At `600^(@)C,K_(P)` for the following reaction is 1 atm.
`X(g) rarr Y(g) + Z(g)`
At equilibrium, 50% of X (g) is dissociated. The total pressure of the equilibrium system is p atm. What is the partial pressure (in atm) of gases at equilibrium ?

Options

A. 1
B. 4
C. 2
D. 0.5

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write down the reaction and the initial conditions.
The reaction given is:
\[ X(g) \rightleftharpoons Y(g) + Z(g) \]

At the start (initially), we assume:
- Initial moles of \( X = 1 \)
- Initial moles of \( Y = 0 \)
- Initial moles of \( Z = 0 \)

### Step 2: Determine the change in moles at equilibrium.
We are told that 50% of \( X \) is dissociated at equilibrium. This means:
- Moles of \( X \) dissociated = \( 0.5 \)
- Remaining moles of \( X = 1 - 0.5 = 0.5 \)

Since the stoichiometry of the reaction shows that for every mole of \( X \) that dissociates, one mole of \( Y \) and one mole of \( Z \) are produced, we have:
- Moles of \( Y = 0.5 \)
- Moles of \( Z = 0.5 \)

### Step 3: Write the equilibrium moles.
At equilibrium, the moles of each gas are:
- Moles of \( X = 0.5 \)
- Moles of \( Y = 0.5 \)
- Moles of \( Z = 0.5 \)

### Step 4: Calculate the total moles at equilibrium.
Total moles at equilibrium:
\[ \text{Total moles} = 0.5 + 0.5 + 0.5 = 1.5 \]

### Step 5: Calculate the mole fractions of each gas.
The mole fraction of each gas is given by:
\[ \text{Mole fraction of } X = \frac{0.5}{1.5} = \frac{1}{3} \]
\[ \text{Mole fraction of } Y = \frac{0.5}{1.5} = \frac{1}{3} \]
\[ \text{Mole fraction of } Z = \frac{0.5}{1.5} = \frac{1}{3} \]

### Step 6: Use Dalton's Law to find partial pressures.
According to Dalton's Law of Partial Pressures:
\[ P_x = \text{Mole fraction of } X \times P \]
\[ P_y = \text{Mole fraction of } Y \times P \]
\[ P_z = \text{Mole fraction of } Z \times P \]

Where \( P \) is the total pressure of the system.

Thus, we have:
\[ P_x = \frac{1}{3}P \]
\[ P_y = \frac{1}{3}P \]
\[ P_z = \frac{1}{3}P \]

### Step 7: Relate \( K_p \) to the partial pressures.
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{P_y \cdot P_z}{P_x} \]

Substituting the expressions for the partial pressures:
\[ K_p = \frac{\left(\frac{1}{3}P\right) \cdot \left(\frac{1}{3}P\right)}{\frac{1}{3}P} \]
\[ K_p = \frac{\frac{1}{9}P^2}{\frac{1}{3}P} = \frac{1}{3}P \]

### Step 8: Solve for \( P \) using the given \( K_p \).
We know from the problem statement that \( K_p = 1 \):
\[ 1 = \frac{1}{3}P \]
\[ P = 3 \text{ atm} \]

### Step 9: Calculate the partial pressures.
Now substituting \( P \) back into the equations for partial pressures:
\[ P_x = \frac{1}{3} \times 3 = 1 \text{ atm} \]
\[ P_y = \frac{1}{3} \times 3 = 1 \text{ atm} \]
\[ P_z = \frac{1}{3} \times 3 = 1 \text{ atm} \]

### Final Answer:
The partial pressures of gases at equilibrium are:
- \( P_x = 1 \text{ atm} \)
- \( P_y = 1 \text{ atm} \)
- \( P_z = 1 \text{ atm} \)

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