For the reaction at `298 `K
`A(g) + B(g) hArr C(g) + D(g)`
If `DeltaH^(@) =29.8` Kcal and `DeltaS^(@)=0.1 Kcal K^(-1)` then calculate reaction constant (k)

Options

A. 1
B. 0
C. 2
D. 4

Explanation

To calculate the reaction constant \( K \) for the reaction \( A(g) + B(g) \rightleftharpoons C(g) + D(g) \) at \( 298 \, K \) with given values of \( \Delta H^\circ = 29.8 \, \text{Kcal} \) and \( \Delta S^\circ = 0.1 \, \text{Kcal K}^{-1} \), we will follow these steps:

### Step 1: Calculate the Standard Gibbs Free Energy Change (\( \Delta G^\circ \))

The formula for the standard Gibbs free energy change is:

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

Substituting the given values into the equation:

\[ \Delta G^\circ = 29.8 \, \text{Kcal} - (298 \, \text{K} \times 0.1 \, \text{Kcal K}^{-1}) \]

Calculating the second term:

\[ 298 \, \text{K} \times 0.1 \, \text{Kcal K}^{-1} = 29.8 \, \text{Kcal} \]

Now substituting back into the Gibbs free energy equation:

\[ \Delta G^\circ = 29.8 \, \text{Kcal} - 29.8 \, \text{Kcal} = 0 \, \text{Kcal} \]

### Step 2: Relate \( \Delta G^\circ \) to the Reaction Constant \( K \)

The relationship between Gibbs free energy and the reaction constant is given by:

\[ \Delta G^\circ = -RT \ln K \]

Where:
- \( R \) is the universal gas constant, which is \( 1.987 \, \text{cal K}^{-1} \text{mol}^{-1} \) or \( 0.001987 \, \text{Kcal K}^{-1} \text{mol}^{-1} \)
- \( T \) is the temperature in Kelvin

Since we have \( \Delta G^\circ = 0 \):

\[ 0 = -RT \ln K \]

### Step 3: Solve for \( K \)

From the equation above, we can see that:

\[ -RT \ln K = 0 \]

This implies:

\[ \ln K = 0 \]

Taking the exponential of both sides:

\[ K = e^0 = 1 \]

### Conclusion

Thus, the reaction constant \( K \) is:

\[ \boxed{1} \]