For the reaction at 298 k,
`A(g) +B(g) hArr C(g) +D(g)`
`DeltaH_0=-29.8 kcal, DeltaS^@=-0.100 kcal K^(-1)`
Find the value of equilibrium constant at the same temp.

Answer

To find the equilibrium constant \( K \) for the reaction \( A(g) + B(g) \rightleftharpoons C(g) + D(g) \) at 298 K, we can use the relationship between Gibbs free energy change (\( \Delta G \)), enthalpy change (\( \Delta H \)), and entropy change (\( \Delta S \)). Here are the steps to solve the problem:

### Step 1: Write the Gibbs Free Energy Equation
The Gibbs free energy change at standard conditions is given by:
\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]

### Step 2: Substitute Given Values
We know:
- \( \Delta H^\circ = -29.8 \, \text{kcal} \)
- \( \Delta S^\circ = -0.100 \, \text{kcal K}^{-1} \)
- \( T = 298 \, \text{K} \)

Substituting these values into the Gibbs free energy equation:
\[ \Delta G^\circ = -29.8 \, \text{kcal} - (298 \, \text{K}) \times (-0.100 \, \text{kcal K}^{-1}) \]

### Step 3: Calculate the Entropy Contribution
Calculate the entropy contribution:
\[ 298 \times (-0.100) = -29.8 \, \text{kcal} \]

### Step 4: Calculate \( \Delta G^\circ \)
Now substituting back:
\[ \Delta G^\circ = -29.8 + 29.8 = 0 \, \text{kcal} \]

### Step 5: Relate \( \Delta G^\circ \) to the Equilibrium Constant
We know that:
\[ \Delta G^\circ = -2.303 RT \log K \]
Where \( R \) is the gas constant. In kcal, \( R = 0.001987 \, \text{kcal K}^{-1} \text{mol}^{-1} \).

### Step 6: Substitute \( \Delta G^\circ \) into the Equation
Since \( \Delta G^\circ = 0 \):
\[ 0 = -2.303 \times (0.001987) \times (298) \log K \]

### Step 7: Solve for \( K \)
Since the left side is zero, we have:
\[ \log K = 0 \]
This implies:
\[ K = 10^0 = 1 \]

### Final Answer
The equilibrium constant \( K \) at 298 K is:
\[ K = 1 \]