The partial pressure of `CH_3OH,CO and H_2` in the equilibrium mixture for the reaction `CO+2H_2 hArr CH_3OH "at" 427^@C` are 2.0 ,1.0 and 0.1 atm respectively. The value of `K_p` for the decomposition of `CH_3OH` to CO and `H_2` is

Options

A. `1 xx 10^(-2)` atm
B. `2 xx 10^(2) "atm"^(-1)`
C. `50 "atm"^2`
D. `5 xx 10^(-3) "atm"^2`

Answer

To find the value of \( K_p \) for the decomposition of \( CH_3OH \) to \( CO \) and \( H_2 \), we will follow these steps:

### Step 1: Write the Equilibrium Reaction
The decomposition reaction of methanol (\( CH_3OH \)) can be written as:
\[ CH_3OH \rightleftharpoons CO + 2H_2 \]

### Step 2: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by the expression:
\[ K_p = \frac{P_{CO} \cdot (P_{H_2})^2}{P_{CH_3OH}} \]
where \( P_{CO} \), \( P_{H_2} \), and \( P_{CH_3OH} \) are the partial pressures of carbon monoxide, hydrogen, and methanol at equilibrium, respectively.

### Step 3: Substitute the Given Partial Pressures
From the problem, we have the following partial pressures:
- \( P_{CH_3OH} = 2.0 \, \text{atm} \)
- \( P_{CO} = 1.0 \, \text{atm} \)
- \( P_{H_2} = 0.1 \, \text{atm} \)

Substituting these values into the \( K_p \) expression:
\[ K_p = \frac{(1.0) \cdot (0.1)^2}{2.0} \]

### Step 4: Calculate \( K_p \)
Calculating the values:
\[ K_p = \frac{1.0 \cdot 0.01}{2.0} = \frac{0.01}{2.0} = 0.005 \]

### Step 5: Adjust for the Reaction Direction
Since we are asked for the \( K_p \) of the decomposition of \( CH_3OH \) (which is the reverse of the formation reaction), we need to take the reciprocal of the calculated \( K_p \):
\[ K_p' = \frac{1}{K_p} = \frac{1}{0.005} = 200 \]

### Final Answer
Thus, the value of \( K_p \) for the decomposition of \( CH_3OH \) to \( CO \) and \( H_2 \) is:
\[ K_p = 200 \]

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