For the reaction `C(s) +CO_(2)(g) rarr 2CO(g), k_(p)=63` atm at 100 K. If at equilibrium `p_(CO)=10p_(CO_(2)) ` then the total pressure of the gases at equilibrium is

Options

A. 6.3 atm
B. 6.93 atm
C. 0.63 atm
D. 0.693 atm

Answer

To solve the problem, we need to follow these steps:

### Step 1: Write the equilibrium expression
For the reaction:
\[ C(s) + CO_2(g) \rightleftharpoons 2CO(g) \]

The equilibrium constant \( K_p \) is given by the expression:
\[ K_p = \frac{(P_{CO})^2}{P_{CO_2}} \]

Since carbon (C) is a solid, it does not appear in the expression.

### Step 2: Substitute the relationship between the pressures
We are given that at equilibrium:
\[ P_{CO} = 10 \times P_{CO_2} \]

Let \( P_{CO_2} = x \). Then:
\[ P_{CO} = 10x \]

### Step 3: Substitute into the equilibrium expression
Now substituting these values into the equilibrium expression:
\[ K_p = \frac{(10x)^2}{x} \]
\[ K_p = \frac{100x^2}{x} \]
\[ K_p = 100x \]

### Step 4: Set \( K_p \) equal to the given value
We know from the problem that \( K_p = 63 \) atm. Therefore, we can set up the equation:
\[ 100x = 63 \]

### Step 5: Solve for \( x \)
Now, solve for \( x \):
\[ x = \frac{63}{100} = 0.63 \, \text{atm} \]
This value represents \( P_{CO_2} \).

### Step 6: Calculate \( P_{CO} \)
Now, we can find \( P_{CO} \):
\[ P_{CO} = 10x = 10 \times 0.63 = 6.3 \, \text{atm} \]

### Step 7: Calculate total pressure at equilibrium
The total pressure \( P_{total} \) at equilibrium is the sum of the partial pressures of \( CO \) and \( CO_2 \):
\[ P_{total} = P_{CO} + P_{CO_2} \]
\[ P_{total} = 6.3 + 0.63 = 6.93 \, \text{atm} \]

### Final Answer
Thus, the total pressure of the gases at equilibrium is:
\[ \boxed{6.93 \, \text{atm}} \]

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