Calculate `triangleE^0` for the reaction:
`2A(g)+B(g) Leftrightarrow A_2B (g)` for which `triangleS^(0)=5.0J//K, K=1.0 xx 10^(-10) and T=300K`

Answer

To calculate \(\Delta E^0\) for the reaction \(2A(g) + B(g) \leftrightarrow A_2B(g)\), we will follow these steps:

### Step 1: Identify the reaction and given values
The reaction is:
\[ 2A(g) + B(g) \leftrightarrow A_2B(g) \]

Given values:
- \(\Delta S^0 = 5.0 \, \text{J/K}\)
- \(K = 1.0 \times 10^{-10}\)
- \(T = 300 \, \text{K}\)

### Step 2: Calculate \(\Delta G^0\)
Using the relationship between the equilibrium constant \(K\) and \(\Delta G^0\):
\[ \Delta G^0 = -RT \ln K \]
Where \(R = 8.314 \, \text{J/(mol K)}\).

Substituting the values:
\[ \Delta G^0 = - (8.314 \, \text{J/(mol K)})(300 \, \text{K}) \ln(1.0 \times 10^{-10}) \]

Calculating \(\ln(1.0 \times 10^{-10})\):
\[ \ln(1.0 \times 10^{-10}) = -10 \ln(10) \approx -10 \times 2.303 = -23.03 \]

Now substituting this back into the equation:
\[ \Delta G^0 = - (8.314)(300)(-23.03) \approx 5740.5 \, \text{J/mol} \approx 5.74 \, \text{kJ/mol} \]

### Step 3: Use the Gibbs free energy equation
The Gibbs free energy equation is:
\[ \Delta G^0 = \Delta H^0 - T \Delta S^0 \]

Rearranging gives:
\[ \Delta H^0 = \Delta G^0 + T \Delta S^0 \]

Substituting the known values:
\[ \Delta H^0 = 5740.5 \, \text{J/mol} + (300 \, \text{K})(5.0 \, \text{J/K}) \]
\[ \Delta H^0 = 5740.5 \, \text{J/mol} + 1500 \, \text{J/mol} = 7240.5 \, \text{J/mol} \approx 7.24 \, \text{kJ/mol} \]

### Step 4: Calculate \(\Delta E^0\)
Using the relationship between \(\Delta H^0\) and \(\Delta E^0\):
\[ \Delta H^0 = \Delta E^0 + \Delta n_g RT \]
Where \(\Delta n_g\) is the change in the number of moles of gas.

For the reaction:
\[ \Delta n_g = (1) - (2 + 1) = -2 \]

Now substituting this into the equation:
\[ \Delta H^0 = \Delta E^0 - 2RT \]
\[ 7240.5 \, \text{J/mol} = \Delta E^0 - 2(8.314)(300) \]
Calculating \(2RT\):
\[ 2RT = 2(8.314)(300) = 4988.4 \, \text{J/mol} \]

Now substituting back:
\[ 7240.5 = \Delta E^0 - 4988.4 \]
\[ \Delta E^0 = 7240.5 + 4988.4 = 12228.9 \, \text{J/mol} \approx 12.23 \, \text{kJ/mol} \]

### Final Answer
\[ \Delta E^0 \approx 12.23 \, \text{kJ/mol} \]

---