At `450 K, K_(p)=2.0xx10^(10)//` bar for the given reaction at equilibrium.
`2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)`
What is `K_(c )` at this temperature?

Options

A. `73.8xx10^(10)`
B. `7.38xx10^(11)`
C. Both (1) and (2)
D. Zero

Answer

To find \( K_c \) at 450 K for the reaction

\[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]

given that \( K_p = 2.0 \times 10^{10} \) bar, we will use the relationship between \( K_p \) and \( K_c \).

### Step-by-Step Solution:

1. **Identify the Reaction and Write the Expression for \( K_p \)**:
The reaction is:
\[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]
The expression for \( K_p \) is given by:
\[ K_p = \frac{(P_{SO_3})^2}{(P_{SO_2})^2 \cdot (P_{O_2})} \]

2. **Calculate \( \Delta N_g \)**:
\( \Delta N_g \) is defined as the difference between the number of moles of gaseous products and the number of moles of gaseous reactants.
- Moles of products (SO3): 2
- Moles of reactants (SO2 + O2): 2 + 1 = 3

Thus,
\[ \Delta N_g = 2 - 3 = -1 \]

3. **Use the Relationship Between \( K_p \) and \( K_c \)**:
The relationship is given by:
\[ K_p = K_c \cdot R T^{\Delta N_g} \]
Rearranging this gives:
\[ K_c = \frac{K_p}{R T^{\Delta N_g}} \]

4. **Substitute the Values**:
- Given \( K_p = 2.0 \times 10^{10} \) bar
- The universal gas constant \( R = 0.083 \, \text{L bar K}^{-1} \text{mol}^{-1} \)
- Temperature \( T = 450 \, K \)
- \( \Delta N_g = -1 \)

Now substituting these values into the equation:
\[ K_c = \frac{2.0 \times 10^{10}}{0.083 \times (450)^{-1}} \]

5. **Calculate \( K_c \)**:
First, calculate \( R \cdot T \):
\[ R \cdot T = 0.083 \times 450 = 37.35 \]
Now substituting this back into the equation for \( K_c \):
\[ K_c = \frac{2.0 \times 10^{10}}{37.35} \]
\[ K_c \approx 5.35 \times 10^{11} \, \text{L}^{-1} \text{mol} \]

### Final Result:
Thus, the value of \( K_c \) at 450 K is approximately:
\[ K_c \approx 5.35 \times 10^{11} \, \text{L}^{-1} \text{mol} \]