Calculate `DeltaG_(reaction) ("kJ"//"mol")` for the given reaction at 300 K
`A_(2)(g)+B_(2)(g)hArr2Ab(g)`
and at particle pressure of `10^(-2)`bar and `10^(-4)`
Given :
`Delta H_(f)^(@) AB =180 kJ//mol," "DeltaH_(f)^(@) A_(2)=60 kJ//mol`
`Delta H_(f)^(@) B_(2) = 29.5 kJ//mol," "DeltaS_(f)^(@) AB=210 J//K-mol`
`Delta S_(f)^(@) A_(2) = 190 kJ//mol," "DeltaS_(f)^(@) B_(2)=205 J//K-mol`
Use :` 2.303 Rxx300=5750 "J"//"mole" `

Answer

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

\[ A_2(g) + B_2(g) \rightleftharpoons 2AB(g) \]

at 300 K, we will follow these steps:

### Step 1: Calculate ΔS (Change in Entropy)
The change in entropy (ΔS) for the reaction can be calculated using the standard molar entropies of the reactants and products.

\[ \Delta S_{reaction} = 2 \Delta S_f^0(AB) - \Delta S_f^0(A_2) - \Delta S_f^0(B_2) \]

Substituting the values:

\[ \Delta S_{reaction} = 2 \times 210 \, \text{J/K·mol} - 190 \, \text{J/K·mol} - 205 \, \text{J/K·mol} \]

Calculating this gives:

\[ \Delta S_{reaction} = 420 - 190 - 205 = 25 \, \text{J/K·mol} \]

### Step 2: Calculate ΔH (Change in Enthalpy)
Next, we calculate the change in enthalpy (ΔH) for the reaction:

\[ \Delta H_{reaction} = 2 \Delta H_f^0(AB) - \Delta H_f^0(A_2) - \Delta H_f^0(B_2) \]

Substituting the values:

\[ \Delta H_{reaction} = 2 \times 180 \, \text{kJ/mol} - 60 \, \text{kJ/mol} - 29.5 \, \text{kJ/mol} \]

Calculating this gives:

\[ \Delta H_{reaction} = 360 - 60 - 29.5 = 270.5 \, \text{kJ/mol} \]

### Step 3: Calculate ΔG (Change in Gibbs Free Energy)
Now we can calculate ΔG using the formula:

\[ \Delta G = \Delta H - T \Delta S + RT \ln Q \]

Where:
- \( R = 8.314 \, \text{J/K·mol} \) (or \( 0.008314 \, \text{kJ/K·mol} \))
- \( T = 300 \, \text{K} \)
- \( Q \) is the reaction quotient.

### Step 4: Calculate Q
The reaction quotient \( Q \) is given by:

\[ Q = \frac{(P_{AB})^2}{(P_{A_2})(P_{B_2})} \]

Given the partial pressures:
- \( P_{AB} = 10^{-4} \, \text{bar} \)
- \( P_{A_2} = 10^{-2} \, \text{bar} \)
- \( P_{B_2} = 10^{-2} \, \text{bar} \)

Substituting these values:

\[ Q = \frac{(10^{-4})^2}{(10^{-2})(10^{-2})} = \frac{10^{-8}}{10^{-4}} = 10^{-4} \]

### Step 5: Calculate \( RT \ln Q \)
Now, we calculate \( RT \ln Q \):

\[ RT = 8.314 \times 300 = 2494.2 \, \text{J/mol} = 2.4942 \, \text{kJ/mol} \]

Calculating \( \ln Q \):

\[ \ln(10^{-4}) = -4 \ln(10) \approx -4 \times 2.303 = -9.212 \]

Now substituting into \( RT \ln Q \):

\[ RT \ln Q = 2.4942 \times (-9.212) \approx -22.94 \, \text{kJ/mol} \]

### Step 6: Final Calculation of ΔG
Now we can substitute everything into the ΔG equation:

\[ \Delta G = 270.5 \, \text{kJ/mol} - 300 \times \frac{25}{1000} + (-22.94) \]

Calculating:

\[ \Delta G = 270.5 - 7.5 - 22.94 = 240.06 \, \text{kJ/mol} \]

### Final Answer
Thus, the change in Gibbs free energy for the reaction at 300 K is:

\[ \Delta G_{reaction} \approx 240.06 \, \text{kJ/mol} \]

---