Caculate `DeltaG(kJ//"mole") "for the reaction at 300K"`
`N_(2)(g)+O_(2)(g)Leftrightarrow 2NO(g)`
at constant where partical pressure of `N_(2), O_(2)` and `NO are 10^(-1)"bar", 10^(-3)"bar"`. `DeltaH_(f)^(@)NO(g) at 300K =90.5kJ//"mole"and DeltaS_(f)^(@),NO(g) at 300K =12.5J//K"mole"and DeltaS_(f)^(@)NO(g)at 300k=12.5J//K"mole" "` `[2.303xxRxx300=5750J//"mole"]`

Options

A. `173.5kJ//"mole"`
B. `185kJ//"mole"`
C. `162 Kj//"mole"`
D. `84.5kJ//"mole"`

Answer

To calculate the Gibbs free energy change (ΔG) for the reaction

\[ \text{N}_2(g) + \text{O}_2(g) \leftrightarrow 2\text{NO}(g) \]

at 300 K with given partial pressures, we will follow these steps:

### Step 1: Calculate the reaction quotient (Q)

The reaction quotient \( Q \) is calculated using the formula:

\[ Q = \frac{P_{\text{products}}^{\text{stoichiometry}}}{P_{\text{reactants}}^{\text{stoichiometry}}} \]

For the reaction, we have:

- Products: \( 2 \text{NO} \) with a partial pressure of \( 10^{-3} \, \text{bar} \)
- Reactants: \( \text{N}_2 \) with a partial pressure of \( 10^{-1} \, \text{bar} \) and \( \text{O}_2 \) with a partial pressure of \( 10^{-3} \, \text{bar} \)

Thus, we can write:

\[ Q = \frac{(10^{-3})^2}{(10^{-1})(10^{-3})} = \frac{10^{-6}}{10^{-4}} = 10^{-2} \]

### Step 2: Calculate ΔG° for the formation of NO

The standard Gibbs free energy change for the formation of NO can be calculated using the following equation:

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

Given:
- \( \Delta H_f^\circ(\text{NO}) = 90.5 \, \text{kJ/mol} \)
- \( \Delta S_f^\circ(\text{NO}) = 12.5 \, \text{J/(K mol)} = 0.0125 \, \text{kJ/(K mol)} \) (conversion from J to kJ)

Substituting the values:

\[ \Delta G^\circ = 2 \times 90.5 - 300 \times 0.0125 \]

Calculating:

\[ \Delta G^\circ = 181 - 3.75 = 177.25 \, \text{kJ/mol} \]

### Step 3: Calculate ΔG using the equation

Now we can calculate the actual Gibbs free energy change (ΔG) using the equation:

\[ \Delta G = \Delta G^\circ + RT \ln Q \]

Where:
- \( R = 8.314 \, \text{J/(K mol)} = 0.008314 \, \text{kJ/(K mol)} \)
- \( T = 300 \, \text{K} \)
- \( Q = 10^{-2} \)

Calculating \( RT \ln Q \):

\[ RT = 0.008314 \times 300 = 2.4942 \, \text{kJ/mol} \]

Now, we need to calculate \( \ln(10^{-2}) \):

\[ \ln(10^{-2}) = -2 \ln(10) \approx -2 \times 2.303 = -4.606 \]

Thus,

\[ RT \ln Q = 2.4942 \times (-4.606) \approx -11.5 \, \text{kJ/mol} \]

### Step 4: Final calculation of ΔG

Now substituting back into the ΔG equation:

\[ \Delta G = 177.25 - 11.5 \approx 165.75 \, \text{kJ/mol} \]

### Final Answer

Therefore, the Gibbs free energy change \( \Delta G \) for the reaction at 300 K is approximately:

\[ \Delta G \approx 165.75 \, \text{kJ/mol} \]