`Delta G^(@)` for the reaction `X+Y hArr Z` is -4.606 kcal. The value of equilibrium constant of the reaction at `227^(@)C` is (R = 2 Cal/mol K)

Options

A. 100
B. 10
C. 2
D. `0.01`

Answer

To find the equilibrium constant \( K_{eq} \) for the reaction \( X + Y \rightleftharpoons Z \) given that \( \Delta G^\circ = -4.606 \) kcal, we can use the relationship between Gibbs free energy and the equilibrium constant:

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

### Step 1: Convert \( \Delta G^\circ \) to calories
Since \( \Delta G^\circ \) is given in kilocalories, we first convert it to calories:

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

### Step 2: Convert temperature to Kelvin
The temperature is given as \( 227^\circ C \). We need to convert this to Kelvin:

\[ T(K) = 227 + 273.15 = 500.15 \text{ K} \approx 500 \text{ K} \]

### Step 3: Use the equation to find \( K_{eq} \)
Now we can substitute the values into the equation:

\[ -4606 = - (2 \text{ cal/mol K}) (500 \text{ K}) \ln K_{eq} \]

### Step 4: Simplify the equation
First, calculate \( -RT \):

\[ -RT = - (2 \times 500) = -1000 \text{ cal/mol} \]

Now, substituting this back into the equation gives:

\[ -4606 = -1000 \ln K_{eq} \]

### Step 5: Solve for \( \ln K_{eq} \)
Dividing both sides by -1000:

\[ \ln K_{eq} = \frac{4606}{1000} = 4.606 \]

### Step 6: Exponentiate to find \( K_{eq} \)
To find \( K_{eq} \), we exponentiate both sides:

\[ K_{eq} = e^{4.606} \]

### Step 7: Convert \( \ln K_{eq} \) to log base 10
Using the conversion from natural logarithm to base 10:

\[ K_{eq} = 10^{\frac{4.606}{2.303}} \approx 10^{2} = 100 \]

Thus, the equilibrium constant \( K_{eq} \) is:

\[ K_{eq} \approx 100 \]

### Final Answer:
The value of the equilibrium constant \( K_{eq} \) at \( 227^\circ C \) is approximately **100**.

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