For the reaction at `300 K`
`A(g)hArrV(g)+S(g)`
`Delta_(r)H^(@)=-30kJ//mol, Delta_(r)S^(@)=-0.1kJK^(-1).mol^(-1)`
What is the value of equilibrium constant ?

Options

A. `0`
B. `1`
C. `10`
D. None of these

Answer

To find the value of the equilibrium constant (K) for the given reaction at 300 K, we will follow these steps:

### Step 1: Calculate ΔG° using the Gibbs free energy equation
The Gibbs free energy change (ΔG°) can be calculated using the formula:
\[ \Delta G° = \Delta H° - T \Delta S° \]
Where:
- ΔH° = -30 kJ/mol (given)
- T = 300 K (given)
- ΔS° = -0.1 kJ/K·mol (given)

### Step 2: Substitute the values into the equation
Substituting the values into the equation:
\[ \Delta G° = (-30 \text{ kJ/mol}) - (300 \text{ K})(-0.1 \text{ kJ/K·mol}) \]
Calculating the second term:
\[ 300 \times -0.1 = -30 \text{ kJ/mol} \]
Now substituting this back into the equation:
\[ \Delta G° = -30 \text{ kJ/mol} + 30 \text{ kJ/mol} = 0 \text{ kJ/mol} \]

### Step 3: Use the relationship between ΔG° and K
We can relate ΔG° to the equilibrium constant (K) using the formula:
\[ \Delta G° = -2.303 RT \log K \]
Where:
- R = 8.314 J/(mol·K) = 0.008314 kJ/(mol·K)
- T = 300 K

### Step 4: Substitute ΔG° into the equation
Since we found that ΔG° = 0 kJ/mol, we substitute this into the equation:
\[ 0 = -2.303 \times (0.008314 \text{ kJ/(mol·K)}) \times (300 \text{ K}) \log K \]

### Step 5: Simplify the equation
Since the left side is 0, we can conclude:
\[ 0 = -2.303 \times (0.008314) \times (300) \log K \]
This implies that:
\[ \log K = 0 \]

### Step 6: Solve for K
From the equation \(\log K = 0\), we find:
\[ K = 10^0 = 1 \]

### Final Answer
The value of the equilibrium constant (K) is **1**.

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