The partial pressures of `CH_3OH, CO and H_2` in the equilibrium mixture for the reaction
` CO + 2H_2 hArr CH_3OH`
at ` 27^(@) C " are " 2.0,1.0 and 0.1 ` atm respectively . The value of `K_p ` for the decompositions of `CH_3OH ` to CO and ` H_2 `is

Options

A. ` 1 xx 10 ^(2) ` atm
B. ` 20 xx 10 ^(2)` atm
C. `50 atm ^(2)`
D. ` 5 xx 10 ^(-3) " atm"^(2)`

Explanation

To find the value of \( K_p \) for the decomposition of methanol (\( CH_3OH \)) into carbon monoxide (\( CO \)) and hydrogen (\( H_2 \)), we follow these steps:

### Step 1: Write the balanced equation for the decomposition of methanol
The balanced equation for the decomposition of methanol is:
\[ CH_3OH (g) \rightleftharpoons CO (g) + 2H_2 (g) \]

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

### Step 3: Substitute the given values into the \( K_p \) expression
From the problem, we have the following partial pressures at equilibrium:
- \( P_{CO} = 1.0 \, \text{atm} \)
- \( P_{H_2} = 0.1 \, \text{atm} \)
- \( P_{CH_3OH} = 2.0 \, \text{atm} \)

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

### Step 4: Calculate \( K_p \)
Now we calculate \( K_p \):
\[ K_p = \frac{1.0 \cdot 0.01}{2.0} = \frac{0.01}{2.0} = 0.005 \]
This can also be expressed in scientific notation:
\[ K_p = 5 \times 10^{-3} \, \text{atm}^2 \]

### Final Answer
Thus, the value of \( K_p \) for the decomposition of methanol is:
\[ K_p = 5 \times 10^{-3} \, \text{atm}^2 \]

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