Calculate the equilibrium constant for the reaction : `NO(g) + 1/2 O_(2)(g) iff NO_(2)(g)`
Given, `Delta_(f)H^(@) at 298 K : NO(g) = 90.4 kJ "mol"^(-1), NO_(2)(g)` = `33.8 kJ "mol"^(-1)` and `DeltaS^(@)` at `298 K = -70.8 J K^(-1) "mol"^(-1)`

Explanation

To calculate the equilibrium constant \( K_c \) for the reaction

\[ \text{NO}(g) + \frac{1}{2} \text{O}_2(g) \iff \text{NO}_2(g) \]

we will follow these steps:

### Step 1: Calculate \( \Delta H^\circ \) for the reaction

Using the standard enthalpy of formation values provided:

- \( \Delta_f H^\circ \) for \( \text{NO}(g) = 90.4 \, \text{kJ/mol} \)
- \( \Delta_f H^\circ \) for \( \text{NO}_2(g) = 33.8 \, \text{kJ/mol} \)
- \( \Delta_f H^\circ \) for \( \text{O}_2(g) = 0 \, \text{kJ/mol} \) (since it is in its standard state)

The formula for \( \Delta H^\circ \) of the reaction is:

\[ \Delta H^\circ = \Delta_f H^\circ (\text{products}) - \Delta_f H^\circ (\text{reactants}) \]

Substituting the values:

\[ \Delta H^\circ = \Delta_f H^\circ (\text{NO}_2) - \left( \Delta_f H^\circ (\text{NO}) + \frac{1}{2} \Delta_f H^\circ (\text{O}_2) \right) \]

\[ \Delta H^\circ = 33.8 \, \text{kJ/mol} - \left( 90.4 \, \text{kJ/mol} + \frac{1}{2} \times 0 \right) \]

\[ \Delta H^\circ = 33.8 \, \text{kJ/mol} - 90.4 \, \text{kJ/mol} = -56.6 \, \text{kJ/mol} \]

### Step 2: Convert \( \Delta H^\circ \) to Joules

To use in the Gibbs free energy equation, we convert \( \Delta H^\circ \) to Joules:

\[ \Delta H^\circ = -56.6 \times 10^3 \, \text{J/mol} = -56600 \, \text{J/mol} \]

### Step 3: Use the given \( \Delta S^\circ \)

Given:

\[ \Delta S^\circ = -70.8 \, \text{J/K} \cdot \text{mol} \]

### Step 4: Calculate \( \Delta G^\circ \)

Using the Gibbs free energy equation:

\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]

At \( T = 298 \, \text{K} \):

\[ \Delta G^\circ = -56600 \, \text{J/mol} - 298 \, \text{K} \times (-70.8 \, \text{J/K} \cdot \text{mol}) \]

Calculating:

\[ \Delta G^\circ = -56600 \, \text{J/mol} + 21098.4 \, \text{J/mol} \]

\[ \Delta G^\circ = -35501.6 \, \text{J/mol} \]

### Step 5: Calculate \( K_c \)

Using the relationship between \( \Delta G^\circ \) and \( K_c \):

\[ \Delta G^\circ = -2.303 RT \log K_c \]

Where \( R = 8.314 \, \text{J/(K} \cdot \text{mol)} \) and \( T = 298 \, \text{K} \):

Rearranging gives:

\[ \log K_c = -\frac{\Delta G^\circ}{2.303 RT} \]

Substituting the values:

\[ \log K_c = -\frac{-35501.6 \, \text{J/mol}}{2.303 \times 8.314 \, \text{J/(K} \cdot \text{mol)} \times 298 \, \text{K}} \]

Calculating:

\[ \log K_c = \frac{35501.6}{2.303 \times 8.314 \times 298} \]

Calculating the denominator:

\[ 2.303 \times 8.314 \times 298 \approx 5730.6 \]

Thus:

\[ \log K_c \approx 6.22 \]

### Step 6: Find \( K_c \)

To find \( K_c \):

\[ K_c = 10^{6.22} \approx 1.67 \times 10^6 \]

### Final Answer:

\[ K_c \approx 1.67 \times 10^6 \]

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