`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 \( K \) for the reaction \( X + Y \rightleftharpoons Z \) given that the standard Gibbs free energy change \( \Delta G^\circ \) is -4.606 kcal, we will use the relationship between Gibbs free energy and the equilibrium constant:

\[ \Delta G^\circ = -2.303 RT \log K \]

### Step 1: Convert \( \Delta G^\circ \) from kcal to cal
Since the gas constant \( R \) is given in cal/mol K, we need to convert \( \Delta G^\circ \) from kilocalories to calories.

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

### Step 2: Convert the temperature from Celsius to Kelvin
The temperature is given as \( 227^\circ C \). To convert this to Kelvin, we add 273.15.

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

### Step 3: Substitute the values into the Gibbs free energy equation
Now we can substitute \( \Delta G^\circ \), \( R \), and \( T \) into the equation.

\[ -4606 = -2.303 \times 2 \text{ cal/mol K} \times 500 \text{ K} \log K \]

### Step 4: Simplify the equation
First, calculate \( -2.303 \times 2 \times 500 \):

\[ -2.303 \times 2 \times 500 = -2303 \text{ cal} \]

Now the equation becomes:

\[ -4606 = -2303 \log K \]

### Step 5: Solve for \( \log K \)
Dividing both sides by -2303 gives:

\[ \log K = \frac{4606}{2303} \approx 2 \]

### Step 6: Calculate \( K \)
To find \( K \), we take the antilog (base 10) of both sides:

\[ K = 10^2 = 100 \]

### Final Answer
The value of the equilibrium constant \( K \) at \( 227^\circ C \) is \( 100 \).

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