A gaseous reaction `X_2(g) to Y + 1/2 Z(g)`. There is increase in pressure from 100 mm to 120 mm in 5 minutes. The rate of disappearance of `X_2` is

Options

A. `8 mm "min"^(-1)`
B. `2 mm "min"^(-1) `
C. `16mm "min"^(-1)`
D. `4mm "min"^(-1)`

Answer

To solve the problem step by step, we will analyze the given reaction and the changes in pressure.

### Step 1: Understand the Reaction
The reaction is given as:
\[ X_2(g) \rightarrow Y + \frac{1}{2} Z(g) \]

### Step 2: Initial and Final Pressure
Initially, the pressure of \( X_2 \) is 100 mm, and the pressures of \( Y \) and \( Z \) are 0 mm. After 5 minutes, the total pressure increases to 120 mm.

### Step 3: Calculate the Change in Pressure
The increase in pressure is:
\[ 120 \, \text{mm} - 100 \, \text{mm} = 20 \, \text{mm} \]

### Step 4: Relate Pressure Change to Reaction Progress
Let \( p \) be the pressure change due to the conversion of \( X_2 \) into products \( Y \) and \( Z \). From the stoichiometry of the reaction, for every 1 mole of \( X_2 \) that reacts, the total pressure change is:
\[ p + \frac{p}{2} = \frac{3p}{2} \]

### Step 5: Set Up the Equation
The total pressure after 5 minutes can be expressed as:
\[ 100 - p + p + \frac{p}{2} = 120 \]
This simplifies to:
\[ 100 + \frac{p}{2} = 120 \]

### Step 6: Solve for \( p \)
Rearranging gives:
\[ \frac{p}{2} = 120 - 100 \]
\[ \frac{p}{2} = 20 \]
\[ p = 40 \, \text{mm} \]

### Step 7: Calculate the Final Pressure of \( X_2 \)
The final pressure of \( X_2 \) after 5 minutes is:
\[ 100 - p = 100 - 40 = 60 \, \text{mm} \]

### Step 8: Determine the Rate of Disappearance of \( X_2 \)
The rate of disappearance of \( X_2 \) can be calculated as:
\[ \text{Rate} = \frac{\text{Change in Pressure of } X_2}{\text{Time}} \]
\[ \text{Rate} = \frac{100 - 60}{5} = \frac{40}{5} = 8 \, \text{mm/min} \]

### Final Answer
The rate of disappearance of \( X_2 \) is:
\[ \text{Rate} = 8 \, \text{mm/min} \]

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