in Kelvin.

4. Substituting the values, we get:

   \Delta G^\circ = -8.314 \times 298 \ln(1.6 \times 10^{12})

5. We also know the standard free energy change is given by:

   \Delta G^\circ = \sum (\Delta G_f^\circ \, \text{products}) - \sum (\Delta G_f^\circ \, \text{reactants})

   For the reaction: \( 2 \text{NO}(g) + \text{O}_2(g) \rightleftharpoons 2 \text{NO}_2(g) \), it becomes:

   \Delta G^\circ = 2\Delta G_f^\circ (\text{NO}_2) - [2\Delta G_f^\circ (\text{NO}) + \Delta G_f^\circ (\text{O}_2)]

   Since the standard free energy of formation of \( \text{O}_2(g) \) is zero:

   \Delta G^\circ = 2\Delta G_f^\circ (\text{NO}_2) - 2 \times 86600\, \text{J/mol}

6. Equating both expressions for \( \Delta G^\circ \) gives:

   2\Delta G_f^\circ (\text{NO}_2) - 2 \times 86600 = -8.314 \times 298 \ln(1.6 \times 10^{12})

7. Solving for \(\Delta G_f^\circ (\text{NO}_2)\):

   2\Delta G_f^\circ (\text{NO}_2) = 2 \times 86600 - 8.314 \times 298 \ln(1.6 \times 10^{12})

   \Delta G_f^\circ (\text{NO}_2) = \frac{1}{2}[2 \times 86600 - 8.314 \times 298 \ln(1.6 \times 10^{12})]

Hence, the correct answer is:

0.5[2 \times 86600 - R(298) \ln(1.6 \times 10^{12})]