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?

Explanation

To solve the problem of finding \( K_c \) from the given \( K_p \) for the reaction \( 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \) at 450 K, we can follow these steps:

### Step 1: Write the relationship between \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by the equation:
\[ K_p = K_c \cdot R T^{\Delta N_g} \]
where:
- \( R \) is the ideal gas constant,
- \( T \) is the temperature in Kelvin,
- \( \Delta N_g \) is the change in the number of moles of gas, calculated as the moles of products minus the moles of reactants.

### Step 2: Determine \( \Delta N_g \)
For the reaction:
\[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]
- Moles of products (SO3) = 2
- Moles of reactants (SO2 + O2) = 2 + 1 = 3

Thus,
\[ \Delta N_g = \text{moles of products} - \text{moles of reactants} = 2 - 3 = -1 \]

### Step 3: Substitute the known values into the equation
We know:
- \( K_p = 2.0 \times 10^{10} \, \text{bar}^{-1} \)
- \( R = 0.0831 \, \text{L bar K}^{-1} \text{mol}^{-1} \)
- \( T = 450 \, \text{K} \)
- \( \Delta N_g = -1 \)

Substituting these values into the equation:
\[ K_p = K_c \cdot R T^{-1} \]
Rearranging gives:
\[ K_c = K_p \cdot R T \]

### Step 4: Calculate \( K_c \)
Now we can calculate \( K_c \):
\[ K_c = (2.0 \times 10^{10}) \cdot (0.0831) \cdot (450) \]

Calculating this step-by-step:
1. Calculate \( R \cdot T \):
\[ R \cdot T = 0.0831 \cdot 450 = 37.395 \]
2. Now multiply by \( K_p \):
\[ K_c = 2.0 \times 10^{10} \cdot 37.395 \]
\[ K_c = 74.79 \times 10^{10} \, \text{L/mol} \]

### Step 5: Final result
Expressing \( K_c \) in scientific notation:
\[ K_c = 7.48 \times 10^{11} \, \text{L/mol} \]

### Summary
The value of \( K_c \) at 450 K for the given reaction is:
\[ K_c = 7.48 \times 10^{11} \, \text{L/mol} \]