For the reaction `2SO_2(g) + O_2(g) hArr 2SO_3(g)` at 300K, the value of `DeltaG^@` is - 690.9R. The equilibrium constant value for the reaction at that temperature is (R is gas constant)

Options

A. `10 "atm"^(-1)`
B. 10 atm
C. 10
D. 1

Explanation

To find the equilibrium constant \( K \) for the reaction \( 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \) at 300 K given that \( \Delta G^\circ = -690.9R \), we can use the relationship between Gibbs free energy change and the equilibrium constant:

### Step 1: Use the relationship between \(\Delta G^\circ\) and \(K\)
The relationship is given by the equation:
\[ \Delta G^\circ = -RT \ln K \]
where:
- \( \Delta G^\circ \) is the standard Gibbs free energy change,
- \( R \) is the gas constant,
- \( T \) is the temperature in Kelvin,
- \( K \) is the equilibrium constant.

### Step 2: Substitute the known values
We know:
\[ \Delta G^\circ = -690.9R \]
Substituting this into the equation gives:
\[ -690.9R = -RT \ln K \]

### Step 3: Cancel out \( R \) and rearrange
Dividing both sides by \( -R \) (noting that \( R \) is positive):
\[ 690.9 = T \ln K \]
Now, substituting \( T = 300 \, K \):
\[ 690.9 = 300 \ln K \]

### Step 4: Solve for \(\ln K\)
Rearranging gives:
\[ \ln K = \frac{690.9}{300} \]
Calculating the right side:
\[ \ln K = 2.303 \]

### Step 5: Exponentiate to find \( K \)
To find \( K \), we exponentiate both sides:
\[ K = e^{2.303} \]
Calculating \( e^{2.303} \):
\[ K \approx 10 \]

### Final Answer
The equilibrium constant \( K \) for the reaction at 300 K is approximately:
\[ K \approx 10 \]

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