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 \( K_{eq} \) for the reaction \( X + Y \rightleftharpoons Z \) at \( 227^\circ C \) given that \( \Delta G^\circ = -4.606 \) kcal, we can follow these steps:

### Step 1: 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 2: Convert temperature from Celsius to Kelvin
To convert the temperature from Celsius to Kelvin, we add 273:
\[ T = 227^\circ C + 273 = 500 \text{ K} \]

### Step 3: Use the Gibbs free energy equation
The relationship between Gibbs free energy and the equilibrium constant is given by the equation:
\[ \Delta G^\circ = -RT \ln K_{eq} \]
Rearranging this equation to solve for \( K_{eq} \):
\[ \ln K_{eq} = -\frac{\Delta G^\circ}{RT} \]

### Step 4: Substitute the known values
Now we substitute the values we have into the equation. We know \( R = 2.0 \text{ cal/mol K} \), \( T = 500 \text{ K} \), and \( \Delta G^\circ = -4606 \text{ cal} \):
\[ \ln K_{eq} = -\frac{-4606 \text{ cal}}{(2.0 \text{ cal/mol K})(500 \text{ K})} \]
\[ \ln K_{eq} = \frac{4606}{1000} = 4.606 \]

### Step 5: Calculate \( K_{eq} \)
To find \( K_{eq} \), we exponentiate both sides:
\[ K_{eq} = e^{4.606} \]
Using a calculator, we find:
\[ K_{eq} \approx 100.08 \]

### Step 6: Round to significant figures
Since we are asked for the equilibrium constant, we can round this value:
\[ K_{eq} \approx 100 \]

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

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