Calculate the value of equilibrium constant of the reaction at `227^(@)C`, If the `DeltaG^(@)` for the reaction `X+YhArrZ` is `-4.606" kcal"`.
`(R=2.0" cal. Mol"^(-1)K^(-1))`

Options

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

Answer

To calculate the equilibrium constant (K) for the reaction \( X + Y \rightleftharpoons Z \) at \( 227^\circ C \) given that \( \Delta G^\circ = -4.606 \, \text{kcal} \), we can use the relationship between Gibbs free energy and the equilibrium constant:

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

### Step 1: Convert the temperature to Kelvin
The temperature in Celsius needs to be converted to Kelvin using the formula:

\[ T(K) = T(°C) + 273.15 \]

For \( 227^\circ C \):

\[ T = 227 + 273.15 = 500.15 \, K \approx 500 \, K \]

### Step 2: Convert \( \Delta G^\circ \) from kcal to cal
Since the value of \( R \) is given in cal, 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 3: Substitute values into the equation
We can now substitute \( \Delta G^\circ \), \( R \), and \( T \) into the equation:

\[ -4606 = -2.303 \times R \times T \times \log K \]

Substituting \( R = 2.0 \, \text{cal/mol K} \) and \( T = 500 \, K \):

\[ -4606 = -2.303 \times 2.0 \times 500 \times \log K \]

### Step 4: Simplify the equation
Calculate the right side:

\[ -4606 = -2.303 \times 1000 \times \log K \]

This simplifies to:

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

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

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

Calculating this gives:

\[ \log K \approx 2 \]

### Step 6: Convert from log to K
To find \( K \), we convert from logarithmic form:

\[ K = 10^{\log K} = 10^2 = 100 \]

### Final Answer
Thus, the equilibrium constant \( K \) for the reaction at \( 227^\circ C \) is:

\[ \boxed{100} \]