Find out `InK_(eq)` for the formation of `NO_(2)` from `NO` and `O_(2)` at `298 K`
`NO_(g)+(1)/(2)O_(2)hArrNO_(2)g`
Given: `DeltaG_(f)^(@)(NO_(2))=52.0KJ//mol e`
`Delta_(f)^(@)(NO)=87.0KJ//mol e`
`Delta_(f)^(@)(O_(2))=0KJ//mol e`

Options

A. `(35xx10^(3))/(8.314xx298)`
B. `-(35xx10^(3))/(8.314xx298)`
C. `(35xx10^(3))/(2.303xx8.314xx298)`
D. `(35xx10^(3))/(2xx298)`

Explanation

To find the equilibrium constant \( K_{eq} \) for the formation of \( NO_2 \) from \( NO \) and \( O_2 \) at 298 K, we can follow these steps:

### Step 1: Write the Reaction
The reaction for the formation of nitrogen dioxide (\( NO_2 \)) from nitrogen monoxide (\( NO \)) and oxygen (\( O_2 \)) is:
\[ NO(g) + \frac{1}{2} O_2(g) \rightleftharpoons NO_2(g) \]

### Step 2: Write Down the Given Data
We are given the standard Gibbs free energy of formation (\( \Delta G_f^\circ \)) for the substances involved:
- \( \Delta G_f^\circ (NO_2) = 52.0 \, \text{kJ/mol} \)
- \( \Delta G_f^\circ (NO) = 87.0 \, \text{kJ/mol} \)
- \( \Delta G_f^\circ (O_2) = 0 \, \text{kJ/mol} \)

### Step 3: Calculate \( \Delta G^\circ_{reaction} \)
Using the formula for the standard Gibbs free energy change of the reaction:
\[ \Delta G^\circ_{reaction} = \sum \Delta G_f^\circ (products) - \sum \Delta G_f^\circ (reactants) \]
Substituting the values:
\[ \Delta G^\circ_{reaction} = \Delta G_f^\circ (NO_2) - \left( \Delta G_f^\circ (NO) + \frac{1}{2} \Delta G_f^\circ (O_2) \right) \]
\[ = 52.0 \, \text{kJ/mol} - \left( 87.0 \, \text{kJ/mol} + \frac{1}{2} \times 0 \right) \]
\[ = 52.0 \, \text{kJ/mol} - 87.0 \, \text{kJ/mol} \]
\[ = -35.0 \, \text{kJ/mol} \]

### Step 4: Convert \( \Delta G^\circ_{reaction} \) to Joules
Since we will use the gas constant \( R \) in Joules, we need to convert \( \Delta G^\circ_{reaction} \) to Joules:
\[ \Delta G^\circ_{reaction} = -35.0 \, \text{kJ/mol} = -35.0 \times 10^3 \, \text{J/mol} \]

### Step 5: Use the Relationship Between \( \Delta G^\circ \) and \( K_{eq} \)
At equilibrium, we know:
\[ \Delta G^\circ = -RT \ln K_{eq} \]
Rearranging gives:
\[ \ln K_{eq} = -\frac{\Delta G^\circ}{RT} \]

### Step 6: Substitute Values
Substituting \( R = 8.314 \, \text{J/(mol K)} \) and \( T = 298 \, \text{K} \):
\[ \ln K_{eq} = -\frac{-35.0 \times 10^3 \, \text{J/mol}}{8.314 \, \text{J/(mol K)} \times 298 \, \text{K}} \]
Calculating the denominator:
\[ = 8.314 \times 298 \approx 2478.572 \, \text{J/mol} \]
Now substituting:
\[ \ln K_{eq} = \frac{35.0 \times 10^3}{2478.572} \approx 14.12 \]

### Step 7: Final Result
Thus, the value of \( \ln K_{eq} \) is approximately \( 14.12 \).

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