Consider
(i) `C(s)+O_(2)hArrCO_(2)(g) K_(P_(2)=(7)/(8)`
(ii) `2C(s)+O_(2)hArr2CO(g) K_(P_(2)=12.5atm`
As `100 L` of air (`80% N_(2),20% O_(2)` by volume) is pased over excess heated coke to establish these equilibrium the equilibrium mixture is found to measure `105 L` at constnat temperature & pressure (`105atm`). Assuming no other reaction, find the sum of partial pressure of `CO "and" CO_(2)` in the final equilibrium mixture.

Explanation

To solve the problem step by step, we will analyze the given reactions and the conditions provided.

### Step 1: Analyze the reactions and equilibrium constants
We have two reactions:
1. \( C(s) + O_2(g) \rightleftharpoons CO_2(g) \) with \( K_{P1} = \frac{7}{8} \)
2. \( 2C(s) + O_2(g) \rightleftharpoons 2CO(g) \) with \( K_{P2} = 12.5 \, \text{atm} \)

### Step 2: Determine the initial conditions
We start with 100 L of air, which consists of:
- 80% \( N_2 \)
- 20% \( O_2 \)

Calculating the volume of \( O_2 \):
\[ V_{O_2} = 0.20 \times 100 \, \text{L} = 20 \, \text{L} \]

### Step 3: Establish the final volume
After the reaction, the equilibrium mixture is found to measure 105 L. This means that the volume change due to the reactions must be accounted for.

### Step 4: Calculate the partial pressure of \( O_2 \)
Using the ideal gas law, we can find the partial pressure of \( O_2 \) in the final state. Since the total pressure is given as 105 atm, we can calculate the partial pressure of \( O_2 \) as follows:
\[ P_{O_2} = \frac{V_{O_2}}{V_{final}} \times P_{total} = \frac{20 \, \text{L}}{105 \, \text{L}} \times 105 \, \text{atm} = 20 \, \text{atm} \]

### Step 5: Relate the partial pressures using \( K_{P2} \)
From the second reaction, we know:
\[ K_{P2} = \frac{(P_{CO})^2}{P_{O_2}} = 12.5 \]
Rearranging gives us:
\[ P_{CO}^2 = K_{P2} \cdot P_{O_2} = 12.5 \cdot 20 = 250 \]
Taking the square root gives:
\[ P_{CO} = \sqrt{250} \approx 15.81 \, \text{atm} \]

### Step 6: Relate the partial pressures using \( K_{P1} \)
From the first reaction, we know:
\[ K_{P1} = \frac{P_{CO_2}}{P_{O_2}} = \frac{7}{8} \]
Rearranging gives us:
\[ P_{CO_2} = K_{P1} \cdot P_{O_2} = \frac{7}{8} \cdot 20 = 17.5 \, \text{atm} \]

### Step 7: Calculate the sum of the partial pressures
Now we can find the total partial pressure of \( CO \) and \( CO_2 \):
\[ P_{CO} + P_{CO_2} = 15.81 \, \text{atm} + 17.5 \, \text{atm} \approx 33.31 \, \text{atm} \]

### Final Answer
The sum of the partial pressures of \( CO \) and \( CO_2 \) in the final equilibrium mixture is approximately \( 33.31 \, \text{atm} \).