Calculate (a) `DeltaG^(Θ)` and (b) the equilibrium constant for the formation of `NO` and `O_(2)` at `298 K`
`NO(g)+1//2 O_(2)(g) hArr NO_(2)(g)`
where
`Delta_(f)G^(Θ)(NO_(2))=52.0 kJ mol^(-1)`
`Delta_(f)G^(Θ)(NO)=87.0 kJ mol^(-1)`
`Delta_(f)G^(Θ)(O_(2))=0 kJ mol^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the Reaction and Identify Given Data
The reaction is:
\[ \text{NO(g)} + \frac{1}{2} \text{O}_2(g) \rightleftharpoons \text{NO}_2(g) \]

Given data:
- \( \Delta_f G^\circ (\text{NO}_2) = 52.0 \, \text{kJ/mol} \)
- \( \Delta_f G^\circ (\text{NO}) = 87.0 \, \text{kJ/mol} \)
- \( \Delta_f G^\circ (\text{O}_2) = 0 \, \text{kJ/mol} \)

### Step 2: Calculate \( \Delta G^\circ \) for the Reaction
Using the formula:
\[ \Delta G^\circ = \Delta G_f^\circ (\text{products}) - \Delta G_f^\circ (\text{reactants}) \]

Substituting the values:
\[ \Delta G^\circ = \Delta_f G^\circ (\text{NO}_2) - \left( \Delta_f G^\circ (\text{NO}) + \frac{1}{2} \Delta_f G^\circ (\text{O}_2) \right) \]

\[ \Delta G^\circ = 52.0 \, \text{kJ/mol} - \left( 87.0 \, \text{kJ/mol} + \frac{1}{2} \times 0 \, \text{kJ/mol} \right) \]

\[ \Delta G^\circ = 52.0 \, \text{kJ/mol} - 87.0 \, \text{kJ/mol} \]

\[ \Delta G^\circ = -35.0 \, \text{kJ/mol} \]

### Step 3: Convert \( \Delta G^\circ \) to Joules
Since we need to use the value in Joules for the next calculation:
\[ \Delta G^\circ = -35.0 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = -35000 \, \text{J/mol} \]

### Step 4: Calculate the Equilibrium Constant \( K_c \)
Using the relationship between \( \Delta G^\circ \) and \( K_c \):
\[ \Delta G^\circ = -2.303RT \log K_c \]

Rearranging for \( \log K_c \):
\[ \log K_c = -\frac{\Delta G^\circ}{2.303RT} \]

Substituting the values:
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 298 \, \text{K} \)

\[ \log K_c = -\frac{-35000 \, \text{J/mol}}{2.303 \times 8.314 \, \text{J/(mol K)} \times 298 \, \text{K}} \]

Calculating the denominator:
\[ 2.303 \times 8.314 \times 298 \approx 5730.4 \]

Now substituting:
\[ \log K_c = \frac{35000}{5730.4} \approx 6.11 \]

### Step 5: Calculate \( K_c \)
To find \( K_c \):
\[ K_c = 10^{\log K_c} = 10^{6.11} \]

Calculating \( K_c \):
\[ K_c \approx 1.36 \times 10^6 \]

### Final Answers
(a) \( \Delta G^\circ = -35.0 \, \text{kJ/mol} \)

(b) \( K_c \approx 1.36 \times 10^6 \)

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