Find `K_(p)` for `SO_(2)(g) +CO_(2)(g) hArr SO_(3)(g)+CO(g)`
`" " K_(p)` for `SO_(2)(g)+1//2O_(2)(g) hArr SO_(3)(g)` is 1.80 atm. units and
`" "K_(p)` for `CO_(2)(g)hArr CO(g)+1//2 O_(2)(g)` is `5xx10^(-10)` atm. units.

Answer

To find the equilibrium constant \( K_p \) for the reaction:

\[ \text{SO}_2(g) + \text{CO}_2(g) \rightleftharpoons \text{SO}_3(g) + \text{CO}(g) \]

we will use the given reactions and their \( K_p \) values.

### Step 1: Write down the given reactions and their \( K_p \) values.

1. For the reaction:
\[ \text{SO}_2(g) + \frac{1}{2} \text{O}_2(g) \rightleftharpoons \text{SO}_3(g) \quad K_{p1} = 1.80 \, \text{atm} \]

2. For the reaction:
\[ \text{CO}_2(g) \rightleftharpoons \text{CO}(g) + \frac{1}{2} \text{O}_2(g) \quad K_{p2} = 5 \times 10^{-10} \, \text{atm} \]

### Step 2: Manipulate the reactions to find the desired reaction.

To derive the desired reaction, we can add the first reaction and the reverse of the second reaction.

- Reverse the second reaction:
\[ \text{CO}(g) + \frac{1}{2} \text{O}_2(g) \rightleftharpoons \text{CO}_2(g) \quad K_{p2}^{-1} = \frac{1}{5 \times 10^{-10}} = 2 \times 10^{9} \, \text{atm} \]

### Step 3: Add the two reactions.

Now, we add the two reactions:

1. \(\text{SO}_2(g) + \frac{1}{2} \text{O}_2(g) \rightleftharpoons \text{SO}_3(g) \)
2. \(\text{CO}(g) + \frac{1}{2} \text{O}_2(g) \rightleftharpoons \text{CO}_2(g) \)

Adding these gives:

\[ \text{SO}_2(g) + \text{CO}_2(g) \rightleftharpoons \text{SO}_3(g) + \text{CO}(g) \]

### Step 4: Calculate the new \( K_p \).

When adding reactions, the equilibrium constants multiply:

\[ K_p = K_{p1} \times K_{p2}^{-1} = 1.80 \times (2 \times 10^{9}) \]

Calculating this:

\[ K_p = 1.80 \times 2 \times 10^{9} = 3.60 \times 10^{9} \, \text{atm} \]

### Final Answer:

\[ K_p = 3.60 \times 10^{9} \, \text{atm} \]

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