`DeltaG^(ɵ)` for the reaction `X+YhArrC` is `-4.606 kcal at `1000 K`. The equilibrium constant for the reverse mode of the reaction will be:

Options

A. `100`
B. `10`
C. `0.01`
D. `0.1`

Explanation

To find the equilibrium constant for the reverse reaction of the given reaction \(X + Y \rightleftharpoons C\), we can follow these steps:

### Step 1: Understand the relationship between \(\Delta G^{\circ}\) and the equilibrium constant \(K\)

The relationship between the standard Gibbs free energy change (\(\Delta G^{\circ}\)) and the equilibrium constant (\(K\)) at a given temperature \(T\) is given by the equation:

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

Where:
- \(R\) is the universal gas constant (1.987 cal/(mol·K))
- \(T\) is the temperature in Kelvin
- \(K\) is the equilibrium constant

### Step 2: Convert \(\Delta G^{\circ}\) to appropriate units

Given \(\Delta G^{\circ} = -4.606 \text{ kcal}\), we need to convert this to calories:

\[ \Delta G^{\circ} = -4.606 \text{ kcal} \times 1000 \text{ cal/kcal} = -4606 \text{ cal} \]

### Step 3: Substitute values into the equation

Now we can substitute the values into the equation. We know \(R = 1.987 \text{ cal/(mol·K)}\) and \(T = 1000 \text{ K}\):

\[ -4606 = - (1.987) \times (1000) \ln K \]

### Step 4: Solve for \(\ln K\)

Rearranging the equation to solve for \(\ln K\):

\[ \ln K = \frac{4606}{1987} \]

Calculating the right side:

\[ \ln K \approx 2.316 \]

### Step 5: Calculate \(K\)

To find \(K\), we exponentiate both sides:

\[ K = e^{2.316} \]

Calculating \(K\):

\[ K \approx 10.1 \]

### Step 6: Find the equilibrium constant for the reverse reaction

For the reverse reaction, the equilibrium constant \(K'\) is the reciprocal of \(K\):

\[ K' = \frac{1}{K} = \frac{1}{10.1} \approx 0.099 \]

### Conclusion

Thus, the equilibrium constant for the reverse reaction \(Y + C \rightleftharpoons X\) is approximately \(0.1\).

### Final Answer

The equilibrium constant for the reverse reaction is \(0.1\).

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